Definition of Matrices
A matrix is an ordered rectangular array of numbers or functions. The numbers or functions are called the elements or the entries of the matrix.
A matrix having m rows and n columns is called a matrix of order m × n or simply m × n matrix (read as an m by n matrix).
The number of elements in an m × n matrix will be equal to mn.
Q. If a matrix has 8 elements, what are the possible orders it can have?
We know that if a matrix is of order m × n, it has mn elements. Thus, to find all possible orders of a matrix with 8 elements, we will find all ordered pairs of natural numbers, whose product is 8.
Thus, all possible ordered pairs are (1, 8), (8, 1), (4, 2), (2, 4)
Hence, possible orders are 1 × 8, 8 ×1, 4 × 2, 2 × 4
Types of Matrices
A matrix is said to be a column matrix if it has only one column.
A matrix is said to be a row matrix if it has only one row.
A matrix in which the number of rows are equal to the number of columns, is said to be a square matrix. Thus an m × n matrix is said to be a square matrix if m = n and is known as a square matrix of order ‘n’.
A square matrix B = [bij]m × m is said to be a diagonal matrix if all its non diagonal elements are zero, that is a matrix [bij]m × m is said to be a diagonal matrix if [bij] = 0, when i ≠ j. m × m
A diagonal matrix is said to be a scalar matrix if its diagonal elements are equal
A square matrix in which elements in the diagonal are all 1 and rest are all zero is called an identity matrix.
A matrix is said to be zero matrix or null matrix if all its elements are zero.
Two matrices A = [aij] and B = [bij] are said to be equal if they are of the same order each element of A is equal to the corresponding element of B, that is [aij] = [bij] for all i and j.
The sum of two matrices is a matrix obtained by adding the corresponding elements of the given matrices. The two matrices have to be of the same order.
The negative of a matrix is denoted by – A. We define –A = (– 1) A.
If A = [aij] and B = [bij] be two matrices of the same order, say m × n, and k and l are scalars, then
k(A +B) = k A + kB
(k + l)A = k A + l A
if AB and BA are both defined, it is not necessary that AB = BA. Thus matrix multiplication is not commutative.
if the product of two matrices is a zero matrix, it is not necessary that one of the matrices is a zero matrix.
Properties of multiplication of matrices
The associative law For any three matrices A, B and C. We have (AB) C = A (BC), whenever both sides of the equality are defined.
The distributive law For three matrices A, B and C. A (B+C) = AB + AC AND (A+B) C = AC + BC, whenever both sides of equality are defined.
The existence of multiplicative identity For every square matrix A, there exist an identity matrix of same order such that IA = AI = A.
If A = [aij] be an m × n matrix, then the matrix obtained by interchanging the rows and columns of A is called the transpose of A. Transpose of the matrix A is denoted by A' or AT.
If A = [aij]m*n then A' = [aji]n*m
(A′)′ = A, (kA)′ = kA′ (where k is any constant)
(A + B)′ = A′ + B′
(A B)′ = B′ A′
A square matrix A = [aij] is said to be symmetric if A′ = A, that is, [aij] = [aji] for all possible values of i and j.
A square matrix A = [aij] is said to be skew symmetric matrix if A' = -A
This means that all the diagonal elements of a skew symmetric matrix are zero.
For any square matrix A with real number entries, A + A′ is a symmetric matrix and A – A′ is a skew symmetric matrix
Any square matrix can be expressed as the sum of a symmetric and a skew symmetric matrix.
EXERCISE – 3.1
Q-1: In a matrix given below:
A = \(\begin{bmatrix} 3 & 6 & 20 & -8 \\ 36 & -3 & \frac{ 7 }{ 2 } & 13 \\ \sqrt{ 5 } & 2 & -6 & 18 \end{bmatrix}\), write
(a) the order of the matrix.
(b) the number of element.
(c) Write the elements of a13, a21, a23, a24, a33
Solution: Based on the formula Matrices
(a) In the matrix given in the question, there are 3 rows and 4 columns. Hence, the order of the matrix will be 3 × 4.
(b) As the order of the given matrix is 3 × 4, there are 3 × 4 = 12 elements in the matrix.
(c) a13 = 20, a21 = 35, a23 = \(\frac{ 7 }{ 2 }\), a24 = 13, a33 = -5
Q-2: Let us consider a matrix A having 24 elements. Then, find the possible order for the matrix. What if there are 15 element of the same matrix?
Solution: Based on the formula Matrices
If the order of the matrix is m × n, then the number of elements in the matrix will be m*n. Then, to get all the possible orders of the matrix which having 24 elements, we need to find all of the possible pairs of the natural numbers having product 24.
Thus,
The possible ordered pairs will be given by:
( 1, 24 ), ( 2, 12 ) , ( 3, 8 ) , ( 4, 6 ) , ( 6, 4 ) , ( 8, 3 ), ( 12, 2 ), ( 24, 1 )
Therefore, the possible orders for the matrix with 24 elements are given as:
( 1 × 24 ), ( 2 × 12 ) , ( 3 × 8 ) , ( 4 × 6 ) , ( 6 × 4 ) , ( 8 × 3 ), ( 12 × 2 ), ( 24 × 1 )
If the number of elements in the matrix = 15, then the possible ordered pairs will be:
( 1, 15 ), ( 3, 5 ), ( 5, 3 ), ( 15, 1 )
Therefore, the possible orders for the matrix with 13 elements are:
( 1 × 15 ), ( 3 × 5 ) , ( 5 × 3 ) , ( 15 × 1 )
Q-3: Let us consider a matrix A having 21 elements. Then, find the possible order for the matrix. What if there are 10 element of the same matrix?
Solution: Based on the formula Matrices
If the order of the matrix is m × n, then the number of elements in the matrix will be m*n. Then, to get all the possible orders of the matrix which having 21 elements, we need to find all of the possible pairs of the natural numbers having product 21.
Thus,
The possible ordered pairs will be given by:
( 1, 21 ), ( 3, 7 ) , ( 7, 3 ), ( 21, 1 )
Therefore, the possible orders for the matrix with 24 elements are:
( 1 × 21 ), ( 3 × 7 ), ( 7 × 3 ), ( 21 × 1 )
If the number of elements in the matrix is 10, then the possible ordered pairs will be:
( 1, 10 ), ( 2, 5 ), ( 5, 2 ), ( 10, 1 )
Therefore, the possible orders for the matrix with 13 elements are:
( 1 × 10 ), ( 2 × 5 ) , ( 5 × 2 ) , ( 10 × 1 )
Q-4: Construct a 2 × 2 matrix, A = [ mij ], whose elements will be given by
(a) mij = \(\frac{ \left ( i + j \right )^{ 2 } }{ 2 }\)
(b) mij = \(\frac{ i }{ j }\)
(c) mij = \(\frac{ \left ( i + 2j \right )^{ 2 } }{ 2 }\)
Solution:
(a) Based on the formula Matrices:
mij = \(\frac{ \left ( i + j \right )^{ 2 } }{ 2 }\)
A ( 2 × 2 ) matrix is usually given by,
A = \(\begin{bmatrix} a_{ 11 } & a_{ 12 } \\ a_{ 21 } & a_{ 22 } \end{bmatrix}\)
Then,
a11 = \(\frac{ \left ( 1 + 1 \right )^{ 2 } }{ 2 } = \frac{ 4 }{ 2 } = 2\)
a12 = \(\frac{ \left ( 1 + 2 \right )^{ 2 } }{ 2 } = \frac{ 9 }{ 2 }\)
a21 = \(\frac{ \left ( 2 + 1 \right )^{ 2 } }{ 2 } = \frac{ 9 }{ 2 }\)
a22 = \(\frac{ \left ( 2 + 2 \right )^{ 2 } }{ 2 } = \frac{ 16 }{ 2 } = 8\)
Hence, the required matrix is A = \(\begin{bmatrix} 2 & \frac{ 9 }{ 2 } \\ \frac{ 9 }{ 2 } & 8 \end{bmatrix}\).
