A = \(\begin{vmatrix} a_1 & b_1\\ a_2 & b_2 \end{vmatrix}\) then det A or △ A is called determinant of A. Then then the system of linear equations has a unique solution \( a_1b_2 - a_2b_1 \neq 0 \)
For matrix A, | A| is read as determinant of A and not modulus of A.
Only square matrices have determinants.
The value of the determinant remains unchanged if its rows and columns are interchanged.
It follows from above property that if A is a square matrix, then det (A) = det (A′), where A′ = transpose of A.
If any two rows (or columns) of a determinant are interchanged, then sign of determinant changes.
If any two rows (or columns) of a determinant are identical (all corresponding elements are same), then value of determinant is zero.
If each element of a row (or a column) of a determinant is multiplied by a constant k, then its value gets multiplied by k.
If corresponding elements of any two rows (or columns) of a determinant are proportional (in the same ratio), then its value is zero
If some or all elements of a row or column of a determinant are expressed as sum of two (or more) terms, then the determinant can be expressed as sum of two (or more) determinants.
If, to each element of any row or column of a determinant, the equi-multiples of corresponding elements of other row (or column) are added, then value of determinant remains the same, i.e., the value of determinant remain same if we apply the operation \(R_i -> R_i + kR_j \) OR \( C_i -> C_i + kC_j \)
Exercise 4.1
Q1: Evaluate the determinants.
\(\begin{vmatrix} 2 & 4\\ -5 & -1 \end{vmatrix}\)
Ans: Based on the rules and formulae given in Determinant tricks
\(\begin{vmatrix} 2 & 4\\ -5 & -1 \end{vmatrix}\) = 2 ( -1 ) – 4 ( -5 ) = – 2 + 20 = 18
Q2: Evaluate the determinant.
(i) \(\begin{vmatrix} cos \Theta & -sin \Theta \\ sin \Theta & cos \Theta \end{vmatrix}\\\)
(ii) \(\\\begin{vmatrix} x^{ 2 } – x + 1 & x – 1 \\ x + 1 & x + 1 \end{vmatrix}\)
Ans: Based on the rules and formulae given in Determinant tricks
(i) (cos Ɵ )(cos Ɵ ) – ( -sin Ɵ ) (sin Ɵ ) = cos2 Ɵ + sin2 Ɵ = 1
(ii) = ( x2 – x + 1 ) ( x + 1 ) – ( x – 1 ) ( x + 1 )
= x3 – x2 + x + x2 – x + 1 – ( x2 – 1 )
= x3 + 1 – x2 + 1
= x3 – x2 + 2
Q3: If A = \(\begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}\) , then show that |2A| = 4 |A|
Ans: Based on the rules and formulae given in Determinant tricksThe given matrix is A = \(\begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}\)
Therefore, 2A = \(2 \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix}\)
Therefore we have, L.H.S. = |2A| = \(\begin{vmatrix} 2 & 4 \\ 8 & 4 \end{vmatrix}\) = 2 × 4 – 4 × 8 = 8 – 32 = -24
Now, |A| = \(\begin{vmatrix} 1 & 2 \\ 4 & 2 \end{vmatrix}\) =
1 × 2 – 2 × 4 = 2 – 8 = -6
Therefore, R.H.S. = 4|A| = 4 × ( -6) = -24
Therefore, L.H.S. = R.H.S.
Q.4: If A = \(\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix}\) , then show that |3A| = 27|A|.
Ans: Based on the rules and formulae given in Determinant tricks The given matrix is A = \(\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix}\)
It can be observed that in the first column, two entries are zero. Thus, we expand along the first column (C1) for easier calculation.
|A| = \(1 \begin{vmatrix} 1 & 2 \\ 0 & 4 \end{vmatrix} – 0 \begin{vmatrix} 0 & 1 \\ 0 & 4 \end{vmatrix} + 0 \begin{vmatrix} 0 & 1 \\ 1 & 2 \end{vmatrix}\) = 1 ( 4 – 0 ) – 0 + 0 = 4
Therefore, 27|A| = 27 (4) = 108 …(i)
Now, 3A = \(3 \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix} = \begin{bmatrix} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{bmatrix}\)
Therefore, |3A| = \(3 \begin{vmatrix} 3 & 6 \\ 0 & 12 \end{vmatrix} – 0 \begin{vmatrix} 0 & 3 \\ 0 & 12 \end{vmatrix} + 0 \begin{vmatrix} 0 & 3 \\ 3 & 6 \end{vmatrix}\)
= 3 ( 36 – 0 ) = 3 ( 36 ) = 108 …(ii)
From equations (i) and (ii), we have:
|3A| = 27|A|
Therefore, the given result is proved above.
Q5: Evaluate the determinants
(i) \(\begin{vmatrix} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{vmatrix}\\\)
(ii) \(\begin{vmatrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{vmatrix}\\\)
(iii) \(\begin{vmatrix} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{vmatrix}\\\)
(iv) \(\begin{bmatrix} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{bmatrix}\)
Ans: Based on the rules and formulae given in Determinant tricks
(i) Let A = \(\begin{vmatrix} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{vmatrix}\)
It can be observed that in the second row, two entries are zero. Thus, we expand along the second row for easier calculations.
|A| = \(0 \begin{vmatrix} -1 & -2 \\ -5 & 0 \end{vmatrix} + 0 \begin{vmatrix} 3 & -2 \\ 3 & 0 \end{vmatrix} – (-1) \begin{vmatrix} 3 & -1 \\ 3 & -5 \end{vmatrix}\) = ( -15 + 3 ) = -12
(ii) Let \(\begin{bmatrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{bmatrix}\)
By expanding along the first row, we have:
|A| = \(3 \begin{vmatrix} 1 & -2 \\ 3 & 1 \end{vmatrix} + 4 \begin{vmatrix} 1 & -2 \\ 2 & 1 \end{vmatrix} + 5 \begin{vmatrix} 1 & 1 \\ 2 & 3 \end{vmatrix}\)
= 3 ( 1 + 6 ) + 4 ( 1 + 4 ) + 5 ( 3 – 2 )
= 3 ( 7 ) + 4 ( 5 ) + 5 ( 1 )
= 21 + 20 + 5 = 46
(iii) Let A = \(\begin{bmatrix} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{bmatrix}\)
By expanding along the first row, we have:
|A| = \(0 \begin{vmatrix} 0 & -3 \\ 3 & 0 \end{vmatrix} – 1 \begin{vmatrix} -1 & -3 \\ -2 & 0 \end{vmatrix} + 2 \begin{vmatrix} -1 & 0 \\ -2 & 3 \end{vmatrix}\)
= 0 – 1( 0 – 6 ) + 2 ( -3 – 0 )
= – 1 ( -6 ) + 2 ( -3 )
= 6 – 6 = 0
(iv) Let A = \(\begin{bmatrix} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{bmatrix}\)
By expanding along the first column, we have:
|A| = \(2 \begin{vmatrix} 2 & -1 \\ -5 & 0 \end{vmatrix} – 0 \begin{vmatrix} -1 & -2 \\ -5 & 0 \end{vmatrix} + 3 \begin{vmatrix} -1 & -2 \\ 2 & -1 \end{vmatrix}\)
= 2 ( 0 – 5 ) – 0 + 3 ( 1 + 4 )
= -10 + 15 = 5
Q6: If A = \(\begin{bmatrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{bmatrix}\) , find |A|.
Ans: Based on the rules and formulae given in Determinant tricks
Let A = \(\begin{bmatrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{bmatrix}\)
By expanding along the first row, we have:
|A| = \(1 \begin{vmatrix} 1 & -3 \\ 4 & -9 \end{vmatrix} – 1 \begin{vmatrix} 2 & -3 \\ 5 & -9 \end{vmatrix} – 2 \begin{vmatrix} 2 & 1 \\ 5 & 4 \end{vmatrix}\)
= 1 ( -9 + 12 ) – 1 ( -18 + 15 ) – 2 ( 8 – 5 )
= 1 ( 3 ) – 1 ( -3 ) – 2 ( 3 )
= 3 + 3 – 6
= 6 – 6
= 0
Q7: Find values of x, if
(i) \(\begin{vmatrix} 2 & 4 \\ 2 & 1 \end{vmatrix} = \begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}\\\)
(ii) \(\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = \begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}\)
Ans: Based on the rules and formulae given in Determinant tricks
(i) \(\begin{vmatrix} 2 & 4 \\ 2 & 1 \end{vmatrix} = \begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}\)
\(\boldsymbol{\Rightarrow }\) 2 × 1 – 5 × 4 = 2x × x – 6 × 4
\(\boldsymbol{\Rightarrow }\) 2 – 20 = 2x2 – 24
\(\boldsymbol{\Rightarrow }\) 2x2 = 6
\(\boldsymbol{\Rightarrow }\) x2 = 3
\(\boldsymbol{\Rightarrow }\) x = ± \(\sqrt{ 3 }\)
(ii) \(\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = \begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}\)
\(\boldsymbol{\Rightarrow }\) 2 × 5 – 3 × 4 = x × 5 – 3 × 2x
\(\boldsymbol{\Rightarrow }\) 10 -12 = 5x – 6x
\(\boldsymbol{\Rightarrow }\) -2 = -x
\(\boldsymbol{\Rightarrow }\) x = 2
Q8: If \(\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = \begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix}\) , then x is equal to
(A) 6
(B) ± 6
(C) – 6
(D) 0
Ans: Based on the rules and formulae given in Determinant tricks
\(\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = \begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix}\\\)\(\\\boldsymbol{\Rightarrow }\) x2 – 36 = 36 – 36
\(\boldsymbol{\Rightarrow }\) x2 – 36 = 0
\(\boldsymbol{\Rightarrow }\) x = ± 6
Therefore, the correct answer is B
Q.1: Using the property of determinants and without expanding, prove that:
\(\begin{vmatrix} x & a & x + a \\ y & b & y + b \\ z & c & z + c \end{vmatrix} = 0\)
Sol: Based on the rules and formulae given in Determinant tricks
\(\begin{vmatrix} x & a & x + a \\ y & b & y + b \\ z & c & z + c \end{vmatrix} = \begin{vmatrix} x & a & x \\ y & b & y \\ z & c & z \end{vmatrix} + \begin{vmatrix} x & a & a \\ y & b & b \\ z & c & c \end{vmatrix} = 0 + 0 = 0\\\)(The two columns of the determinants are identical is given)
Q2: Using the property of determinants and without expanding, prove that:
\(\begin{vmatrix} a – b & b – c & c – a \\ b – c & c – a & a – b \\ c – a & a – b & b – c \end{vmatrix} = 0\)
Sol: Based on the rules and formulae given in Determinant tricks
\(\begin{vmatrix} a – b & b – c & c – a \\ b – c & c – a & a – b \\ c – a & a – b & b – c \end{vmatrix} = 0\\\)Applying \(R_{1} \rightarrow R_{1} + R_{2}\), we have:
\(\boldsymbol{\Rightarrow }\) \(\Delta = \begin{vmatrix} a – b & b – a & c – b \\ b – c & c – a & a – b \\ -(a – c) & -(b – a) & -(c – b) \end{vmatrix} = 0\\\)
\(\\\boldsymbol{\Rightarrow }\) \(– \begin{vmatrix} a – c & b – a & c – b \\ b – c & c – a & a – b \\ a – c & b – a & c – b \end{vmatrix} = 0\)
The two rows R1 and R3 are identical is given.