(b) As per the data :
mij = \(\frac{ i }{ j }\)
A ( 2 × 2 ) matrix is given by:
A = \(\begin{bmatrix} a_{ 11 } & a_{ 12 } \\ a_{ 21 } & a_{ 22 } \end{bmatrix}\)
Then,
a11 = \(\frac{ 1 }{ 1 } = 1\)
a12 = \(\frac{ 1 }{ 2 }\)
a21 = \(\frac{ 2 }{ 1 } = \frac{ 2 }{ 1 }\)
a22 = \(\frac{ 2 }{ 2 } = 1\)
Hence Based on the formula Matrices, the required matrix is A = \(\begin{bmatrix} 1 & \frac{ 1 }{ 2 } \\ 2 & 1 \end{bmatrix}\)
(c) As per the data given:
mij = \(\frac{ \left ( i + 2j \right )^{ 2 } }{ 2 }\)
A ( 2 × 2 ) matrix is given by,
A = \(\begin{bmatrix} a_{ 11 } & a_{ 12 } \\ a_{ 21 } & a_{ 22 } \end{bmatrix}\)
Then,
a11 = \(\frac{ \left ( 1 + 2 \times 1 \right )^{ 2 } }{ 2 } = \frac{ 9 }{ 2 }\)
a12 = \(\frac{ \left ( 1 + 2 \times 2 \right )^{ 2 } }{ 2 } = \frac{ 25 }{ 2 }\)
a21 = \(\frac{ \left ( 2 + 2 \times 1 \right )^{ 2 } }{ 2 } = \frac{ 16 }{ 2 } = 8\)
a22 = \(\frac{ \left ( 2 + 2 \times 2 \right )^{ 2 } }{ 2 } = \frac{ 36 }{ 2 } = 18\)
Hence Based on the formula Matrices, the required matrix is A = \(\begin{bmatrix} \frac{ 9 }{ 2 } & \frac{ 25 }{ 2 } \\ 8 & 18 \end{bmatrix}\)
Q-5: Construct a 3 × 4 matrix, A = [ mij ], whose elements will be given by
(a) mij = \(\frac{ 1 }{ 2 }| -3i + j |\)
(b) mij = \(2i – j\)
Solution:
(a) As per the data given:
mij = \(\frac{ 1 }{ 2 }| -3i + j |\)
A ( 3 × 4 ) matrix is given by:
A = \(\begin{bmatrix} a_{ 11 } & a_{ 12 } & a_{ 13 } & a_{ 14 } \\ a_{ 21 } & a_{ 22 } & a_{ 23 } & a_{ 24 } \\ a_{ 31 } & a_{ 32 } & a_{ 33 } & a_{ 34 } \end{bmatrix}\)Then,
a11 = \(\frac{ 1 }{ 2 }| -3 \times 1 + 1 | = 1\)
a12 = \( \frac{ 1 }{ 2 }| -3 \times 1 + 2 |= \frac{ 1 }{ 2 }\)
a13 = \( \frac{ 1 }{ 2 }| -3 \times 1 + 3 |= 0\)
a14 = \( \frac{ 1 }{ 2 }| -3 \times 1 + 4 | = \frac{ 1 }{ 2 }\)
a21 = \(\frac{ 1 }{ 2 }| -3 \times 2 + 1 | = \frac{ 5 }{ 2 }\)
a22 = \( \frac{ 1 }{ 2 }| -3 \times 2 + 2 |= 2\)
a23 = \( \frac{ 1 }{ 2 }| -3 \times 2 + 3 |= \frac{ 3 }{ 2 }\)
a24 = \( \frac{ 1 }{ 2 }| -3 \times 2 + 4 | = 1\)
a31 = \(\frac{ 1 }{ 2 }| -3 \times 3 + 1 | = 4\)
a32 = \( \frac{ 1 }{ 2 }| -3 \times 3 + 2 |= \frac{ 7 }{ 2 }\)
a33 = \( \frac{ 1 }{ 2 }| -3 \times 3 + 3 |= 3\)
a34 = \( \frac{ 1 }{ 2 }| -3 \times 3 + 4 | = \frac{ 5 }{ 2 }\)
Hence Based on the formula Matrices, the required matrix is A = \(\begin{bmatrix} 1 & \left ( \frac{ 1 }{ 2 } \right ) & 0 & \left ( \frac{ 1 }{ 2 } \right ) \\ \\ \left ( \frac{ 5 }{ 2 } \right ) & 2 & \left ( \frac{3 }{ 2 } \right ) & 1 \\ \\ 4 & \left ( \frac{ 7 }{ 2 } \right ) & 3 & \left ( \frac{ 5 }{ 2 } \right ) \end{bmatrix}\)
(b) As per the data given:
mij = \(2i + j\)
A ( 3 × 4 ) matrix is given by,
A = \(\begin{bmatrix} a_{ 11 } & a_{ 12 } & a_{ 13 } & a_{ 14 } \\ a_{ 21 } & a_{ 22 } & a_{ 23 } & a_{ 24 } \\ a_{ 31 } & a_{ 32 } & a_{ 33 } & a_{ 34 } \end{bmatrix}\)Then,
a11 = \(2 \times 1 – 1 = 1\)
a12 = \(2 \times 1 – 2 = 0\)
a13 = \( 2 \times 1 – 3 = -1\)
a14 = \( 2 \times 1 – 4 = -2\)
a21 = \( 2 \times 2 – 1 = 3\)
a22 = \( 2 \times 2 – 2 = 2\)
a23 = \( 2 \times 2 – 3 = 1\)
a24 = \( 2 \times 2 – 4 = 0\)
a31 = \( 2 \times 3 – 1 = 5\)
a32 = \( 2 \times 3 – 2 = 4\)
a33 = \( 2 \times 3 – 3 = 3\)
a34 = \( 2 \times 3 – 4 = 2\)
Hence, the required matrix is A = \(\begin{bmatrix} 1 & 0 & -1 & -2 \\ 3 & 2 & 1 & 0 \\ 5 & 4 & 3 & 2 \end{bmatrix}\)
Q-6: What will be the value of a, b and c in the following equation:
(a) \(\begin{bmatrix} 6 & 3 \\ a & 7 \end{bmatrix} = \begin{bmatrix} b & c \\ 1 & 7 \end{bmatrix}\)
(b) \(\begin{bmatrix} a + b & 2 \\ 5 + c & ab \end{bmatrix} = \begin{bmatrix} 9 & 2 \\ 6 & 8\end{bmatrix}\)
(c) \(\begin{bmatrix} a + b + c \\ a + c \\ b + c \end{bmatrix} = \begin{bmatrix} 9 \\ 5 \\ 7 \end{bmatrix}\)
Solution: Based on the formula Matrices
(a) \(\begin{bmatrix} 6 & 3 \\ a & 7 \end{bmatrix} = \begin{bmatrix} b & c \\ 1 & 7 \end{bmatrix}\)
Since, the given matrices are equal, then their corresponding elements will also be equal.
Now,
Compare the corresponding elements, we will get:
a = 1, b = 6 and c = 3
(b) \(\begin{bmatrix} a + b & 2 \\ 5 + c & ab \end{bmatrix} = \begin{bmatrix} 9 & 2 \\ 6 & 8\end{bmatrix}\)
Since, the given matrices are equal, then their corresponding elements will also be equal.
Now,
By comparing the corresponding elements, we will get
a + b = 9 ………………… (i)
5 + c = 6 ………………… (ii)
And,
ab = 8 ………………… (iii)
From (ii), we will get:
c = 6 – 5 = 1
From (i), we have
a = 9 – b ………………… (iv)
Substituting (iv) in equation (iii), we get:
( 9 – b )b = 8
⟹ 9b – b2 = 8
⟹ b2 – 9b + 8 = 0
⟹ b2 – 8b – b + 8 = 0
⟹ b( b – 8 ) -1( b – 8 ) = 0
⟹ ( b – 8 )( b – 1 ) = 0
⟹ b = 1, 8
For b = 1, a = 9 – 1 = 8
For b = 8, a = 9 – 8 = 1
Hence , a = 1, b = 8 and c = 1 or, a = 8, b = 1 and c = 1
(c) \(\begin{bmatrix} a + b + c \\ a + c \\ b + c \end{bmatrix} = \begin{bmatrix} 9 \\ 5 \\ 7 \end{bmatrix}\)
Since, the given matrices are equal, then their corresponding elements will also be equal Based on the formula Matrices.
Now,
By comparing the corresponding elements, we will get
a + b + c = 9 ………………(i)
a + c = 5 .………………(ii)
b + c = 7.………………(iii)
From equation (i), we have:
a + ( b + c ) = 9 ………………..(iv)
From equation (iii) and (iv), we will get:
a + 7 = 9
⟹ a = 2
Putting the value of an in equation (ii), we will get:
2 + c = 5
⟹ c = 3
Putting the value of c in equation (iii), we will get:
b + 3 = 7
⟹ b = 4
Hence the answer is, a = 2, b = 4 and c= 3.
Q-7: What will be the value of p, q, r and s in the following equation:
\(\begin{bmatrix} p – q & 2p – r \\ 2p – q & 3r + s \end{bmatrix} = \begin{bmatrix} -2 & 6 \\ 0 & 14 \end{bmatrix}\)
Solution: Based on the formula Matrices
\(\begin{bmatrix} p – q & 2p – r \\ 2p – q & 3r + s \end{bmatrix} = \begin{bmatrix} -2 & 6 \\ 0 & 14 \end{bmatrix}\)Since, the given matrices are equal, then their corresponding elements will also be equal.
Now,
Compare the corresponding elements, we will get:
p – q = -2 …………(i)
2p – r = 0 …………(ii)
2p + r = 6 …………(iii)
3r + s = 14 …………(iv)
From equation (ii), we have:
r = 2p
Putting it in equation (iii), we will get:
2p + 2p = 6
⟹ 4p = 6
⟹ p = \(\frac{ 6 }{ 4 }\)
⟹ p = \(\frac{ 3 }{ 2 }\)
Putting the value of p in equation (ii), we will get:
r = 2 × \(\frac{ 3 }{ 2 }\) = 3
Putting the value of r in equation (iv), we will get:
3 × 3 + s = 14
⟹ s = 5
Put the value of p in equation (i), we will get:
\(\frac{ 3 }{ 2 }\) – q = -2
⟹ q = \(\frac{ 3 }{ 2 }\) + 2
⟹ q = \(\frac{ 3 + 2 \times 2 }{ 2 } = \frac{ 7 }{ 2 }\)
Hence Based on the formula Matrices, p = \(\frac{ 3 }{ 2 }\), q = \(\frac{ 7 }{ 2 }\), r = 3 and s = 5.
Q-8: Let us consider a matrix A = [aij]m × n. The matrix will be a square matrix, if
(a) m < n (b) m > n (c) m = n (d) None of these
Solution:
A matrix can be a square matrix if and only if the number of rows and number of the columns of the matrix.
i.e., m = n
Hence, A = [aij]m × n will be a square matrix when m = n
Therefore, (c) is the correct answer.
Q-9. What will be the value of a and b in the following equation:
\(\begin{bmatrix} 3a + 7 & 5 \\ b + 1 & 2 – 3a \end{bmatrix} = \begin{bmatrix} 0 & b – 2 \\ 8 & 4 \end{bmatrix}\)
(a) a = \(\frac{ -1 }{ 3 }\), b = 7
(b) Not possible to find
(c) y = 7, x \(\frac{ -2 }{ 3 }\)
(d) x = \(\frac{ -1 }{ 3 }, y = \frac{ -2 }{ 3 }\)
Solution:
We have:
3a + 7 = 0
⟹ a = \(\frac{ -7 }{ 3 }\)
b + 1 = 8
⟹ b = 7
Also,
5 = b – 2
⟹ b = 3
2 – 3a = 4
⟹ 3a = -2
⟹ a = \(\frac{ -2 }{ 3 }\)
Hence Based on the formula Matrices, the correct answer is (b).
Q-10. The number of all the possible matrix of order 3 × 3 having each entry as 0 and 1 is-
(a) 512 (b) 81 (c) 18 (d) 27
Solution:
Given matrix has order 3 × 3, i.e., the matrix have 9 elements and each of the element of this matrix will be either 0 or 1.
As we have only two digits and there are 9 elements in the matrix, so, the matrix will be filled in two possible ways only.
Hence, by the principle of multiplication, required number for the possible matrices will be 29 = 512
Therefore, the correct answer is (a).