Therefore, \(\bigtriangleup = 0\)
Q3: Using the property of determinants and without expanding, prove that:
\(\begin{vmatrix} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{vmatrix} = 0\)
Sol: Based on the rules and formulae given in Determinant tricks
\(\boldsymbol{\Rightarrow }\) \(\begin{vmatrix} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{vmatrix} = \) \(\begin{vmatrix} 2 & 7 & 63 + 2 \\ 3 & 8 & 72 + 3 \\ 5 & 9 & 81 + 5 \end{vmatrix}\\\)
\(\\\boldsymbol{\Rightarrow }\) \(\begin{vmatrix} 2 & 7 & 63 \\ 3 & 8 & 72 \\ 5 & 9 & 81 \end{vmatrix} + \begin{vmatrix} 2 & 7 & 2 \\ 3 & 8 & 3 \\ 5 & 9 & 5 \end{vmatrix}\\\)
\(\\\boldsymbol{\Rightarrow }\) \(\begin{vmatrix} 2 & 7 & 9(7) \\ 3 & 8 & 9(8) \\ 5 & 9 & 9(9) \end{vmatrix} + 0\) (Two columns are identical)
\(\\9 \begin{vmatrix} 2 & 7 & 7 \\ 3 & 8 & 8 \\ 5 & 9 & 9 \end{vmatrix} + 0\) (Two columns are identical)
= 0
Q4: Using the property of determinants and without expanding, prove that:
\(\begin{vmatrix} 1 & bc & a(b + c) \\ 1 & ca & b(c + a) \\ 1 & ab & c(a + b) \end{vmatrix} = 0\)
Sol: Based on the rules and formulae given in Determinant tricks
\(\Delta = \begin{vmatrix} 1 & bc & a(b + c) \\ 1 & ca & b(c + a) \\ 1 & ab & c(a + b) \end{vmatrix} = 0\\\)By applying \(C_{3} \rightarrow C_{3} + C_{2}\)
\(\\\boldsymbol{\Rightarrow }\) \(\Delta = \begin{vmatrix} 1 & bc & ab + bc + ca \\ 1 & ca & ab + bc + ca \\ 1 & ab & ab + bc + ca \end{vmatrix} = 0\\\)
\(\\\boldsymbol{\Rightarrow }\) \((ab + bc + ca ) \begin{vmatrix} 1 & bc & 1 \\ 1 & ca & 1 \\ 1 & ab & 1 \end{vmatrix} = 0\)
Here, two columns C1 and C3 are proportional.
\(\boldsymbol{\Rightarrow }\) \(\Delta = 0\)
Q5: Using the property of determinants and without expanding, prove that:
\(\begin{vmatrix} b + c & q + r & y + z \\ c + a & r + p & z + x \\ a + b & p + q & x + y \end{vmatrix} = 2 \begin{vmatrix} a & p & x\\ b & q & y\\ c & r & z \end{vmatrix}\)
Sol:
\(\Delta = \begin{vmatrix} b + c & q + r & y + z \\ c + a & r + p & z + x \\ a + b & p + q & x + y \end{vmatrix}\\\) \(= \begin{vmatrix} b + c & q + r & y + z \\ c + a & r + p & z + x \\ a & p & x \end{vmatrix} + \begin{vmatrix} b + c & q + r & y + z \\ c + a & r + p & z + x \\ b & q & y \end{vmatrix}\\\)= \(\Delta _{1} + \Delta _{2}\) ……………………….(i)
Now, \(\Delta _{1} = \begin{vmatrix} b + c & q + r & y + z \\ c + a & r + p & z + x \\ a & p & x \end{vmatrix}\)
Applying \(R_{2}\rightarrow R_{2} – R_{3}\), we have:
\(\Delta _{1} = \begin{vmatrix} b + c & q + r & y + z \\ c & r & z \\ a & p & x \end{vmatrix}\\\)Applying \(R_{1}\rightarrow R_{1} – R_{2}\), we have:
\(\Delta _{1} = \begin{vmatrix} b & q & y \\ c & r & z \\ a & p & x \end{vmatrix}\\\)Applying \(R_{1} \leftrightarrow R_{3} \;\; and R_{2} \leftrightarrow R_{3}\) we have:
\(\Delta _{1} = (-1)^{2} \begin{vmatrix} a & p & x \\ b & q & y \\ c & r & z \end{vmatrix} = \begin{vmatrix} a & p & x \\ b & q & y \\ c & r & z \end{vmatrix}\) …………………..(ii)
\(\Delta _{2} = \begin{vmatrix} b + c & q + r & y + z \\ c + a & r + p & z + x \\ b & q & y \end{vmatrix}\\\)Applying \(R_{1} \rightarrow R_{1} – R_{3}\), we have:
\(\Delta _{2} = \begin{vmatrix} c & r & z \\ c + a & r + p & z + x \\ b & q & y \end{vmatrix}\\\)Applying \(R_{2} \rightarrow R_{2} – R_{1}\), we have:
\(\Delta _{2} = \begin{vmatrix} c & r & z \\ a & p & x \\ b & q & y \end{vmatrix}\\\)Applying \(R_{1} \leftrightarrow R_{2} \;\; and R_{2} \leftrightarrow R_{3}\) we have:
\(\Delta _{2} = (-1)^{2} \begin{vmatrix} a & p & x \\ b & q & y \\ c & r & z \end{vmatrix} = \begin{vmatrix} a & p & x \\ b & q & y \\ c & r & z \end{vmatrix}\) ………………………(iii)
From (i), (ii) and (iii), we have:
\(\Delta _{2} = 2 \begin{vmatrix} a & p & x \\ b & q & y \\ c & r & z \end{vmatrix}\)
Q6: By using properties of determinants, show that:
\(\begin{vmatrix} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{vmatrix} = 0\)
Sol: Based on the rules and formulae given in Determinant tricks
\(\Delta = \begin{vmatrix} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{vmatrix}\\\)Applying \(R_{1} = c R_{1}\), we have:
\(\Delta = \frac{1}{c} \begin{vmatrix} 0 & ac & -bc \\ -a & 0 & -c \\ b & c & 0 \end{vmatrix}\\\)Applying \(R_{1} \rightarrow R_{1} – b R_{2}\), we have:
\(\Delta = \frac{1}{c} \begin{vmatrix} ab & ac & 0 \\ -a & 0 & -c \\ b & c & 0 \end{vmatrix}\) \(\Delta = \frac{a}{c} \begin{vmatrix} b & c & 0 \\ -a & 0 & -c \\ b & c & 0 \end{vmatrix}\)Here, the two rows \(R_{1} \;\; and \;\; R_{3}\) are identical.
\(\boldsymbol{\Rightarrow }\) \(\Delta = 0\)
Q7: By using the properties of determinants, show that:
\(\begin{vmatrix} -a^{2} & ab & ac \\ ba & -b^{2} & bc \\ ca & cb & -c^{2} \end{vmatrix} = 4a^{2}b^{2}c^{2}\)
Sol: Based on the rules and formulae given in Determinant tricks
\(\begin{vmatrix} -a^{2} & ab & ac \\ ba & -b^{2} & bc \\ ca & cb & -c^{2} \end{vmatrix}\\\)\(\\\begin{vmatrix} -a^{2} & ab & ac \\ ba & -b^{2} & bc \\ ca & cb & -c^{2} \end{vmatrix}\) (Taking out factors a, b, c from R1 , R2 and R3)
\(\Delta = a^{2}b^{2}c^{2} \begin{vmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix}\) (Taking out factors a,b,c from C1, C2 and C3)
Applying \(R_{2} \rightarrow R_{2} + R_{1} \;\; and \;\; R_{3} \rightarrow R_{3} + R_{1}\) we have:
\(\Delta = a^{2}b^{2}c^{2} \begin{vmatrix} -1 & 1 & 1 \\ 0 & 0 & 2 \\ 0 & 2 & 0 \end{vmatrix}\\\)\(\\\boldsymbol{\Rightarrow }\) \(\Delta = a^{2}b^{2}c^{2} (-1) \begin{vmatrix} 0 & 2 \\ 2 & 0 \end{vmatrix}\\\)
\(\\\boldsymbol{\Rightarrow }\) \(-a^{2} b^{2} c^{2} (0 – 4) = 4a^{2} b^{2} c^{2}\)
Q8: By using the properties of determinants, show that:
(i) \(\begin{vmatrix} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{vmatrix} = (a – b) (b – c) (c – a)\)
(ii) \(\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3} \end{vmatrix} = (a – b) (b – c) (c – a)(a + b + c)\)
Sol:
Let \(\Delta = \begin{vmatrix} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{vmatrix}\)
Applying \(R_{1} \rightarrow R_{1} – R_{3}\) and \(R_{2} \rightarrow R_{2} – R_{3}\), we have:
\(\Delta = \begin{vmatrix} 0 & a – c & a^{2} – c^{2} \\ 0 & b – c & b^{2} – c^{2} \\ 1 & c & c^{2} \end{vmatrix}\) \(= (c-a) (b – c)\begin{vmatrix} 0 & a – c & a^{2} – c^{2} \\ 0 & b – c & b^{2} – c^{2} \\ 1 & c & c^{2} \end{vmatrix}\\\)Applying \(R_{1}\rightarrow R_{1}+R_{2}\) we have:
= \((b – c) (c-a) \begin{vmatrix} 0 & 0 & a – b \\ 0 & 1 – c & b + c \\ 1 & c & c^{2} \end{vmatrix}\\\)
= \(\\(a – b) (b – c) (c-a) \begin{vmatrix} 0 & 0 & 1\\ 0 & 1 – c & b + c \\ 1 & c & c^{2} \end{vmatrix}\)
Expanding along \(C_{1}\) we have:
\(\\\Delta = (a – b) (b – c) (c-a) \begin{vmatrix} 0 & -1 0 & b + c \end{vmatrix} = (a – b) (b – c) (c – a)\)Therefore, the given result is proved.
(ii) Let \(\Delta = \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3} \end{vmatrix}\)
Applying \(C_{1}\rightarrow C_{1}-C_{3} \;\; and C_{2}\rightarrow C_{2}- C_{3}\) we have:
= \(\begin{vmatrix} 0 & 0 & 1 \\ a-c & b-c & c \\ a^{3}- c^{3} & b^{3}-c^{3} & c^{3} \end{vmatrix}\\\)
= \(\\\begin{vmatrix} 0 & 0 & 1 \\ a-c & b-c & c \\ (a-c)(a^{2}+ac+c^{2}) & (b-c)(b^{2}+bc+c^{2}) & c^{3} \end{vmatrix}\)
= \(\\(c-a)(b-c)\begin{vmatrix} 0 & 0 & 1 \\ -1 & 1 & c \\ -(a^{2}+ac+c^{2}) & (b^{2}+bc+c^{2}) & c^{3} \end{vmatrix}\\\)
Applying \(C_{1}\rightarrow C_{1}+C_{2}\) we have:
= \(\\(c-a)(b-c)\begin{vmatrix} 0 & 0 & 1 \\ 0 & 1 & c \\ (b^{2}-a^{2})+ (bc-ac) & (b^{2}+bc+c^{2}) & c^{3} \end{vmatrix}\\\)
= \(\\(a-b)(b-c)(c-a)\begin{vmatrix} 0 & 0 & 1 \\ 0 & 1 & c \\ -(a+b+c) & (b^{2}+bc+c^{2}) & c^{3} \end{vmatrix}\\\)
= \(\\(a-b)(b-c)(c-a)(a+b+c)\begin{vmatrix} 0 & 0 & 1 \\ 0 & 1 & c \\ -1 & (b^{2}+bc+c^{2}) & c^{3} \end{vmatrix}\\\)
Expanding along \(C_{1}\) we have:
\(\Delta = (a-b)(b-c)(c-a)(a+b+c)(-1)\begin{vmatrix} 0 & 1 \\ 1 & c \end{vmatrix}\\\)= \(\\(a-b)(b-c)(c-a)(c-a)(a+b+c)\)
Therefore, the given result is proved above.