Q-1: A = \(\begin{bmatrix} 3 & 5 \\ 4 & 3 \end{bmatrix}\), B = \(\begin{bmatrix} 2 & 4 \\ -3 & 6 \end{bmatrix}\), C = \(\begin{bmatrix} -3 & 6 \\ 4 & 5 \end{bmatrix}\)
Find the value of each of the following from the matrices given:
(a) A + B (b) A – B (c) 3A – C (d) AB (e) BA
Solution:
(a) A + B = \(\begin{bmatrix} 3 & 5 \\ 4 & 3 \end{bmatrix}\) + \(\begin{bmatrix} 2 & 4 \\ -3 & 6 \end{bmatrix}\)
= \(\begin{bmatrix} 3 + 2 & 5 + 4 \\ 4 – 3 & 3 + 6 \end{bmatrix} \)
= \( \begin{bmatrix} 5 & 9 \\ 1 & 9 \end{bmatrix}\)
(b) A – B = \(\begin{bmatrix} 3 & 5 \\ 4 & 3 \end{bmatrix}\) – \(\begin{bmatrix} 2 & 4 \\ -3 & 6 \end{bmatrix}\)
= \(\begin{bmatrix} 3 – 2 & 5 – 4 \\ 4 + 3 & 3 – 6 \end{bmatrix} \)
= \( \begin{bmatrix} 1 & 1 \\ 7 & -3 \end{bmatrix}\)
(c) 3A – C = \(3\begin{bmatrix} 3 & 5 \\ 4 & 3 \end{bmatrix} – \begin{bmatrix} -3 & 6 \\ 4 & 5 \end{bmatrix}\)
= \(\begin{bmatrix} 3 \times 3 & 5 \times 3 \\ 4 \times 3 & 3 \times 3 \end{bmatrix} – \begin{bmatrix} -3 & 6 \\ 4 & 5 \end{bmatrix}\)
= \(\begin{bmatrix} 9 & 15 \\ 12 & 9 \end{bmatrix} – \begin{bmatrix} -3 & 6 \\ 4 & 5 \end{bmatrix}\)
= \(\begin{bmatrix} 9 + 3 & 15 – 6 \\ 12 – 4 & 9 – 5 \end{bmatrix}\)
= \(\begin{bmatrix} 12 & 9 \\ 8 & 4 \end{bmatrix}\)
(d) The matrix A and B, both are a square matrix as the number of rows and the number of columns is the same in both of the matrices, then
Multiplication of the two matrices A and B will be defined as:
AB = \(\begin{bmatrix} 3 & 5 \\ 4 & 3 \end{bmatrix}\)\(\begin{bmatrix} 2 & 4 \\ -3 & 6 \end{bmatrix}\)
= \(\begin{bmatrix} 3( 2 ) + 5( -3 ) & 3( 4 ) + 5( 6 ) \\ 4( 2 ) + 3( -3 ) & 4( 4 ) + 3( 6 ) \end{bmatrix}\)
= \(\begin{bmatrix} 6 – 15 & 12 + 30 \\ 8 – 9 & 16 + 18 \end{bmatrix}\)
= \(\begin{bmatrix} -9 & 42 \\ -1 & 34 \end{bmatrix}\)
Based on the formula Matrices
(e) The matrix A and B, both are a square matrix as the number of rows and the number of columns is the same in both of the matrices, then
Multiplication of the two matrices B and A will be defined as:
BA = \(\begin{bmatrix} 2 & 4 \\ -3 & 6 \end{bmatrix}\) \(\begin{bmatrix} 3 & 5 \\ 4 & 3 \end{bmatrix}\)
= \(\begin{bmatrix} 2( 3 ) + 3( 4 ) & 2( 5 ) + 4( 3 ) \\ -3( 3 ) + 6( 4 ) & -3( 5 ) + 6( 3 ) \end{bmatrix}\)
= \(\begin{bmatrix} 6 + 12 & 10 + 12 \\ – 9 + 24 & -15 + 18 \end{bmatrix}\)
= \(\begin{bmatrix} 18 & 22 \\ 15 & 3 \end{bmatrix}\)
Q-2: Solve the following:
(i) \(\begin{bmatrix} x & y \\ -y & x \end{bmatrix} + \begin{bmatrix} x & y \\ y & x \end{bmatrix}\)
(ii) \(\begin{bmatrix} x^{ 2 } + y^{ 2 } & y^{ 2 } + z^{ 2 } \\ x^{ 2 } + z^{ 2 } & x^{ 2 } + y^{ 2 } \end{bmatrix} + \begin{bmatrix} 2xy & 2yz \\ -2xz & -2xy \end{bmatrix}\)
(iii) \(\begin{bmatrix} -2 & 5 & -7 \\ 9 & 6 & 17 \\ 3 & 9 & 6 \end{bmatrix} + \begin{bmatrix} 13 & 8 & 7\\ 9 & 1 & 6 \\ 4 & 3 & 5 \end{bmatrix}\)
(iv) \(\begin{bmatrix} cos^{ 2 }a & sin^{ 2 }a \\ sin^{ 2 }a & cos^{ 2 }a \end{bmatrix} + \begin{bmatrix} sin^{ 2 }a & cos^{ 2 }a \\ cos^{ 2 }a & sin^{ 2 }a \end{bmatrix}\)
Solution:
(i) \(\begin{bmatrix} x & y \\ -y & x \end{bmatrix} + \begin{bmatrix} x & y \\ y & x \end{bmatrix}\)
= \(\begin{bmatrix} x + x & y + y \\ -y + y & x + x \end{bmatrix}\)
= \(\begin{bmatrix} 2x & 2y \\ 0 & 2x \end{bmatrix}\)
(ii) \(\begin{bmatrix} x^{ 2 } + y^{ 2 } & y^{ 2 } + z^{ 2 } \\ x^{ 2 } + z^{ 2 } & x^{ 2 } + y^{ 2 } \end{bmatrix} + \begin{bmatrix} 2xy & 2yz \\ -2xz & -2xy \end{bmatrix}\)
= \(\begin{bmatrix} x^{ 2 } + y^{ 2 } + 2xy & y^{ 2 } + z^{ 2 } + 2yz \\ x^{ 2 } + z^{ 2 } – 2xz & x^{ 2 } + y^{ 2 } – 2ab \end{bmatrix}\)
= \(\begin{bmatrix} \left ( x + y \right )^{ 2 } & \left ( y + z \right )^{ 2 } \\ \left ( x – y \right )^{ 2 } & \left ( x – y \right )^{ 2 } \end{bmatrix}\)
(iii) \(\begin{bmatrix} -2 & 5 & -7 \\ 9 & 6 & 17 \\ 3 & 9 & 6 \end{bmatrix} + \begin{bmatrix} 13 & 8 & 7\\ 9 & 1 & 6 \\ 4 & 3 & 5 \end{bmatrix}\)
= \(\begin{bmatrix} -2 + 13 & 5 + 8 & -7 + 7 \\ 9 + 9 & 6 + 1 & 17 + 6 \\ 3 + 4 & 9 + 3 & 6 + 5 \end{bmatrix}\)
= \(\begin{bmatrix} 11 & 13 & 0 \\ 18 & 6 & 23 \\ 7 & 12 & 11 \end{bmatrix}\)
(iv) \(\begin{bmatrix} cos^{ 2 }a & sin^{ 2 }a \\ sin^{ 2 }a & cos^{ 2 }a \end{bmatrix} + \begin{bmatrix} sin^{ 2 }a & cos^{ 2 }a \\ cos^{ 2 }a & sin^{ 2 }a \end{bmatrix}\)
= \(\begin{bmatrix} cos^{ 2 }a + sin^{ 2 }a & sin^{ 2 }a + cos^{ 2 }a \\ sin^{ 2 }a + cos^{ 2 }a & cos^{ 2 }a + sin^{ 2 }a \end{bmatrix}\)
= \(\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \;\;\;\;\;\;\;\;\left ( as sin^{ 2 }a + cos^{ 2 }a = 1 \right )\)
Q-3: Compute the following:
(i) \(\begin{bmatrix} x & y \\ -y & x \end{bmatrix}\begin{bmatrix} x & -y \\ y & a \end{bmatrix}\)
(ii) \(\begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \end{bmatrix}\)
(iii) \(\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 2 & 3 & 4 \\ 3 & 4 & 2 \end{bmatrix}\)
(iv) \(\begin{bmatrix} 3 & 4 & 5 \\ 4 & 5 & 6 \\ 5 & 6 & 7 \end{bmatrix}\begin{bmatrix} 2 & -4 & 6 \\ 0 & 3 & 4 \\ 3 & 0 & 5 \end{bmatrix}\)
(v) \(\begin{bmatrix} 3 & 2 \\ 4 & 3 \\ -2 & 2 \end{bmatrix}\begin{bmatrix} 1 & 0 & 1 \\ -1 & 3 & 2 \end{bmatrix}\)
(vi) \(\begin{bmatrix} 4 & -2 & 4 \\ -2 & 0 & 3 \end{bmatrix} \begin{bmatrix} 3 & -4 \\ 2 & 1 \\ 4 & 2 \end{bmatrix}\)
Solution:
(i) \(\begin{bmatrix} x & y \\ -y & x \end{bmatrix}\begin{bmatrix} x & -y \\ y & a \end{bmatrix}\)
= \(\begin{bmatrix} x( x )+ y( y ) & x( -y ) + y( x ) \\ -y( x ) + x( y ) & -y( -y ) + x( x ) \end{bmatrix}\)
= \(\begin{bmatrix} x^{ 2 } + y^{ 2 } & -xy + xy \\ -yx + xy & y^{ 2 } + x^{ 2 } \end{bmatrix}\)
= \(\begin{bmatrix} x^{ 2 } + y^{ 2 } & 0 \\ 0 & y^{ 2 } + x^{ 2 } \end{bmatrix}\)
(ii) \(\begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \end{bmatrix}\)
= \(\begin{bmatrix} 2( 1 ) & 2( 2 ) & 2( 3 )\\ 3( 1 ) & 3( 2 ) & 3( 3 )\\ 4( 1 ) & 4( 2 ) & 4( 3 ) \end{bmatrix}\)
= \(\begin{bmatrix} 2 & 4 & 6 \\ 3 & 6 & 9 \\ 4 & 8 & 12 \end{bmatrix}\)
(iii) \(\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 2 & 3 & 4 \\ 3 & 4 & 2 \end{bmatrix}\)
= \(\begin{bmatrix} 2( 2 ) – 3( 3 ) & 2( 3 ) – 3( 4 ) & 2( 4 ) – 3( 2 )\\ 3( 2 ) + 4( 3 ) & 3( 3 ) + 4( 4 ) & 3( 4 ) + 4( 2 ) \end{bmatrix}\)
= \(\begin{bmatrix} 4 – 9 & 6 – 12 & 8 – 6 \\ 6 + 12 & 9 + 16 & 12 + 8 \end{bmatrix}\)
= \(\begin{bmatrix} -5 & -6 & -2 \\ 18 & 25 & 20 \end{bmatrix}\)
(iv) \(\begin{bmatrix} 3 & 4 & 5 \\ 4 & 5 & 6 \\ 5 & 6 & 7 \end{bmatrix}\begin{bmatrix} 2 & -4 & 6 \\ 0 & 3 & 4 \\ 3 & 0 & 5 \end{bmatrix}\)
= \(\begin{bmatrix} 3( 2 ) + 4( 0 ) + 5( 3 ) & 3( -4 ) + 4( 3 ) + 5( 0 ) & 3( 6 ) + 4( 4 ) + 5( 5 ) \\ 4( 2 ) + 5( 0 ) + 6( 3 ) & 4( -4 ) + 5( 3 ) + 6( 0 ) & 4( 6 ) + 5( 4 ) + 6( 5 ) \\ 5( 2 ) + 6( 0 ) + 7( 3 ) & 5( -4 ) + 6( 3 ) + 7( 0 ) & 5( 6 ) + 6( 4 ) + 7( 5 ) \end{bmatrix}\)
= \(\begin{bmatrix} 6 + 0 + 15 & -12 + 12 + 0 & 18 + 16 + 25 \\ 8 + 0 + 18 & -16 + 15 + 0 & 24 + 20 + 30 \\ 10 + 0 + 21 & -20 + 18 + 0 & 30 + 24 + 35 \end{bmatrix}\)
= \(\begin{bmatrix} 21 & 0 & 59 \\ 26 & -1 & 74 \\ 31 & -2 & 89 \end{bmatrix}\)
(v) \(\begin{bmatrix} 3 & 2 \\ 4 & 3 \\ -2 & 2 \end{bmatrix}\begin{bmatrix} 1 & 0 & 1 \\ -1 & 3 & 2 \end{bmatrix}\)
= \(\begin{bmatrix} 3( 1 ) + 2( -1 ) & 3( 0 ) + 2( 3 ) & 3( 1 ) + 2( 2 ) \\ 4( 1 ) + 3( -1 ) & 4( 0 ) + 3( 3 ) & 4( 1 ) + 3( 2 ) \\ -2( 1 ) + 2( -1 ) & -2( 0 ) + 2( 3 ) & -2( 1 ) + 2( 2 ) \end{bmatrix}\)
= \(\begin{bmatrix} 3 – 2 & 0 + 6 & 3 + 4 \\ 4 – 3 & 0 + 9 & 4 + 6 \\ -2 – 2 & 0 + 6 & -2 + 4 \end{bmatrix}\)