Q9: By using the properties of determinants, show that:
\(\begin{vmatrix} x & x^{2} & yz \\ y & y^{2} & zx \\ z & z^{2} xy & \end{vmatrix} = (x – y)(y-z)(z-x)(xy+yz+zx)\)
Sol: Based on the rules and formulae given in Determinant tricks
Let \(\Delta = \begin{vmatrix} x & x^{2} & yz \\ y & y^{2} & zx \\ z & z^{2} xy & \end{vmatrix}\)
Applying \(R_{2}\rightarrow R_{2}-R_{1} \;\; and R_{3}\rightarrow R_{3}-R_{1}\) we have:
\(\\\boldsymbol{\Rightarrow }\) \(\Delta = \begin{vmatrix} x & x^{2} & yz \\ y-x & y^{2}-x^{2} & zx-yz \\ z-x & z^{2}-x^{2} xy-yz & -y(z-x) \end{vmatrix}\\\)
= \(\\\begin{vmatrix} x & x^{2} & yz \\ -(x-y) & -(x-y)(x+y) & z(x-y) \\ z-x & (z-x)(z+x) & -y(z-x) \end{vmatrix}\\\)
= \(\\(x-y)(z-x)\begin{vmatrix} x & x^{2} & yz \\ -1 & -x-y & z \\ 1 & (z+x) & -y \end{vmatrix}\\\)
Applying \(R_{3}\rightarrow R_{3}+R_{2} \) we have:
\(\boldsymbol{\Rightarrow }\) \(\Delta = (x-y)(z-x)\begin{vmatrix} x & x^{2} & yz \\ -1 & -x-y & z \\ 0 & z-y & z-y \end{vmatrix}\\\)
= \(\\(x-y)(z-x)(z-y)\begin{vmatrix} x & x^{2} & yz \\ -1 & -x-y & z \\ 0 & 1 & 1 \end{vmatrix}\\\)
Expanding along \(R_{3}\) we have:
\(\boldsymbol{\Rightarrow }\) \(\Delta = \left [ (x-y)(z-x)(z-y) \right ] \left [ (-1) \begin{vmatrix} x & yz \\ -1 & z \end{vmatrix} + 1 \begin{vmatrix} x & x^{2} \\ -1 & -x-y \end{vmatrix} \right ]\\\)
= (x-y)(z-x)(z-y)[(-xz-zy)\(+(x^{2}-xy+x^{2})\)]
= -(x-y)(y-z)(z-x)(xy+yz+zx)
Therefore, the given result is proved above.
Q10: By using properties of determinants, show that:
(i) \(\begin{vmatrix} x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{vmatrix} = (5x+4)(4-x)^{2}\\\)
(ii) \(\\\begin{vmatrix} y+k & y & y \\ y & y+k & y \\ y & y & y+k \end{vmatrix} = k^{2}(3y+k)\)
Sol: Based on the rules and formulae given in Determinant tricks
(i) \(\begin{vmatrix} x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{vmatrix} \\\)
Applying \(R_{1}\rightarrow R_{1}+R_{2}+R_{3}\) we have:
\(\Delta = \begin{vmatrix} 5x+4 & 5x+4 & 5x+4 \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{vmatrix}\) \(= (5x+4) \begin{vmatrix} 1 & 1 & 1 \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{vmatrix}\\\)Applying \(C_{2}\rightarrow C_{2}-C_{1}, C_{3}\rightarrow C_{3}-C_{1}\) we have:
\(\Delta = (5x+4) \begin{vmatrix} 1 & 0 & 0 \\ 2x & -x+4 & 0 \\ 2x & 0 & -x+4 \end{vmatrix}\\\)= \(\\(5x+4)(4-x)(4-x) \begin{vmatrix} 1 & 0 & 0 \\ 2x & 1 & 0 \\ 2x & 0 & 1 \end{vmatrix}\)
Expanding along \(C_{3}\) we have:
\(\Delta = (5x+4)(4-x)^{2} \begin{vmatrix} 1 & 0 \\ 2x & 1\end{vmatrix}\) \(= (5x+4)(4-x)^{2}\\\)Therefore, the given result is proved.
(ii) \(\begin{vmatrix} y+k & y & y \\ y & y+k & y \\ y & y & y+k \end{vmatrix} \)
Applying \(R_{1}\rightarrow R_{1}+R_{2}+R_{3}\) we have:
\(\\\Delta = \begin{vmatrix} 3y+k & 3y+k & 3y+k \\ y & y+k & y \\ y & y & y+k \end{vmatrix}\) \( = (3y+k) \begin{vmatrix} 1 & 1 & 1 \\ y & y+k & y \\ y & y & y+k \end{vmatrix}\\\)Applying \(C_{2}\rightarrow C_{2}-C_{1}, C_{3}\rightarrow C_{3}-C_{1}\) we have:
\(\Delta = (3y+k) \begin{vmatrix} 1 & 0 & 0 \\ y & k & 0 \\ y & 0 & k \end{vmatrix}\) \(= k^{2} (3y+k) \begin{vmatrix} 1 & 0 & 0 \\ y & 1 & 0 \\ y & 0 & 1 \end{vmatrix}\\\)Expanding along \(C_{3}\) we have:
\(\Delta = k^{2} (3y+k) \begin{vmatrix} 1 & 0 \\ y & 1 \end{vmatrix} = k^{2} (3y+k)\)Therefore, the given result is proved.
Q.11: By using properties of determinants, show that:
(i) \( \begin {vmatrix} a – b – c & 2a & 2a \\ 2b & b – c – a & 2b \\ 2c & 2c & c – a – b \end {vmatrix} = \left ( a + b + c \right )^{3}\\ \)
(ii) \( \\\begin {vmatrix} x + y + 2z & x & y \\ z & x + z + 2x & y \\ z & x & z + x + 2y \end {vmatrix} = 2 \left ( x + y + z \right )^{3} \)
Sol: Based on the rules and formulae given in Determinant tricks
\( \Delta = \begin {vmatrix} a – b – c & 2a & 2a \\ 2b & b – c – a & 2b \\ 2c & 2c & c – a – b \end {vmatrix} \)(i)
Applying \( R_{1} \rightarrow R_{1} + R_{2} + R_{3} \), we have:
\( \Delta = \begin {vmatrix} a + b + c & a + b + c & a + b + c \\ 2b & b – c – a & 2b \\ 2c & 2c & c – a – b \end {vmatrix} \\\)= \( \left ( a + b + c \right ) \begin {vmatrix} 1 & 1 & 1 \\ 2b & b – c – a & 2b\\ 2c & 2c & c – a – b \end {vmatrix} \)
Applying \( C_{2} \rightarrow C_{2} – C_{1} , C_{3} \rightarrow C_{3} – C_{1} \), we have:
\( \Delta = \left ( a + b + c \right ) \begin {vmatrix} 1 & 0 & 0 \\ 2b & – \left ( a + b + c \right ) & 0 \\ 2c & 0 & – \left ( a + b + c \right ) \end {vmatrix} \\\)= \( \\\left ( a + b + c \right ) ^{3} \begin{vmatrix} 1 & 0 & 0 \\ 2b & -1 & 0\\ 2c & 0 & -1 \end {vmatrix} \)
Expanding along C3, we have:
∆ = (a + b + c)3 (-1) (-1) = (a + b + c)3
Therefore, the given result is proved.
\( \Delta = \begin {vmatrix} x + y + 2z & x & y \\ z & y + z + 2x & y \\ z & x & z + x + 2y \end {vmatrix} \)
(ii) Applying C1 → C1 + C2 + C3, we have:
\( \\\Delta = \begin{vmatrix} 2 \left ( x + y + z \right ) & x & y \\ 2 \left ( x + y + z \right ) & y + z + 2x & y \\ 2 \left ( x + y + z \right ) & x & z + x + 2y \end {vmatrix} \)= \(\\ 2 \left ( x + y + z \right ) \begin {vmatrix} 1 & x & y \\ 1 & y + z + 2x & y \\ 1 & x & z + x + 2y \end {vmatrix} \)
Applying R2 → R2 – R1 and R3 → R3 – R1, we have below equations:
\( \Delta = 2 \left ( x + y + z \right ) \begin {vmatrix} 1 & x & y \\ 0 & x + y + z & 0 \\ 0 & 0 & x + y + z \end {vmatrix}\\ \)= \(\\ 2 \left ( x + y + z \right ) ^{3} \begin {vmatrix} 1 & x & y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end {vmatrix} \)
Expanding along R3, we have:
∆ = 2(x + y + z)3 (1) (1 – 0) = 2(x + y + z)3
Therefore, the given results are proved Based on the rules and formulae given in Determinant tricks.
Q.12: By using properties of determinants, show that:
\( \begin {vmatrix} 1 & x & x ^ {2} \\ x ^ {2} & 1 & x \\ x & x ^ {2} & 1 \end {vmatrix} = \left ( 1 – x ^{3} \right ) ^ {2} \)
Sol:
\( \Delta = \begin {vmatrix} 1 & x & x ^ {2} \\ x ^ {2} & 1 & x \\ x & x ^ {2} & 1 \end {vmatrix} \\\)Applying R1 → R1 + R2 + R3, we have:
\( \\\Delta = \begin{vmatrix} 1 + x + x ^ {2} & 1 + x + x ^ {2} & 1 + x + x ^ {2} \\ x ^ {2} & 1 & x \\ x & x ^ {2} & 1 \end {vmatrix} \\\)= \( \\\left ( 1 + x + x ^ {2} \right ) \begin {vmatrix} 1 & 1 & 1 \\ x ^ {2} & 1 & x \\ x & x ^ {2} & 1 \end {vmatrix} \)
Applying C2 → C2 – C1 and C3 → C3 – C1, we have:
\(\\ \Delta = \left ( 1 + x + x ^ {2} \right ) \begin {vmatrix} 1 & 0 & 0 \\ x ^ {2} & 1 – x ^ {2} & x – x ^ {2} \\ x & x ^ {2} – x & 1 – x \end {vmatrix} \)= \( \\\left ( 1 + x + x ^ {2} \right ) \left ( 1 – x \right ) \left ( 1 – x \right ) \begin {vmatrix} 1 & 0 & 0 \\ x ^ {2} & 1 + x & x \\ x & -x & 1 \end {vmatrix}\\ \)
= \( \\\left ( 1 – x ^ {3} \right ) \left ( 1 – x \right ) \begin {vmatrix} 1 & 0 & 0 \\ x ^ {2} & 1 + x & x \\ x & -x & 1 \end {vmatrix}\\ \)
Expanding along R1, we have:
\( \\\Delta = \left ( 1 – x ^ {3} \right ) \left ( 1 – x \right ) \left ( x \right ) \begin {vmatrix} 1 + x & x \\ -x & 1 \end {vmatrix}\\ \)= (1 – x3) (1 – x) (1 + x + x2)
= (1 – x3) (1 – x3)
= (1 – x3)2
Therefore, the given result is proved Based on the rules and formulae given in Determinant tricks.