= \(\begin{bmatrix} 1 & 6 & 7 \\ 1 & 9 & 10 \\ -4 & 6 & 2 \end{bmatrix}\)
(vi) \(\begin{bmatrix} 4 & -2 & 4 \\ -2 & 0 & 3 \end{bmatrix} \begin{bmatrix} 3 & -4 \\ 2 & 1 \\ 4 & 2 \end{bmatrix}\)
= \(\begin{bmatrix} 4( 3 ) – 2( 2 ) + 4( 4 ) & 4( -4 ) – 2( 1 ) + 4( 2 )\\ -2( 3 ) + 0( 2 ) + 3( 4 ) & -2( -4 ) + 0( 1 ) + 4( 2 ) \end{bmatrix}\)
= \(\begin{bmatrix} 12 – 4 + 16 & -16 – 2 + 8 \\ -6 + 0 + 12 & 8 + 0 + 6 \end{bmatrix}\)
= \(\begin{bmatrix} 24 & -10 \\ 6 & 14 \end{bmatrix}\)
Q-4: If, A = \(\begin{bmatrix} 2 & 3 & -4 \\ 5 & 0 & 3 \\ 2 & -2 & 2 \end{bmatrix}\), B = \(\begin{bmatrix} 4 & -2 & 3 \\ 5 & 3 & 6 \\ 3 & 1 & 4 \end{bmatrix}\) and C = \(\begin{bmatrix} 5 & 2 & 3 \\ 1 & 4 & 3 \\ 2 & -3 & 4 \end{bmatrix}\), then
Find ( A + B ) and ( B – C ). Also, check whether A + ( B – C ) = ( A + B ) – C
Solution: Based on the formula Matrices
A + B = \(\begin{bmatrix} 2 & 3 & -4 \\ 5 & 0 & 3 \\ 2 & -2 & 2 \end{bmatrix}\) + \(\begin{bmatrix} 4 & -2 & 3 \\ 5 & 3 & 6 \\ 3 & 1 & 4 \end{bmatrix}\)
= \(\begin{bmatrix} 2 + 4 & 3 – 2 & -4 + 3 \\ 5 + 5 & 0 + 3 & 3 + 6 \\ 2 + 3 & -2 + 0 & 2 + 4 \end{bmatrix}\)
= \(\begin{bmatrix} 6 & 1 & -1 \\ 10 & 3 & 9 \\ 5 & -2 & 6 \end{bmatrix}\)
B – C = \(\begin{bmatrix} 4 & -2 & 3 \\ 5 & 3 & 6 \\ 3 & 1 & 4 \end{bmatrix}\) – \(\begin{bmatrix} 5 & 2 & 3 \\ 1 & 4 & 3 \\ 2 & -3 & 4 \end{bmatrix}\)
= \(\begin{bmatrix} 4 – 5 & -2 – 2 & 3 – 3 \\ 5 – 1 & 3 – 4 & 6 – 3 \\ 3 – 2 & 1 – ( -3 ) & 4 – 4 \end{bmatrix}\)
= \(\begin{bmatrix} -1 & -4 & 0 \\ 4 & -1 & 3 \\ 1 & 4 & 0 \end{bmatrix}\)
A + ( B – C ) = \(\begin{bmatrix} 2 & 3 & -4 \\ 5 & 0 & 3 \\ 2 & -2 & 2 \end{bmatrix}\) + \(\begin{bmatrix} -1 & -4 & 0 \\ 4 & -1 & 3 \\ 1 & 4 & 0 \end{bmatrix}\)
= \(\begin{bmatrix} 2 – 1 & 3 – 4 & -4 + 0 \\ 5 + 4 & 0 – 1 & 3 + 3 \\ 2 + 1 & -2 + 4 & 2 + 0 \end{bmatrix}\)
= \(\begin{bmatrix} 1 & -1 & -4 \\ 9 & -1 & 6 \\ 3 & 2 & 2 \end{bmatrix}\)
( A + B ) – C = \(\begin{bmatrix} 6 & 1 & -1 \\ 10 & 3 & 9 \\ 5 & -2 & 6 \end{bmatrix}\) – \(\begin{bmatrix} 5 & 2 & 3 \\ 1 & 4 & 3 \\ 2 & -3 & 4 \end{bmatrix}\)
= \(\begin{bmatrix} 6 – 5 & 1 – 2 & -1 – 3 \\ 10 – 1 & 3 – 4 & 9 – 3 \\ 5 – 2 & -2 + 3 & 6 – 4 \end{bmatrix}\)
= \(\begin{bmatrix} 1 & -1 & -4 \\ 9 & -1 & 6 \\ 3 & 1 & 2 \end{bmatrix}\)
Therefore, A + ( B – C ) = ( A + B ) – C = \(\begin{bmatrix} 1 & -1 & -4 \\ 9 & -1 & 6 \\ 3 & 1 & 2 \end{bmatrix}\)
Hence, proved.
Q-5: If, A = \(\begin{bmatrix} \frac{ 2 }{ 3 } & 2 & \frac{ 5 }{ 3 } \\ \\ \frac{ 1 }{ 3 } & \frac{ 2 }{ 3 } & \frac{ 4 }{ 3 } \\ \\ \frac{ 7 }{ 3 } & 3 & \frac{ 2 }{ 3 } \end{bmatrix}\) and B = \(\begin{bmatrix} \frac{ 2 }{ 5 } & \frac{ 3 }{ 5 } & 2 \\ \\ \frac{ 1 }{ 5 } & \frac{ 2 }{ 5 } & \frac{ 4 }{ 5 } \\ \\ \frac{ 7 }{ 5 } & \frac{ 6 }{ 5 } & \frac{ 2 }{ 5 } \end{bmatrix}\), then
Find the value of 3A – 5B.
Solution:
3A – 5B = 3 \(\begin{bmatrix} \frac{ 2 }{ 3 } & 2 & \frac{ 5 }{ 3 } \\ \\ \frac{ 1 }{ 3 } & \frac{ 2 }{ 3 } & \frac{ 4 }{ 3 } \\ \\ \frac{ 7 }{ 3 } & 3 & \frac{ 2 }{ 3 } \end{bmatrix}\) – 5 \(\begin{bmatrix} \frac{ 2 }{ 5 } & \frac{ 3 }{ 5 } & 2 \\ \\ \frac{ 1 }{ 5 } & \frac{ 2 }{ 5 } & \frac{ 4 }{ 5 } \\ \\ \frac{ 7 }{ 5 } & \frac{ 6 }{ 5 } & \frac{ 2 }{ 5 } \end{bmatrix}\)
= \(\begin{bmatrix} 2 & 6 & 5 \\ \\ 1 & 2 & 4 \\ \\ 7 & 9 & 2 \end{bmatrix}\) – \(\begin{bmatrix} 2 & 3 & 10 \\ \\ 1 & 2 & 4 \\ \\ 7 & 6 & 2 \end{bmatrix}\)
= \(\begin{bmatrix} 0 & 3 & -5 \\ \\ 0 & 0 & 0 \\ \\ 0 & 3 & 0 \end{bmatrix}\) Based on the formula Matrices
Q-6: Find the value of:
\(cos\Theta \begin{bmatrix} cos\Theta & sin\Theta \\ -sin\Theta & cos\Theta \end{bmatrix}\) + \(sin\Theta \begin{bmatrix} sin\Theta & -cos\Theta \\ cos\Theta & sin\Theta \end{bmatrix}\).