Q.13: By using properties of determinants, show that:
\( \begin {vmatrix} 1 + a ^ {2} – b ^ {2} & 2ab & -2b \\ 2ab & 1 – a ^ {2} + b ^ {2} & 2a \\ 2b & -2a & 1 – a ^ {2} – b ^ {2} \end {vmatrix} = \left ( 1 + a ^ {2} + b ^ {2} \right ) ^ {3} \)
Sol: Based on the rules and formulae given in Determinant tricks
\( \Delta = \begin {vmatrix} 1 + a ^{ 2} – b ^ {2} & 2ab & -2b \\ 2ab & 1 – a ^ {2} + b ^ {2} & 2a \\ 2b & -2a & 1 – a ^ {2} – b ^ {2} \end {vmatrix}\\ \)Applying R1 → R1 + bR3 and R2 → R2 – aR3, we have:
\( \\\Delta = \begin {vmatrix} 1 + a ^ {2} + b ^ {2} & 0 & -b \left ( 1 + a ^ {2} + b ^ {2} \right ) \\ 0 & 1 + a ^ {2} + b ^ {2} & a \left ( 1 + a ^ {2} + b ^ {2} \right ) \\ 2b & -2a & 1 – a ^ {2} – b ^ {2} \end {vmatrix}\\ \)= \(\\ \left ( 1 + a ^ {2} + b ^ {2} \right ) \begin {vmatrix} 1 & 0 & -b \\ 0 & 1 & a \\ 2b & -2a & 1 – a ^ {2} – b ^ {2} \end {vmatrix}\\ \)
Expanding along R1, we have:
\(\\ \Delta = \left ( 1 + a ^ {2} + b ^ {2} \right ) ^ {2} \begin {bmatrix} (1) \begin {vmatrix} 1 & a \\ -2a & 1 – a^ {2} – b^ {2} \end {vmatrix} \\ -b \begin {vmatrix} 0 & 1 \\ 2b & -2a \end {vmatrix} \end {bmatrix} \\\)= (1 + a2 + b2)2 [1 – a2 – b2 + 2a2 –b (-2b)]
= (1 + a2 + b2)2 (1 + a2 + b2)
= (1 + a2 + b2)3
Q.14: By using properties of determinants, show that:
\( \begin {vmatrix} a ^ {2} + 1 & ab & ac \\ ab & b ^ {2} + 1 & bc \\ ca & cb & c ^ {2} + 1 \end {vmatrix} = 1 + a ^ {2} + b ^ {2} + c ^ {2} \)
Sol: Based on the rules and formulae given in Determinant tricks
\( \Delta = \begin {vmatrix} a ^ {2} + 1 & ab & ac \\ ab & b ^ {2} + 1 & bc \\ ca & cb & c ^ {2} + 1 \end {vmatrix} \)Taking out common factors a, b, c from R1, R2 and R3 respectively, we have:
\( \Delta = abc \begin {vmatrix} a + \frac {1} {b} & b & c \\ a & b + \frac {1} {b} & c \\ a & b & c + \frac {1} {c} \end {vmatrix}\\ \)Applying R2 → R2 – R1 and R3 → R3 – R1, we have:
\(\\ \Delta = abc \begin {vmatrix} a + \frac {1} {b} & b & c \\ – \frac {1} {a} & \frac {1} {b} & 0 \\ – \frac {1} {a} & 0 & \frac {1} {c} \end {vmatrix}\\ \)Applying C1 → aC1, C2 → bC2, and C3 → cC3, we have:
\( \\\Delta = abc \times \frac {1} {abc} \begin{vmatrix} a ^{2} + 1 & b^ {2} & c ^{2} \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end {vmatrix} \\\)= \(\\ \begin {vmatrix} a ^ {2} + 1 & b ^ {2} & c ^ {2} \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end {vmatrix}\\ \)
Expanding along R3, we have:
\( \\\Delta = -1 \begin {vmatrix} b ^ {2} & c ^ {2} \\ 1 & 0 \end {vmatrix} + 1 \begin {vmatrix} a ^ {2} + 1 & b ^ {2} \\ -1 & 1 \end {vmatrix} \\\)= -1(-c2) + (a2 + 1 + b2) = 1 + a2 + b2 + c2
Therefore, the given result is proved.
Q.15: Choose the correct answer:
Let A be a square matrix of order 3 × 3, KA is equal to
(1). k|A|
(2). k2|A|
(3). k3|A|
(4). 3k|A|
Answer: Based on the rules and formulae given in Determinant tricks
(3)
A is a square matrix of order 3 × 3
Let A = \( \begin {vmatrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end {vmatrix} \\\)
Then, kA = \( \begin{vmatrix} ka_{1} & kb_{1} & kc_{1} \\ ka_{2} & kb_{2} & kc_{2} \\ ka_{3} & kb_{3} & kc_{3} \end {vmatrix} \)
Therefore, \(\\|kA| = \begin {vmatrix} ka_{1} & kb_{1} & kc_{1} \\ ka_{2} & kb_{2} & kc_{2} \\ ka_{3} & kb_{3} & kc_{3} \end {vmatrix}\\ \)
\(\\ k ^ {3} = \begin {vmatrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end {vmatrix} \; \; \; \; \; \; \; \; \; \; \left ( Taking \; out \; common \; factors \; k \; from \; each \; row \right ) \)= k3|A|
Therefore, |kA| = k3|A|
Therefore C is right.
Q.16: Which of the following is correct?
(1) Determinant is a square matrix.
(2) Determinant is a number associated to a matrix.
(3) Determinant is a number associated to a square matrix.
(4) None of these answer
Sol: Based on the rules and formulae given in Determinant tricks
(3)
We know that as per given data , to every square matrix, A = [aij] of order n. we can associate a number called the determinant of a square matrix A, where aij = (i, j)th element of A.
Thus, the determinant is a number associated to a square matrix.
Therefore, the correct answer is C Based on the rules and formulae given in Determinant tricks.
Q.1: Find area of the triangle with vertices at the point given in each of the following:
(i) (1, 0), (6, 0), (4, 3)
(ii) (2, 7), (1, 1), (10, 8)
(iii) (−2, −3), (3, 2), (−1, −8)
Sol:
(i) The area of the triangle with vertices (1, 0), (6, 0), (4, 3) is given by the relation,
\(\Delta =\frac{1}{2} \begin{vmatrix} 1 & 0 & 1 \\ 6 & 0 & 1 \\ 4 & 3 & 1 \end{vmatrix}\) \(=\frac{1}{2}\left[1(0-3)-0(6-4)+1(18-0) \right ]\\\)\(\\=\frac{1}{2}\left[-3+18 \right ]\) = \(\frac{15}{2}\) unit2
(ii) The area of the triangle with vertices (2, 7), (1, 1), (10, 8) is given by the relation,
\(\Delta =\frac{1}{2} \begin{vmatrix} 2 & 7 & 1 \\ 1 & 1 & 1 \\ 10 & 8 & 1 \end{vmatrix}\) \(=\frac{1}{2}\left[2(1-8)-7(1-10)+1(8-0) \right ]\\\)= \(\\\frac{1}{2}\left[-14+63-2 \right ]=\frac{1}{2}\left[-16+63 \right ]=\frac{47}{2} \) unit2
(iii) The area of the triangle with vertices (−2, −3), (3, 2), (−1, −8) is given by the relation,
\(\Delta =\frac{1}{2} \begin{vmatrix} -2 & -3 & 1 \\ 3 & 2 & 1 \\ -1 & -8 & 1 \end{vmatrix}\) \(=\frac{1}{2}\left[-2(2+8)+3(3+1)+1(-24+2) \right ]\\\)=\(\\\frac{1}{2}\left[-20+12-22 \right ]\;=\;-\frac{30}{2}=-15\) unit2
Therefore, the area of the triangle is \(\begin{vmatrix} -15 \end{vmatrix}=15\) unit2 Based on the rules and formulae given in Determinant tricks
Q.2: Show that points A(a,b+c),B(b,c+a),C(c,a+b) are collinear
Sol: Based on the rules and formulae given in Determinant tricks
Area of ∆ABC is given by the relation:
\(\\\Delta=\frac{1}{2}\begin{vmatrix} a & b+c & 1\\ b & c+a & 1\\ c & a+b & 1 \end{vmatrix}\\\)= \(\\\frac{1}{2}\begin{vmatrix} a & b+c & 1\\ b-a & a-b & 0\\ c-a & a-b & 0 \end{vmatrix}\\\)
= \(\frac{1}{2}(a-b)(c-a)\begin{vmatrix} a & b+c & 1\\ -1 & 1 & 0\\ a & b+c & 1 \end{vmatrix}\\\)
= \(\\\frac{1}{2}(a-b)(c-a)\begin{vmatrix} a & b+c & 1\\ -1 & 1 & 0\\ a & b+c & 1 \end{vmatrix}\)
=0
Thus, the area of the triangle formed by points A, B, and C is zero.The points A, B, and C are collinear is proved.
Q.3: Find values of k if area of triangle is 4 square units and vertices are
(i) (k, 0), (4, 0), (0, 2)
(ii) (−2, 0), (0, 4), (0, k)
Sol: Determinants
We know that the area of a triangle whose vertices are\((x_{1}, y_{1}), (x_{2}, y_{2}), and (x_{3}, y_{3})\) is the absolute value of the determinant (∆), where
\(\Delta=\frac{1}{2}\begin{vmatrix} x_{1} & y_{1} & 1\\ x_{2} & y_{2} & 1\\ x_{3} & y_{3} & 1 \end{vmatrix}\)It is given that the area of triangle is 4 square units.
Therefore \(\Delta=\pm 4\)
(i) The area of the triangle with vertices (k, 0), (4, 0), (0, 2) is given by the relation:
\(\Delta=\frac{1}{2}\begin{vmatrix} k & 0 & 1\\ 4 & 0 & 1\\ 0 & 2 & 1 \end{vmatrix}\) \(=\frac{1}{2}[k(0-2)-0(4-0)+1(8-0)]\\\)= \(\\\frac{1}{2}[-2k+8]\;=\;-k+4 \;=\;-k+4=\pm4\)
When −k + 4 = − 4, k = 8.
When −k + 4 = 4, k = 0.
Therefore, k = 0, 8.
(ii) The area of the triangle with vertices (−2, 0), (0, 4), (0, k) is given by the relation:
i.e. \(\\k-4 = \pm 4\)
When k-4 = − 4, k = 0.
When k-4 = 4, k = 8.
Therefore, k = 0, 8.
Q.4: (i) Find equation of line joining (1, 2) and (3, 6) using determinants
(ii) Find equation of line joining (3, 1) and (9, 3) using determinants
Sol:
(i) Let P (x, y) be any point on the line joining points A (1, 2) and B (3, 6). Then, the
points A, B, and P are collinear. Therefore, the area of triangle ABP will be zero.
6-y-6+2x+3y-6x=0
2y-4x=0
y=2x
Therefore, the equation of the line joining the given points is y = 2x.
(ii) Let P (x, y) be any point on the line joining points A (3, 1) and
B (9, 3). Then, the points A, B, and P are collinear. Therefore, the area of triangle ABP
will be zero.
= \(\\\frac{1}{2}[3(3-y)-1(9-x)+1(9y-3x)]=0\\\)
9 – 3y – 9 + x + 9y – 3x = 0
6y – 2x = 0
x – 3y = 0
Therefore, the equation of the line joining the given points is x − 3y = 0.
Q.5: If area of triangle is 35 square units with vertices (2, −6), (5, 4), and (k, 4). Then k is
A. 12 B. −2 C. −12, −2 D. 12, −2
Sol:
D
The area of the triangle with vertices (2, −6), (5, 4), and (k, 4) is:
\(\Delta=\frac{1}{2}\begin{vmatrix} 2 & -6 & 1\\ 5 & 4 & 1\\ k & 4 & 1 \end{vmatrix}=0\) \(=\frac{1}{2}[2(4-4)+6(5-k)+1(20-4k)]\) \(=\frac{1}{2}[30-6k+20-4k]\)\(=\frac{1}{2}[50-10k]\) = 25 – 5k
It is given that the area of the triangle is ±35.
Therefore, we have:
\(25-5k=\pm35\) \(5(5-k)=\pm35\) \(5-k=\pm7\)When 5-k = −7, k = 5+7 = 12.
When 5-k = 7, k = 5-7 = −2.
Therefore, k = 12, −2.
The correct Sol: is D.
Q1: Write Minors and Cofactors of the elements of following determinants:
(i) \(\begin{vmatrix} 2 & -4 \\ 0 & 3 \end{vmatrix}\)
(ii) \(\begin{vmatrix} a & c \\ b & d \end{vmatrix}\)
Sol:
(i) The given determinant is \(\begin{vmatrix} 2 & -4 \\ 0 & 3 \end{vmatrix}\)
Minor of element aij is Mij .
Therefore, M11 = minor of element a11 = 3
M12 = minor of element a12 = 0
M21 = minor of element a21 = -4
M22 = minor of element a22 = 2
Cofactor of aij is Aij = (-1) i+j Mij .
Therefore, A11 = (-1)1+1 M11 = (-1)2 (3) = 3
A12 = ( -1 )1 + 2 M12 = ( -1 )3 ( 0 ) = 0
A21 = ( – 1 )2 + 1 M21 = ( -1 )3 ( -4 ) = 4
A22 = ( -1 )2 + 2 M22 ( -1 )4 ( 2 ) = 2
(ii) The given determinant is \(\begin{vmatrix} a & c \\ b & d \end{vmatrix}\)
Minor of element aij is Mij .
Therefore, M11 = minor of element a11 is Mij .
M12 = minor of element a12 = b
M21 = minor of element a21 = c
M22 = minor of element a22 = a
Cofactor of aij is Aij = (-1)I + j Mij .
Therefore, A11 = (-1)1 + 1 M11 = (-1)2 (d) = d
A12 = ( -1 )1 + 2 M12 = ( -1 )3 ( b ) = -b
A21 = ( -1 )2 + 1 M21 = ( -1 )3 ( c ) = -c
A22 = ( -1 ) 2 + 2 M22 = ( -1 )4 ( a ) = a
Q2: (i) \(\begin{vmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}\)
(ii) \(\begin{vmatrix} 1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2 \end{vmatrix}\)
Sol: Determinants
(i) The given determinant is \(\begin{vmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}\)
By the definition of minors and cofactors, we have :
M11 = minor of a11 = \(\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}\) = 1
M12 = minor of a12 = \(\begin{vmatrix} 0 & 0 \\ 0 & 1 \end{vmatrix}\) = 0
M13 = minor of a13 = \(\begin{vmatrix} 0 & 1 \\ 0 & 0 \end{vmatrix}\) = 0
M21 = minor of a21 = \(\begin{vmatrix} 0 & 0 \\ 0 & 1 \end{vmatrix}\) = 0
M22 = minor of a22 = \(\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}\) = 1
M23 = minor of a23 = \(\begin{vmatrix} 1 & 0 \\ 0 & 0 \end{vmatrix}\) = 0
M31 = minor of a31 = \(\begin{vmatrix} 0 & 0 \\ 1 & 0 \end{vmatrix}\) = 0
M32 = minor of a32 = \(\begin{vmatrix} 1 & 0 \\ 0 & 0 \end{vmatrix}\) = 0
M33 = minor of a33 = \(\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}\) = 1
A11 = cofactor of a11 = ( -1 )1 + 1 M11 = 1
A12 = cofactor of a12 = ( -1 )1 + 2 M12 = 0
A13 = cofactor of a13 = ( -1 )1 + 3 M13 = 0
A21 = cofactor of a21 = ( -1 )2 + 1 M21 = 0
A22 = cofactor of a22 = ( -1 )2 + 2 M22 = 1
A23 = cofactor of a23 = ( -1 )2 + 3 M23 = 0
A31 = cofactor of a31 = ( -1 )3 + 1 M31 = 0
A32 = cofactor of a32 = ( -1 )3 + 2 M32 = 0
A33 = cofactor of a33 = ( -1 )3 + 3 M33 = 1
(ii) The given determinant is \(\begin{vmatrix} 1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2 \end{vmatrix}\)
By definition of minors and cofactors , we have:
M11 = minor of a11 = \(\begin{vmatrix} 5 & -1 \\ 1 & 2 \end{vmatrix}\) = 10 + 1 = 11
M12 = minor of a12 = \(\begin{vmatrix} 3 & -1 \\ 0 & 2 \end{vmatrix}\) = 6 – 0 = 6
M13 = minor of a13 = \(\begin{vmatrix} 3 & 5 \\ 0 & 1 \end{vmatrix}\) = 3 – 0 = 3
M21 = minor of a21 = \(\begin{vmatrix} 0 & 4 \\ 1 & 2 \end{vmatrix}\) = 0 – 4 = -4
M22 = minor of a22 = \(\begin{vmatrix} 1 & 4 \\ 0 & 2 \end{vmatrix}\) = 2 – 0 = 2
M23 = minor of a23 = \(\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}\) = 1 – 0 = 1
M31 = minor of a31 = \(\begin{vmatrix} 0 & 4 \\ 5 & -1 \end{vmatrix}\) = 0 – 20 = -20
M32 = minor of a32 = \(\begin{vmatrix} 1 & 4 \\ 3 & -1 \end{vmatrix}\) = -1 – 12 = -13
M33 = minor of a33 = \(\begin{vmatrix} 1 & 0 \\ 3 & 5 \end{vmatrix}\) = 5 – 0 = 5
A11 = cofactor of a11 = ( -1 )1 + 1 M11 = 11
A12 = cofactor of a12 = ( -1 )1 + 2 M12 = -6
A13 = cofactor of a13 = ( -1 )1 + 3 M13 = 3
A21 = cofactor of a21 = ( -1 )2 + 1 M21 = 4
A22 = cofactor of a22 = ( -1 )2 + 2 M22 = 2
A23 = cofactor of a23 = ( -1 )2 + 3 M23 = -1
A31 = cofactor of a31 = ( -1 )3 + 1 M31 = -20
A32 = cofactor of a32 = ( -1 )3 + 2 M32 = 13
A33 = cofactor of a33 = ( -1 )3 + 3 M33 = 5
Q.3: Using Cofactors of elements of second row, evaluate \(\Delta = \begin{vmatrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{vmatrix}\) .
Sol:
The given determinant is \(\Delta = \begin{vmatrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{vmatrix}\)
We have:
M21 = \(\begin{vmatrix} 3 & 8\\ 2 & 3 \end{vmatrix}\) = 9 – 16 = -7
Therefore, A21 = cofactor of a21 = ( -1 )2 + 1 M21 = 7
M22 = \(\begin{vmatrix} 5 & 8\\ 1 & 3 \end{vmatrix}\) = 15 – 8 = 7
Therefore, A22 = cofactor of a22 = ( -1 )2 + 2 M22 = 7
M23 = \(\begin{vmatrix} 5 & 3\\ 1 & 2 \end{vmatrix}\) = 10 – 3 = 7
Therefore, A23 = cofactor of a23 = ( -1 )2 + 3 M23 = -7
We know that \(\Delta\) is equal to the sum of the product of the elements of the second row with their corresponding cofactors.
Therefore, \(\Delta\) = a21A21 + a22A22 + a23A23 =
2 ( 7 ) + 0 ( 7 ) + 1 ( -7 ) = 14 – 7 = 7
Q4: Using Cofactors of elements of third column, evaluate \(\Delta = \begin{vmatrix} 1 & x & yz \\ 1 & y & zx \\ 1 & z & xy \end{vmatrix}\) .
Sol: Determinants
The given determinant is \(\begin{vmatrix} 1 & x & yz\\ 1 & y & zx \\ 1 & z & xy \end{vmatrix}\) .
We have:
M13 = \(\begin{vmatrix} 1 & y \\ 1 & z \end{vmatrix}\) = z – y
M23 = \(\begin{vmatrix} 1 & x \\ 1 & z \end{vmatrix}\) = z – x
M33 = \(\begin{vmatrix} 1 & x \\ 1 & \end{vmatrix}\) = y – x
Therefore, A13 = cofactor of a13 = ( -1 )1 + 3 M13 = ( z – y )
A23 = cofactor of a23 = ( -1 )2 + 3 M23 = – ( z – x ) = ( x – z )
A33 = cofactor of a33 = ( -1 )3 + 3 M33 = ( y – x )
We know that \(\Delta\) is equal to the sum of the product of the elements of the second row with their corresponding cofactors.
Therefore, \(\Delta\) = a13A13 + a23A23 + a33A33
= yz ( z – y ) + zx ( x – z ) + xy ( y – x )
= yz2 – y2 z + x2 z + x2 z – xz2 + xy2 – x2 y
= ( x2 z – y2 z ) + ( yz2 – xz2 ) + ( xy2 – x2 y )
= z ( x2 – y2 ) + z2 ( y – x ) + xy ( y – x )
= z ( x – y ) ( x + y ) + z2 ( y – x ) + xy ( y – x )
= ( x – y ) [ zx + zy – z2 – xy ]
= ( x – y ) [ z ( x – z ) + y ( z – x ) ]
= ( x – y ) ( z – x ) [ -z + y ]
= ( x – y ) ( y – z ) ( z – x )
Therefore, \(\Delta\) = ( x – y ) ( y – z ) ( z – x )
Q5: For the matrices A and B, verify that (AB)’ = B’A’ where
(i) A = \(\begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix}\) ,
B = \(\begin{bmatrix} -1 & 2 & 1 \end{bmatrix}\)
(ii) A = = \(\begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}\) ,
B = \(\begin{bmatrix} 1 & 5 & 7 \end{bmatrix}\)
Sol: Determinants
(i) AB = \(\begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix} \begin{bmatrix} -1 & 2 & 1 \end{bmatrix} = \begin{bmatrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3 \end{bmatrix}\)
Therefore, ( AB )’ = \(\begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}\)
Now, A’ = \(\begin{bmatrix} -1 & -4 & 3 \end{bmatrix}\) ,
B’ = \(\begin{bmatrix} -1\\ 2 \\ 1 \end{bmatrix}\)
Therefore, B’A’= \(\begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix} \begin{bmatrix} 1 & -4 & 3 \end{bmatrix} = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}\)
Therefore, we have verified that ( AB )’ = B’A’ .
(ii) AB = \(\begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix} \begin{bmatrix} 1 & 5 & 7 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 5 & 7 \\ 2 & 10 & 14 \end{bmatrix}\)
Therefore, ( AB )’ = \(\begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix}\)
Now, A’ = \(\begin{bmatrix} 0 & 1 & 2 \end{bmatrix}\) ,
B’ = \(\begin{bmatrix} 1 \\ 5 \\ 7 \end{bmatrix}\)
Therefore, B’A’= \(\begin{bmatrix} 1 \\ 5 \\ 7 \end{bmatrix} \begin{bmatrix} 0 & 1 & 2 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix}\)
Therefore, we have verified that (AB)’ = B’A’
Q.1: Find the adjoint of each of the matrices
\(\begin{bmatrix} 2 & 4 \\ 5 & 7 \end{bmatrix}\)
Sol: Determinants
Suppose A = \(\begin{bmatrix} 2 & 4 \\ 5 & 7 \end{bmatrix}\)
For X = \(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\\\)
Then adj (X) = \(\begin{bmatrix} d & – b \\ – c & a \end{bmatrix}\)
We have,
A 11 = 7, A12 = – 4, A21 = -5, A22 = 2
Therefore, adj (A) = \(\begin{bmatrix} 7 & – 4 \\ – 5 & 2 \end{bmatrix}\)
Q.2: Find the adjoint of each of the matrices
\(\begin{bmatrix} 1 & – 1 & 2 \\ 2 & 3 & 5 \\ – 2 & 0 & 1 \end{bmatrix}\)
Sol: Determinants
Suppose, D = \(\begin{bmatrix} 1 & – 1 & 2 \\ 2 & 3 & 5 \\ – 2 & 0 & 1 \end{bmatrix}\)
We have,
\(D_{11} = \begin{vmatrix} 3 & 5 \\ 0 & 1 \end{vmatrix} = 3 – 0 = 3 \\ D_{12} = – \begin{vmatrix} 2 & 5 \\ – 2 & 1 \end{vmatrix} = – (2 + 10) = – 12 \\ D_{13} = \begin{vmatrix} 2 & 3 \\ – 2 & 0 \end{vmatrix} = 0 + 6 = 6 \\\) \(D_{21} = – \begin{vmatrix} – 1 & 2 \\ 0 & 1 \end{vmatrix} = – (- 1 – 0) = 1 \\ D_{22} = \begin{vmatrix} 1 & 2 \\ – 2 & 1 \end{vmatrix} = 1 + 4 = 5 \\ D_{23} = – \begin{vmatrix} 1 & – 1 \\ – 2 & 0 \end{vmatrix} = – (0 – 2) = 2 \\\) \(D_{31} = \begin{vmatrix} – 1 & 2 \\ 3 & 5 \end{vmatrix} = – 5 – 6 = – 11 \\ D_{32} = – \begin{vmatrix} 1 & 2 \\ 2 & 5 \end{vmatrix} = – (5 – 4) = – 1 \\ D_{33} = \begin{vmatrix} 1 & – 1 \\ 2 & 3 \end{vmatrix} = 3 + 2 = 5\\\)Therefore, adj D = \(\\\begin{bmatrix} D ^{11} & D ^{12} & D ^{13} \\ D ^{21} & D ^{22} & D ^{23} \\ D ^{31} & D ^{32} & D ^{33} \end{bmatrix} = \begin{bmatrix} 3 & 1 & – 11 \\ – 12 & 5 & – 1 \\ 6 & 2 & 5 \end{bmatrix}\)
Q.3: Prove whether D (adj D) = (adj D) D = \(\left | D \right | I\)
\(D = \begin{bmatrix} 2 & 3 \\ – 4 & – 6 \end{bmatrix}\)
Sol:
\(D = \begin{bmatrix} 2 & 3 \\ – 4 & – 6 \end{bmatrix}\)We have,
\(\left | D \right | = – 12 – (- 12) = – 12 + 12 = 0 \\ \left | D \right | I = 0 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)D 11 = – 6, D 12 = 4, D 21 = – 3, D 22 = 2
adj D = \(\begin{bmatrix} – 6 & – 3 \\ 4 & 2 \end{bmatrix}\)
Now,
D (adj D) = \(\begin{bmatrix} 2 & 3 \\ – 4 & – 6 \end{bmatrix} \begin{bmatrix} – 6 & – 3 \\ 4 & 2 \end{bmatrix} \\ = \begin{bmatrix} – 12 + 12 & – 6 + 6 \\ 24 – 24 & 12 – 12 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)
And,
(adj D) D = \(\begin{bmatrix} – 6 & – 3 \\ 4 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ – 4 & – 6 \end{bmatrix} \\ = \begin{bmatrix} – 12 + 12 & – 18 + 18 \\ 8 – 8 & 12 – 12 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)
Thus, D (adj D) = (adj D) D = \(\left | D \right | I\)
Q.4: Obtain the inverse of the matrices if it exists
\(\begin{bmatrix} 2 & – 2 \\ 4 & 3 \end{bmatrix}\)
Sol: Determinants
Suppose D = \(\begin{bmatrix} 2 & – 2 \\ 4 & 3 \end{bmatrix}\)
We have,
\(\left | D \right | = 6 + 8 = 14\)Now,
D 11 = 3, D 12 = – 4, D 21 = 2, D 22 = 2
adj D = \(\begin{bmatrix} 3 & 2 \\ – 4 & 2 \end{bmatrix} \\ D ^{- 1} = \frac{1}{\left | D \right |} = \frac{1}{14}\; \begin{bmatrix} 3 & 2 \\ – 4 & 2 \end{bmatrix}\)
Q.5: Obtain the inverse of the matrices if it exists
\(\begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{bmatrix}\)
Sol: Determinants
Suppose D = \(\begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{bmatrix}\)
We have,
\(\left | D \right | = 1 (10 – 0) – 2 (0 – 0) + 3 (0 – 0) = 10\)Now, D 11 = 10 – 0 = 10, D 12 = – (0 – 0) = 0, D 13 = 0 – 0 = 0, D 21 = – (10 – 0) = – 10, D 22 = 5 – 0 = 5, D 23 = – (0 – 0) = 0, D 31 = 8 – 6 = 2, D 32 = – (4 – 0) = – 4, D 33 = 2 – 0 = 2
adj D = \(\begin{bmatrix} 10 & – 10 & 2 \\ 0 & 5 & – 4 \\ 0 & 0 & 2 \end{bmatrix} \\\)
D-1 = \(\\\frac{1}{\left | D \right |} adj D = \frac{1}{10} \begin{bmatrix} 10 & – 10 & 2 \\ 0 & 5 & – 4 \\ 0 & 0 & 2 \end{bmatrix}\)
Q.6: Find the inverse of each of the matrices (if it exists).