Solution:
\(cos\Theta \begin{bmatrix} cos\Theta & sin\Theta \\ -sin\Theta & cos\Theta \end{bmatrix}\) + \(sin\Theta \begin{bmatrix} sin\Theta & -cos\Theta \\ cos\Theta & sin\Theta \end{bmatrix}\)
= \(\begin{bmatrix} cos^{ 2 }\Theta & cos\Theta sin\Theta \\ -sin\Theta cos\Theta & cos^{ 2 }\Theta \end{bmatrix}\) + \(\begin{bmatrix} sin^{ 2 }\Theta & -sin\Theta cos\Theta \\ sin\Theta cos\Theta & sin^{ 2 }\Theta \end{bmatrix}\)
= \(\begin{bmatrix} cos^{ 2 }\Theta + sin^{ 2 }\Theta & cos\Theta sin\Theta – cos\Theta sin\Theta \\ -sin\Theta cos\Theta + cos\Theta sin\Theta & cos^{ 2 }\Theta + sin^{ 2 }\Theta \end{bmatrix}\)
= \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\) Based on the formula Matrices
Q-7: What will be the value of A and B, if
(i) A + B = \(\begin{bmatrix} 8 & 0 \\ 4 & 6 \end{bmatrix}\) and A – B = \(\begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}\)
(ii) 2A + 3B = \(\begin{bmatrix} 3 & 4 \\ 5 & 0 \end{bmatrix}\) and 3A + 2B = \(\begin{bmatrix} 3 & -3 \\ -2 & 6 \end{bmatrix}\)
Solution:
(i) A + B = \(\begin{bmatrix} 8 & 0 \\ 4 & 6 \end{bmatrix}\) …………………..(a)
A – B = \(\begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}\) …………………..(b)
By adding equation (a) and (b), we will get:
2A = \(\begin{bmatrix} 8 & 0 \\ 4 & 6 \end{bmatrix}\) + \(\begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}\)
⟹ 2A = \(\begin{bmatrix} 8 + 4 & 0 + 0 \\ 0 + 4 & 6 + 4 \end{bmatrix}\)
⟹ 2A = \(\begin{bmatrix} 12 & 0 \\ 4 & 10 \end{bmatrix}\)
Thus,
A = \(\frac{ 1 }{ 2 } \) \(\begin{bmatrix} 12 & 0 \\ 4 & 10 \end{bmatrix}\)
⟹ A = \(\begin{bmatrix} 6 & 0 \\ 2 & 5 \end{bmatrix}\)
A + B = \(\begin{bmatrix} 8 & 0 \\ 4 & 6 \end{bmatrix}\)
⟹ \(\begin{bmatrix} 6 & 0 \\ 2 & 5 \end{bmatrix}\) + B = \(\begin{bmatrix} 8 & 0 \\ 4 & 6 \end{bmatrix}\)
⟹ B = \(\begin{bmatrix} 8 & 0 \\ 4 & 6 \end{bmatrix}\) – \(\begin{bmatrix} 6 & 0 \\ 2 & 5 \end{bmatrix}\)
⟹ B = \(\begin{bmatrix} 8 – 6 & 0 – 0 \\ 4 – 2 & 6 – 5 \end{bmatrix}\)
⟹ B = \(\begin{bmatrix} 2 & 0 \\ 2 & 1 \end{bmatrix}\)
Hence, A = \(\begin{bmatrix} 6 & 0 \\ 2 & 5 \end{bmatrix}\) and, B = \(\begin{bmatrix} 2 & 0 \\ 2 & 1 \end{bmatrix}\)
(ii) 2A + 3B = \(\begin{bmatrix} 3 & 4 \\ 5 & 0 \end{bmatrix}\) …………..(a)
3A + 2B = \(\begin{bmatrix} 3 & -3 \\ -2 & 6 \end{bmatrix}\) ……………(b)
By multiplying equation (a) by 2, we will get:
2( 2A + 3B ) = 2 \(\begin{bmatrix} 3 & 4 \\ 5 & 0 \end{bmatrix}\)
⟹ 4A + 6B = \(\begin{bmatrix} 6 & 8 \\ 10 & 0 \end{bmatrix}\) ……………(c)
By multiplying equation (b) by 3, we will get:
3( 3A + 2B ) = 3 \(\begin{bmatrix} 3 & -3 \\ -2 & 6 \end{bmatrix}\)
⟹ 9A + 6B = \(\begin{bmatrix} 9 & -9 \\ -6 & 12 \end{bmatrix}\) …………….(d)
By subtracting equation (c) and (d), we will get:
( 4A + 6B ) – ( 9A + 6B ) = \(\begin{bmatrix} 6 – 9 & 8 + 9 \\ 10 + 6 & 0 – 12 \end{bmatrix}\)
⟹ -5A = \(\begin{bmatrix} -3 & 17 \\ 16 & -12 \end{bmatrix}\)
⟹ A = – \(\frac{ 1 }{ 5 }\) \(\begin{bmatrix} -3 & 17 \\ 16 & -12 \end{bmatrix}\)
⟹ A = \(\begin{bmatrix} \frac{ 3 }{ 5 }& \frac{ -17 }{ 5 } \\ \frac{ -16 }{ 5 } & \frac{ 12 }{ 5 } \end{bmatrix}\)
Now,
2A + 3B = \(\begin{bmatrix} 3 & 4 \\ 5 & 0 \end{bmatrix}\)
⟹ 2 \(\begin{bmatrix} \frac{ 3 }{ 5 }& \frac{ -17 }{ 5 } \\ \frac{ -16 }{ 5 } & \frac{ 12 }{ 5 } \end{bmatrix}\) + 3B = \(\begin{bmatrix} 3 & 4 \\ 5 & 0 \end{bmatrix}\)
⟹ 3B = \(\begin{bmatrix} 3 & 4 \\ 5 & 0 \end{bmatrix}\) – \(\begin{bmatrix} \frac{ 6 }{ 5 } & \frac{ -34 }{ 5 } \\ \frac{ -32 }{ 5 } & \frac{ 24 }{ 5 } \end{bmatrix}\)
⟹ 3B = \(\begin{bmatrix} 3 – \frac{ 6 }{ 5 } & 4 – \frac{ -34 }{ 5 } \\ 5 – \frac{ -32 }{ 5 } & 0 – \frac{ 24 }{ 5 } \end{bmatrix}\)
⟹ 3B = \(\begin{bmatrix} \frac{ 9 }{ 5 } & \frac{ 54 }{ 5 } \\ \frac{ 57 }{ 5 } & \frac{ 24 }{ 5 } \end{bmatrix}\)
⟹ B = \(\frac{ 1 }{ 3 }\) \(\begin{bmatrix} \frac{ 9 }{ 5 } & \frac{ 54 }{ 5 } \\ \frac{ 57 }{ 5 } & \frac{ 24 }{ 5 } \end{bmatrix}\)
⟹ B = \(\begin{bmatrix} \frac{ 3 }{ 5 } & \frac{ 18 }{ 5 } \\ \frac{ 19 }{ 5 } & \frac{ 8 }{ 5 } \end{bmatrix}\)
Hence, A = \(\begin{bmatrix} \frac{ 3 }{ 5 }& \frac{ -17 }{ 5 } \\ \frac{ -16 }{ 5 } & \frac{ 12 }{ 5 } \end{bmatrix}\) and B = \(\begin{bmatrix} \frac{ 3 }{ 5 } & \frac{ 18 }{ 5 } \\ \frac{ 19 }{ 5 } & \frac{ 8 }{ 5 } \end{bmatrix}\)
Q-8: What will be the value of A if,
B = \(\begin{bmatrix} 4 & 3 \\ 2 & 5 \end{bmatrix}\) and 2A + B = \(\begin{bmatrix} 2 & 1 \\ -4 & 3 \end{bmatrix}\)
Solution:
2A + B = \(\begin{bmatrix} 2 & 1 \\ -4 & 3 \end{bmatrix}\)
⟹ 2A = \(\begin{bmatrix} 2 & 1 \\ -4 & 3 \end{bmatrix}\) – \(\begin{bmatrix} 4 & 3 \\ 2 & 5 \end{bmatrix}\)
⟹ 2A = \(\begin{bmatrix} 2 – 4 & 1 – 3 \\ -4 – 2 & 3 – 5 \end{bmatrix}\)
⟹ 2A = \(\begin{bmatrix} -2 & -2 \\ -6 & -2 \end{bmatrix}\)
⟹ A = \(\frac{ 1 }{ 2 }\) \(\begin{bmatrix} -2 & -2 \\ -6 & -2 \end{bmatrix}\)
⟹ A = \(\begin{bmatrix} -1 & -1 \\ -3 & -1 \end{bmatrix}\)
Hence, A = \(\begin{bmatrix} -1 & -1 \\ -3 & -1 \end{bmatrix}\).