\(\begin{bmatrix} 1 & 0 & 0 \\ 0 & cos \alpha & sin \alpha \\ 0 & sin \alpha & – cos \alpha \end{bmatrix}\)
Sol:
Let A = \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & cos \alpha & sin \alpha \\ 0 & sin \alpha & – cos \alpha \end{bmatrix}\)
We have,
\(\left |A \right | = 1 \left ( – cos ^{2} \alpha – sin ^{2} \alpha\right )\) \(\left |A \right | = – \left ( cos ^{2} \alpha + sin ^{2} \alpha \right ) = – 1\)Now,
\(A _{ 11 } = – cos ^{2 } \alpha – sin ^{2} \alpha = – 1 , A _{ 12 } = 0 , A _{ 13 } = 0\) \(A _{21} = 0 , A _{ 22 } = – cos \alpha , A _{ 23 } = – sin \alpha \\ A _{ 31 } = 0 , A_{32} = – sin \alpha , A_{33} = cos \alpha\)Therefore,
\(adj A = \begin{bmatrix} -1 & 0 & 0 \\ 0 & – cos \alpha & – sin \alpha \\ 0 & – sin \alpha & cos \alpha \end{bmatrix}\)Therefore,
\(A ^{ -1 } = \frac{ 1 }{ \left | A \right |}.adj A\) = \(\begin{bmatrix} -1 & 0 & 0 \\ 0 & – cos \alpha & – sin \alpha \\ 0 & – sin \alpha & cos \alpha \end{bmatrix}\) = \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & cos \alpha & sin \alpha \\ 0 & sin \alpha & – cos \alpha \end{bmatrix}\)
Q.7: Let \(A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}\) and \(B = \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix}\). Verify that \(\left ( AB \right )^{ -1} = B ^{-1} A ^{ -1 }\)
Sol: Determinants
Let \(A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}\)
We have ,
\(\left |A \right | = 15 – 14 = 1\)Now,
\(A _{ 11} = 5 , A _{ 12} = – 2 , A _{ 21} = – 7 , A _{ 22} = 3\)Therefore,
\(adj A = \begin{bmatrix} 5 & – 7 \\ – 2 & 3 \end{bmatrix}\)Therefore,
\(A ^{ – 1 } = \frac{1}{ \left |A \right |}. adj A = \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}\)Now, let \(adj B = \begin{bmatrix} 6 & 8 \\ 7 & 6 \end{bmatrix}\)
We have ,
\(\left |B \right | = 54 – 56 = – 2\)Therefore,
\(adj B = \begin{bmatrix} 9 & – 8 \\ – 7 & 6 \end{bmatrix}\)Therefore,
\(B ^{ – 1 } = \frac{1}{ \left |B \right |}. adj B = 0.5 \begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix}\) \(\begin{bmatrix} \frac{ – 9}{2} & 4 \\ \frac{7}{2} & – 3 \end{bmatrix}\)Now,
\(B^{-1}A^{-1} = \begin{bmatrix} \frac{-9}{2} & 4 \\ \frac{7}{2} & -3 \end{bmatrix}\begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}\) \(B^{-1}A^{-1} = \begin{bmatrix} \frac{-45}{2} – 8 & \frac{63}{2} + 12 \\ \frac{35}{2} + 6 & \frac{-49}{2} – 9 \end{bmatrix}\begin{bmatrix} \frac{ 61}{2} & \frac{87}{2} \\ \frac{47}{2} & \frac{ -67}{2} \end{bmatrix}\)Then,
\(\\\boldsymbol{\Rightarrow }\) \(AB = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}\begin{bmatrix} 6 & 7 \\ 7 & 9 \end{bmatrix}\\\)
\(\\\boldsymbol{\Rightarrow }\) \(AB = \begin{bmatrix} 18 + 49 & 24 + 63 \\ 12 + 35 & 16 + 45 \end{bmatrix}\\\)
\(\\\boldsymbol{\Rightarrow }\) \(AB = \begin{bmatrix} 67 & 87 \\ 47 & 61 \end{bmatrix}\\\)
Therefore, we have \(AB = \begin{bmatrix} 18 + 49 & 24 + 63 \\ 12 + 35 & 16 + 45 \end{bmatrix}\)
Also,
\(adj (AB) = \begin{bmatrix} 61 & – 87 \\ – 47 & 67 \end{bmatrix}\)Therefore, \(\left ( AB \right )^{ -1 } = \frac{1}{ \left |AB \right |} adj\left ( AB \right ) = \frac{-1}{2}\begin{bmatrix} 61 & – 87 \\ – 47 & 67 \end{bmatrix}\) . . . . . . . . . . . . . . . (2)
\(= \begin{bmatrix} \frac{61}{2} & \frac{87}{2}\\ \frac{47}{2} & \frac{-67}{2} \end{bmatrix}\)From (1) and (2), we have: (AB)−1 = B−1 A -1 Therefore, the given result is proved.
Q.8: If \(A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\), show that A 2 – 5 A + 7 I = 0. Therefore find A – 1
Sol:
\(A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\)A 2 = A. A = \(\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\)\(\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\) = \(A = \begin{bmatrix} 9 – 1 & 3 +2 \\ -3 – 2 & – 1 + 4 \end{bmatrix}\) = \(\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}\)
Therefore,
A 2 – 5 A + 7 I = 0
= \(\\\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}\) – 5 \(\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\) + 7 \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\\\)
= \(\\\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}\) – \(\begin{bmatrix} 15 & 5 \\ – 5 & 10 \end{bmatrix}\) + \(\begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}\\\)
= \(\\\begin{bmatrix} – 7 & 0 \\ 0 & – 7 \end{bmatrix}\) – \(\begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}\) = \(\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\\\)
Therefore, A 2 – 5 A + 7 I = 0
Therefore, A.A – 5 A = – 7I
\(\boldsymbol{\Rightarrow }\) A.A ( A -1 ) – 5A A -1 = – 7I A -1 . . . . . . . . . . . . . . . . . [ post – multiplying by A -1 as \(\left | A \right | \neq 0\) ]
\(\boldsymbol{\Rightarrow }\) A. ( A A -1 ) – 5I = – 7 A -1
\(\boldsymbol{\Rightarrow }\) AI – 5I = – 7 A -1
\(\boldsymbol{\Rightarrow }\) A -1 \(= – \frac{ 1 }{ 7 }\left ( A – 5I \right )\)
\(\boldsymbol{\Rightarrow }\) A -1 \(= – \frac{ 1 }{ 7 }\left ( 5I – A \right )\)
=\(\\\frac{1}{7} \left ( \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} – \begin{bmatrix} 3 & 1 \\ – 1 & 2 \end{bmatrix} \right ) = \frac{1}{ 7 } \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}\)
Therefore,
A -1 = \(\frac{1}{ 7 } \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}\)
Q.9: Let A be a nonsingular square matrix of order 3 × 3. Then \(\left | adj A \right |\) is equal to
(A) \(\left | adj A \right |\)
(B) \(\left | adj A \right |^{2}\)
(C) \(\left | adj A \right |^{ 3 }\)
(D) 3 \(\left | adj A \right |\)
Sol: Determinants
The correct option is B
\(\left ( adj A \right )A = \left | A \right |I = \begin{bmatrix} \left | A \right | & 0 & 0 \\ 0 & \left | A \right | & 0 \\ 0 & 0 & \left | A \right | \end{bmatrix}\)\(\boldsymbol{\Rightarrow }\) \(\left ( adj A \right )A = \begin{bmatrix} \left | A \right | & 0 & 0 \\ 0 & \left | A \right | & 0 \\ 0 & 0 & \left | A \right | \end{bmatrix}\)
\(\boldsymbol{\Rightarrow }\) \(\left ( adj A \right )\left |A \right | = \left | A \right |^{3} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \left | A \right |^{3} \left (I \right )\)
Therefore, \(\left |\left ( adj A \right ) \right | = \left | A \right |^{2}\)
Therefore, the correct Sol: is B.
Q.10: If A is an invertible matrix of order 2, then det (A −1 ) is equal to
(A) det (A)
(B) \(\frac{ 1 }{ det \left ( A \right )}\)
(C) 1
(D) 0
Sol:
Since A is an invertible matrix , A -1 exists and A -1 = \(\frac{ 1 }{ \left | a \right |} adj A\)
As matrix A is of order 2, let \(A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\)
Then, \(\left |A \right | = ad – bc and adj A = \begin{bmatrix} d & -b \\ – c & a \end{bmatrix}\)
Now,
\(A^{-1} = \frac{1}{ \left | A \right |} adj A = \begin{bmatrix} \frac{d}{\left | A \right |} & \frac{-b}{ \left | A \right |}\\ \frac{-c}{ \left | A \right |} & \frac{a }{ \left | A \right |} \end{bmatrix}\)Therefore,
\(\left |A^{-1} \right | = \begin{bmatrix} \frac{d}{\left | A \right |} & \frac{-b}{ \left | A \right |}\\ \frac{-c}{ \left | A \right |} & \frac{a }{ \left | A \right |} \end{bmatrix} = \frac{1}{ \left | A \right |^{2}}\begin{bmatrix} d & – b \\ – c & a \end{bmatrix} = \frac{1}{ \left | A \right |^{2}} \left ( ad – bc \right ) = \frac{1}{ \left | A \right |^{2}} . \left |A \right | = \frac{1}{ \left | A \right |}\)Therefore,
\(det \left ( A \right )^{ -1} = \frac{ 1 }{ det \left ( A \right )}\)Therefore, the correct Sol: is B.
Q.11: Suppose D = \(\begin{bmatrix} 2 & – 1 & 1 \\ – 1 & 2 & – 1 \\ 1 & – 1 & 2 \end{bmatrix}\), verify that D 3 – 6 D2 + 9D – 4I = 0 and hence find D – 1
Sol: Determinants
D = \(\begin{bmatrix} 2 & – 1 & 1 \\ – 1 & 2 & – 1 \\ 1 & – 1 & 2 \end{bmatrix}\\\)
D 2 = \(\\\begin{bmatrix} 2 & – 1 & 1 \\ – 1 & 2 & – 1 \\ 1 & – 1 & 2 \end{bmatrix} \begin{bmatrix} 2 & – 1 & 1 \\ – 1 & 2 & – 1 \\ 1 & – 1 & 2 \end{bmatrix}\\\)
\(\\\begin{bmatrix} 4 + 1 + 1 & – 2 – 2 – 1 & 2 + 1 + 2 \\ – 2 – 2 – 1 & 1 + 4 + 1 & – 1 – 2 – 2 \\ 2 + 1 + 2 & – 1 – 2 – 2 & 1 + 1 + 4 \end{bmatrix} \\ = \begin{bmatrix} 6 & – 5 & 5 \\ – 5 & 6 & – 5 \\ 5 & – 5 & 6 \end{bmatrix}\\\)D 3 = D 2 D = \(\\\begin{bmatrix} 6 & – 5 & 5 \\ – 5 & 6 & – 5 \\ 5 & – 5 & 6 \end{bmatrix} \begin{bmatrix} 2 & – 1 & 1 \\ – 1 & 2 & – 1 \\ 1 & – 1 & 2 \end{bmatrix} \\ = \begin{bmatrix} 12 + 5 + 5 & – 6 – 10 – 5 & 6 + 5 + 10 \\ – 10 – 6 – 5 & 5 + 12 + 5 & 5 – 6 – 10 \\ 10 + 5 + 6 & – 5 – 10 – 6 & 5 + 5 + 12 \end{bmatrix} \\ = \begin{bmatrix} 22 & – 21 & 21 \\ – 21 & 22 & – 21 \\ 21 & – 21 & 22 \end{bmatrix}\\\)
Now,
D 3 – 6 D2 + 9D – 4I = 0
= \(\\\begin{bmatrix} 22 & – 21 & 21 \\ – 21 & 22 & – 21 \\ 21 & – 21 & 22 \end{bmatrix} – 6 \begin{bmatrix} 6 & – 5 & 5 \\ – 5 & 6 & – 5 \\ 5 & – 5 & 6 \end{bmatrix} + 9 \begin{bmatrix} 2 & – 1 & 1 \\ – 1 & 2 & – 1 \\ 1 & – 1 & 2 \end{bmatrix} – 4 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\\\)
= \(\\\begin{bmatrix} 22 & – 21 & 21 \\ – 21 & 22 & – 21 \\ 21 & – 21 & 22 \end{bmatrix} – \begin{bmatrix} 36 & – 30 & 30 \\ – 30 & 36 & – 30 \\ 30 & – 30 & 36 \end{bmatrix} + \begin{bmatrix} 18 & – 9 & 9 \\ – 9 & 18 & – 9 \\ 9 & – 9 & 18 \end{bmatrix} – \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix}\\\)
= \(\\ \begin{bmatrix} 40 & – 30 & 30\\ – 30 & 40 & – 30 \\ 30 & – 30 & 40 \end{bmatrix} – \begin{bmatrix} 40 & – 30 & 30\\ – 30 & 40 & – 30 \\ 30 & – 30 & 40 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\)
Therefore, D 3 – 6 D2 + 9D – 4I = 0
Now,
D 3 – 6 D2 + 9D – 4I = 0
(DDD) D – 1 – 6 D + 9D – 4I = 0 [Multiplying by D – 1 on as \(\left | D \right | I\) is not equal to 0]
DD (DD– 1) – 6D (DD– 1) + 9 (DD– 1) = 4 ID– 1
DDI – 6DI + 9I = 4 D– 1
D2 – 6D + 9I = 4 D– 1
D– 1 = \(\frac{1}{4}\) (D2 – 6D + 9I)
D2 – 6D + 9I
= \(\\\begin{bmatrix} 6 & – 5 & 5 \\ – 5 & 6 & – 5 \\ 5 & – 5 & 6 \end{bmatrix} – 6 \begin{bmatrix} 2 & – 1 & 1 \\ – 1 & 2 & – 1 \\ 1 & – 1 & 2 \end{bmatrix} + 9 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\\)
= \(\\\begin{bmatrix} 6 & – 5 & 5 \\ – 5 & 6 & – 5 \\ 5 & – 5 & 6 \end{bmatrix} – \begin{bmatrix} 12 & – 6 & 6 \\ – 6 & 12 & – 6 \\ 6 & – 6 & 12 \end{bmatrix} + \begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix} \\ = \begin{bmatrix} 3 & 1 & – 1 \\ 1 & 3 & 1 \\ – 1 & 1 & 3 \end{bmatrix} From\; equation\; (1), we\; have, D ^{- 1} = \frac{1}{4} \begin{bmatrix} 3 & 1 & – 1 \\ 1 & 3 & 1 \\ – 1 & 1 & 3 \end{bmatrix}\)
Q-1: Check the consistency for the system of two equations given below:
a + 3b = 2
2a + 4b = 3
Sol: Determinants
As per the data given in the question,
The system of the two equations is:
a + 3b = 2
2a + 4b = 3
The system of equations will be written as in the form of MX = N, where
\(M = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}, X = \begin{bmatrix} a \\ b \end{bmatrix} \; and \; N = \begin{bmatrix} 2 \\ 3 \end{bmatrix}\)Thus,
|M| = 1( 4 ) – 3( 2 ) = 3 – 4 = -1 ≠ 0
Therefore, M is non- singular.
Thus, M-1 exists.
Therefore, the given system of two equations will be consistent.
Q-2. Check the consistency of the system of two equations given below:
a + 4b = 5
2a + 7b = 8
Sol: Determinants
The given system of the two equations is:
a + 4b = 5
2a + 7b = 8
The given system of equations will be written as in the form of MX = N, where
\(M = \begin{bmatrix} 1 & 4 \\ 2 & 7 \end{bmatrix}, X = \begin{bmatrix} a \\ b \end{bmatrix} \; and \; N = \begin{bmatrix} 5 \\ 8 \end{bmatrix}\)Thus,
|M| = 1(7) – 4(2) = 7 – 8 = -1 ≠ 0
Therefore, M is a singular matrix.
Thus,
\(\left ( adj M \right ) = \begin{bmatrix} 7 & -4 \\ -2 & 1 \end{bmatrix}\) \(\left ( adj M \right )N = \begin{bmatrix} 7 & -4 \\ -2 & 1 \end{bmatrix}\begin{bmatrix} 5 \\ 8 \end{bmatrix}\)= \(\begin{bmatrix} 35 – 32 \\ -10 + 8 \end{bmatrix} = \begin{bmatrix} 3 \\ -2 \end{bmatrix} \neq 0\)
Therefore, the solution for the given system of the equation does not exist. Thus, the given system of the equations will be inconsistent.
Q-3. Check the consistency of the system of three equations given below:
a + b + c = 1
2a + 3b + 2c = 2
pa + pb + 2pc = 4
Sol: Determinants
The given system of the two equations is:
a + b + c = 1
2a + 3b + 2c = 2
pa + pb + 2pc = 4
The given system of three equations will be written as in the form MX = N, where
\(M = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 3 & 2 \\ p & p & 2p \end{bmatrix}, X = \begin{bmatrix} a \\ b \\ c \end{bmatrix} \; and \; N = \begin{bmatrix} 1 \\ 2 \\ 4 \end{bmatrix}\)|M| = 1( 6p – 2p ) – 1( 4p – 2p ) + 1( 2p – 3p )
= 1( 4p ) – 1( 2p ) + 1( -p )
= 4p – 2p – p
= p ≠ 0
Therefore, M is non- singular.
Thus, M-1 exists.
Therefore, the given system of two equations will be consistent.
Q-4. Check the consistency of the system of three equations given below:
3a – b – 2c = 2
2b – c = -1
3a – 5b = 3
Sol: Determinants
The given system of the two equations is:
3a – b – 2c = 2
2b – c = -1
3a – 5b = 3
The given system of three equations will be written as in the form MX = N, where
\(M = \begin{bmatrix} 3 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{bmatrix}, X = \begin{bmatrix} a \\ b \\ c \end{bmatrix} \; and \; N = \begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}\)|M| = 3[ 2 × 0 – ( -1 ) × ( -5 ) ] – 0[ ( -1 ) × 0 – ( -2 ) × ( -5 ) ] + 3[ ( -1 ) × ( -1 ) – ( -1 ) × ( 4 ) ]
= 3( 0 – 5 ) – 0 + 3(1 + 4)
= 3( -5 ) – 0 + 3( 5 )
= -15 + 15 = 0
Thus,
M is a singular matrix.
Then,
\(\left ( adj M \right ) = \begin{bmatrix} -5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6 \end{bmatrix}\) \(\left ( adj M \right )N = \begin{bmatrix} -5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6 \end{bmatrix}\begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}\)= \(\begin{bmatrix} -10 – 10 + 15 \\ -6 -6 + 9 \\ -12 – 12 + 18 \end{bmatrix} = \begin{bmatrix} -5 \\ -3 \\ -6 \end{bmatrix} \neq 0\)
Therefore, the solution for the given system of the equation does not exist. Thus, the given system of the equations will be inconsistent.
Q-5. Solve the following system of the linear equations by suing the matrix method:
5a + 2b = 4
7a + 3b = 5
Sol: Determinants
The given system of linear equations will be written as in the form of MX = N, where
\(M = \begin{bmatrix} 5 & 2 \\ 7 & 3 \end{bmatrix}, X = \begin{bmatrix} a \\ b \end{bmatrix} \; and \; N = \begin{bmatrix} 4 \\ 5 \end{bmatrix}\)|M| = ( 5 × 3 ) – ( 7 × 2 ) = 15 – 14 = 1 ≠ 0
Therefore, M is non- singular.
Thus, M-1 exists.
Now,
M-1 = \(\frac{ 1 }{ \left | M \right | } \left( adj M \right )\)
⟹ M-1 = \(\frac{ 1 }{ 1 }\begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}\)
Thus,
X = M-1 N = \(\begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}\begin{bmatrix} 4 \\ 5 \end{bmatrix}\)
⟹ \(\begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}\begin{bmatrix} 4 \\ 5 \end{bmatrix}\)
⟹ \(\begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 3 \times 4 -2 \times 5 \\ -7\times 4 + 5 \times 5 \end{bmatrix} = \begin{bmatrix} 12 – 10 \\ -28 + 25 \end{bmatrix} = \begin{bmatrix} 2 \\ -3 \end{bmatrix}\)
⟹ \(\begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 2 \\ -3 \end{bmatrix}\)
Therefore, a = 2 and b = -3.
Q-6. Solve the following system of the linear equations by suing the matrix method:
4a – 3b = 3
3a – 5b = 7
Sol: Determinants
The given system of linear equations will be written as in the form of MX = N, where
\(M = \begin{bmatrix} 4 & -3 \\ 3 & -5 \end{bmatrix}, X = \begin{bmatrix} a \\ b \end{bmatrix} \; and \; N = \begin{bmatrix} 3 \\ 7 \end{bmatrix}\)|M| = [ 4 × (-5) ] – [ ( -3 ) × 3 ] = -20 + 9= -11 ≠ 0
Therefore, M is non- singular.
Thus, M-1 exists.
Now,
M-1 = \(\frac{ 1 }{ \left | M \right | } \left( adj M \right )\)
⟹ M-1 = \(– \frac{ 1 }{ 11 }\begin{bmatrix} -5 & 3 \\ -3 & 4 \end{bmatrix}\) = \(\frac{ 1 }{ 11 }\begin{bmatrix} 5 & -3 \\ 3 & -4 \end{bmatrix}\)
Thus,
X = M-1 N = \( \frac{ 1 }{ 11 }\begin{bmatrix} 5 & -3 \\ 3 & -4 \end{bmatrix}\begin{bmatrix} 3 \\ 7 \end{bmatrix}\)
⟹ \(\begin{bmatrix} a \\ b \end{bmatrix} = \frac{ 1 }{ 11 } \begin{bmatrix} 5 & -3 \\ 3 & -4 \end{bmatrix}\begin{bmatrix} 3 \\ 7 \end{bmatrix}\)
⟹ \(\begin{bmatrix} a \\ b \end{bmatrix} = \frac{ 1 }{ 11 } \begin{bmatrix} 5 \times 3 + \left( -3 \right )\times 7 \\ 3\times 3 + \left( -4 \right ) \times 7 \end{bmatrix} = \frac{ 1 }{ 11 } \begin{bmatrix} 15 – 21 \\ 9 – 28 \end{bmatrix} = \frac{ 1 }{ 11 } \begin{bmatrix} -6 \\ -19 \end{bmatrix} = \begin{bmatrix} \frac{ -6 }{ 11 } \\ -\frac{ 19 }{ 11 } \end{bmatrix} \)
⟹ \(\begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} \frac{ -6 }{ 11 } \\ -\frac{ 19 }{ 11 }\end{bmatrix}\)
Therefore, a = \(– \frac{ 6 }{ 11 }\) and b = \(– \frac{ 19 }{ 11 }\).
Q-7. Solve the following system of the linear equations by suing the matrix method:
2a + b + c = 1
a – 2b – c = \(\frac{ 3 }{ 2 }\)
3b – 5c = 9
Sol: Determinants
The given system of linear equations will be written as in the form of MX = N, where
\(M = \begin{bmatrix} 2 & 1 & 1 \\ 1 & -2 & -1 \\ 0 & 3 & -5 \end{bmatrix}, \; X = \begin{bmatrix} a \\ b \\ c \end{bmatrix} \; and \; N = \begin{bmatrix} 1 \\ \frac{ 3 }{ 2 }\\ 9 \end{bmatrix}\)Thus,
|M| = 2[ (-2) × (-5) – (-1) × 3 ] – 1[ 1 × (-5) – 3 × 1 ] + 0[ (-2) × 1 – 1 × (-1) ] = 2( 10 + 3 ) – 1( -5 – 3 ) + 0
= 2 × 13 – (-8)
= 26 + 8 = 34 ≠ 0
Therefore, M is non- singular.
Thus, M-1 exists.
Now,
M11 = 13, M12 = 5, M13 = 3
M21 = 8, M22 = -10, M23 = -6
M31 = 1, M32 = 3, M33 = -5
Thus,
M-1 = \(\frac{ 1 }{ \left | M \right | } \left( adj M \right )\) = \(\frac{ 1 }{ 34 } \begin{bmatrix} 13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5 \end{bmatrix}\)
Thus,
X = M-1 N = \(\frac{ 1 }{ \left | M \right | } \left( adj M \right )\) = \(\frac{ 1 }{ 34 } \begin{bmatrix} 13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5 \end{bmatrix} \begin{bmatrix} 1 \\ \frac{ 3 }{ 2 }\\ 9 \end{bmatrix}\)
⟹ \(\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \frac{ 1 }{ 34 } \begin{bmatrix} 13\times 1 + 8 \times \frac{ 3 }{ 2 } + 1 \times 9 \\ 5\times 1 + \left ( -10 \right ) \times \frac{ 3 }{ 2 } + 3 \times 9 \\ 3\times 1 + \left ( -6 \right ) \times \frac{ 3 }{ 2 } + \left ( -5 \right ) \times 9 \end{bmatrix}\)
⟹ \(\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \frac{ 1 }{ 34 } \begin{bmatrix} 13 + 12 + 9 \\ 5 – 15 + 27 \\ 3 – 9 – 45 \end{bmatrix}\)
⟹ \(\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \frac{ 1 }{ 34 } \begin{bmatrix} 34 \\ 17 \\ -51 \end{bmatrix}\)
⟹ \(\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 1 \\ \frac{ 1 }{ 2 } \\ -\frac{ 3 }{ 2 } \end{bmatrix}\)
Therefore, a = 1, b = \(\frac{ 1 }{ 2 }\) and c = –\(\frac{ 3 }{ 2 }\)
Q-8. Solve the following system of the linear equations by suing the matrix method:
a – b + c = 4
2a + b – 3c = 0
a + b + c = 2
Sol: Determinants
The given system of linear equations will be written as in the form of MX = N, where
\(M = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}, \; X = \begin{bmatrix} a \\ b \\ c \end{bmatrix} \; and \; N = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}\)Thus,
|M| = 1[ 1 × 1 – (-3) × 1 ] – ( -1 )[ 2 × 1 – ( -3 ) × 1 ] + 1[ 2 × 1 – 1 × 1 ] = 2( 1 + 3 ) + 1( 2 + 3 ) + 1( 2 – 1 )
= 1 × 4 + 1 × 5 + 1 × 1
= 4 + 5 + 1 = 10 ≠ 0
Therefore, M is non- singular.
Thus, M-1 exists.
Now,
M11 = 4, M12 = -5, M13 = 1
M21 = 2, M22 = 0, M23 = -2
M31 = 2, M32 = 5, M33 = 3
Thus,
M-1 = \(\frac{ 1 }{ \left | M \right | } \left( adj M \right )\) = \(\frac{ 1 }{ 10 } \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}\)
Thus,
X = M-1 N = \(\frac{ 1 }{ \left | M \right | } \left( adj M \right )\) = \(\frac{ 1 }{ 10 } \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{bmatrix} \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}\)
⟹ \(\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \frac{ 1 }{ 10 } \begin{bmatrix} 4\times 4 + 2 \times 0 + 2 \times 2 \\ -5\times 4 + 0 \times 0 + 5 \times 2 \\ 1 \times 4 + \left ( -2 \right ) \times 0 + 3 \times 2 \end{bmatrix}\)
⟹ \(\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \frac{ 1 }{ 10 } \begin{bmatrix} 16 + 0 + 4 \\ -20 + 0 + 10 \\ 4 + 0 + 6 \end{bmatrix}\)
⟹ \(\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \frac{ 1 }{ 10 } \begin{bmatrix} 20 \\ -10 \\ 10 \end{bmatrix}\)
⟹ \(\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}\)
Therefore, a = 2, b = -1 and c = 1
Q-9. If from equations,
2a – 3b + 5c = 11, 3a + 2b – 4c = -5 and a + b – 2c = -3
M = \(\begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}\)
Find M-1. By using M-1, solve the system of the linear equations.
Sol: Determinants
M = \(\begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}\)
|M| = 2[ 2 × (-2) – 1 × (-4) ] + 3[ 3 × (-2) – 1 × (-4) ] + 5[ 3 × 1 – 2 × 1 ]
= 2( -4 + 4 ) + 3( -6 + 4 ) + 5( 3 – 2 )
= 2 × 0 + 3 × (-2) + 5 × 1
= 0 – 6 + 5 = -1 ≠ 0
Therefore, M is non- singular.
Thus, M-1 exists.
Now,
M11 = 0, M12 = 2, M13 = 1
M21 = -1, M22 = -9, M23 = -5
M31 = 2, M32 = 23, M33 = 13
Thus,
M-1 = \(\frac{ 1 }{ \left | M \right | } \left( adj M \right )\) = \(\frac{ 1 }{ -1 } \begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix} = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix}\)
The given system of linear equations will be written as in the form of MX = N, where
\(M = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}, \; X = \begin{bmatrix} a \\ b \\ c \end{bmatrix} \; and \; N = \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix}\)The solution of the given system of linear equations will be given by X = M-1 N.
⟹ X = M-1 N = \(\frac{ 1 }{ \left | M \right | } \left( adj M \right )\) = \(\begin{bmatrix} 4 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix} \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix}\)
⟹ \(\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 0\times 11 + 1 \times \left( -5 \right ) + \left( -2 \right ) \times \left( -3 \right ) \\ -2\times 11 + 9 \times \left( -5 \right ) + \left( -23 \right ) \times \left( -3 \right ) \\ \left( -1 \right )\times 11 + 5 \times \left( -5 \right ) + \left( -13 \right ) \times \left( -3 \right ) \end{bmatrix}\)
⟹ \(\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 0 – 5 + 6 \\ -22 – 45 + 69 \\ -11 – 25 + 39 \end{bmatrix}\)
⟹ \(\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}\)
Therefore, a = 1, b = 2 and c = 3.
Q-10. The price of 4 kg of onion, 3 kg of wheat and 2 kg of rice is Rs 60. The price of 2 kg of onion, 4 kg of wheat and 6 kg of rice is Rs 90. The price of 6 kg of onion, 2 kg of wheat and 3 kg of rice is Rs 70. What is the price of each of the items (per kg)? Use matrix method to find the price.
Sol:
Let us consider the cost of onions, wheat, and rice per kg be given by Rs a, Rs b,and Rs c, respectively.
Thus, the situation will be represented by the system of the equations as:
4a + 3b + 2c = 60
2a + 4b + 6c = 90
6a + 2b + 3c = 70
The system of the equations will be written as in the form of MX = N, where
\(M = \begin{bmatrix} 4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3 \end{bmatrix}, \; X = \begin{bmatrix} a \\ b \\ c \end{bmatrix} \; and \; N = \begin{bmatrix} 60 \\ 90 \\ 70 \end{bmatrix}\)Thus,
|M| = 4[ 4 × 3 – 6 × 2 ] – 3[ 2 × 3 – 6 × 6 ] + 2[ 2 × 2 – 6 × 4 ] = 2( 12 – 12 ) – 3( 6 – 36 ) + 2( 4 – 24 )
= 0 + 90 – 40 = 50 ≠ 0
Therefore, M is non- singular.
Thus, M-1 exists.
Now, Determinants
M11 = 0, M12 = 30, M13 = -20
M21 = -5, M22 = 0, M23 = 10
M31 = 10, M32 = -20, M33 = 10
M-1 = \(\frac{ 1 }{ \left | M \right | } \left( adj M \right )\) = \(\frac{ 1 }{ 50 } \begin{bmatrix} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{bmatrix}\)
Thus,
X = M-1 N = \(\frac{ 1 }{ \left | M \right | } \left( adj M \right )\) = \(\frac{ 1 }{ 50 } \begin{bmatrix} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{bmatrix} \begin{bmatrix} 60 \\90 \\ 70 \end{bmatrix}\)
⟹ \(\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \frac{ 1 }{ 50 } \begin{bmatrix} 0\times 60 + \left( -5 \right ) \times 90 + 10 \times 70 \\ 30 \times 60 + 0 \times 90 + \left( -20 \right ) \times 70 \\ \left( -20 \right ) \times 60 + 10 \times 90 + 10 \times 70 \end{bmatrix}\)
⟹ \(\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \frac{ 1 }{ 50 } \begin{bmatrix} 0 – 450 + 700 \\ 1800 + 0 – 1400 \\ -1200 + 900 + 700 \end{bmatrix}\)
⟹ \(\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \frac{ 1 }{ 50 } \begin{bmatrix} 250 \\ 400 \\ 400 \end{bmatrix}\)
⟹ \(\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 5 \\ 8 \\ 8 \end{bmatrix}\)
Therefore, a = 5, b = 8 and c = 8
Thus, the price for onions is Rs. 5 per kg, the price for wheat is Rs. 8 per kg and the price for the rice is Rs. 8 per kg. Based on the formulae given for Determinants