Q-9: Calculate the value of a and b, if
\(2 \begin{bmatrix} 2 & 4 \\ 1 & a \end{bmatrix} + \begin{bmatrix} b & 1 \\ 2 & 2 \end{bmatrix} = \begin{bmatrix} 6 & 7 \\ 2 & 9 \end{bmatrix}\).
Solution:
\(2 \begin{bmatrix} 2 & 4 \\ 1 & a \end{bmatrix} + \begin{bmatrix} b & 1 \\ 2 & 2 \end{bmatrix} = \begin{bmatrix} 6 & 7 \\ 2 & 9 \end{bmatrix}\)⟹ \( \begin{bmatrix} 4 & 8 \\ 2 & 2a \end{bmatrix} + \begin{bmatrix} b & 1 \\ 2 & 2 \end{bmatrix} = \begin{bmatrix} 6 & 7 \\ 2 & 9 \end{bmatrix}\)
⟹ \(\begin{bmatrix} 4 + b & 8 + 1 \\ 2 + 2 & 2a + 2 \end{bmatrix} = \begin{bmatrix} 6 & 7 \\ 2 & 9 \end{bmatrix}\)
⟹ \(\begin{bmatrix} 4 + b & 9 \\ 4 & 2a + 2 \end{bmatrix} = \begin{bmatrix} 6 & 7 \\ 2 & 9 \end{bmatrix}\)
By comparing the corresponding elements of the above two matrices, we will get:
4 + b = 6
⟹ b = 2
2a + 2 = 9
⟹ 2a = 7
⟹ a = \(\frac{ 7 }{ 2 }\)
Hence, a = \(\frac{ 7 }{ 2 }\) and b = 2.
Q-10: Calculate the value of a, b, c and d, if
\(2 \begin{bmatrix} a & b \\ c & d \end{bmatrix} + 3 \begin{bmatrix} 2 & -2 \\ 0 & 3 \end{bmatrix} = 3 \begin{bmatrix} 4 & 6 \\ 5 & 7 \end{bmatrix}\).
Solution: Based on the formula Matrices
\(2\begin{bmatrix} a & b \\ c & d \end{bmatrix} + 3 \begin{bmatrix} 2 & -2 \\ 0 & 3 \end{bmatrix} = 3 \begin{bmatrix} 4 & 6 \\ 5 & 7 \end{bmatrix}\)⟹ \(\begin{bmatrix} 2a & 2b \\ 2c & 2d \end{bmatrix} + \begin{bmatrix} 6 & -6 \\ 0 & 9 \end{bmatrix} = \begin{bmatrix} 12 & 18 \\ 15 & 21 \end{bmatrix}\)
⟹ \(\begin{bmatrix} 2a + 6 & 2b – 6 \\ 2c + 0 & 2d + 9 \end{bmatrix} = \begin{bmatrix} 12 & 18 \\ 15 & 21 \end{bmatrix}\)
By comparing the corresponding elements of the above two matrices, we will get:
2a + 6 = 12
⟹ 2a = 6
⟹ a = 3
2b – 6 = 18
⟹ 2b = 24
⟹ b = 12
2c + 0 = 15
⟹ 2c = 15
⟹ c = \(\frac{ 15 }{ 2 }\)
2d + 9 = 21
⟹ 2d = 12
⟹ d = 6
Hence, a = 3, b = 12, c = \(\frac{ 15 }{ 2 }\) and d = 6.
Q-11: What will be the value of x and y in the matrices given below:
\(x \begin{bmatrix} 3 \\ 4 \end{bmatrix} + y \begin{bmatrix} -2 \\ 2 \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix}\)?
Solution: Based on the formula Matrices
\(x \begin{bmatrix} 3 \\ 4 \end{bmatrix} + y \begin{bmatrix} -2 \\ 2 \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix}\\\)⟹ \(\\\begin{bmatrix} 3x \\ 4x \end{bmatrix} + \begin{bmatrix} -2y \\ 2y \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix}\\\)
⟹ \(\\\begin{bmatrix} 3x – 2y \\ 4x + 2y \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix}\)
By comparing the corresponding elements of the above two matrices, we will get:
3x – 2y = 10………….. (i)
4x + 2y = 5 ……………(ii)
Adding equation (i) and (ii), we will get:
7x = 15
⟹ x = \(\frac{ 15 }{ 7 }\)
Putting the value of x in equation (i), we will get:
3 × \(\frac{ 15 }{ 7 }\) – 2y = 10
⟹ \(\frac{ 45 }{ 7 }\) – 2y = 10
⟹ 2y = \(\frac{ 45 }{ 7 }\) – 10
⟹ 2y = \(\frac{ 45 – 70 }{ 7 }\)
⟹ 2y = \(\frac{ -25 }{ 7 }\)
⟹ y = \(\frac{ -25 }{ 14 }\)
Hence we have proved that, x = \(\frac{ 15 }{ 7 }\) and, y = \(\frac{ -25 }{ 14 }\)
Q-12: What will be the value of a, b, c and d in the following matrix
\(3\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a & 7 \\ -2 & 2d \end{bmatrix} + \begin{bmatrix} 5 & a + b \\ c + d & 4 \end{bmatrix}\)?
Solution: Based on the formula Matrices
\(3\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a & 7 \\ -2 & 2d \end{bmatrix} + \begin{bmatrix} 5 & a + b \\ c + d & 4 \end{bmatrix}\)⟹ \( \begin{bmatrix} 3a & 3b \\ 3c & 3d \end{bmatrix} = \begin{bmatrix} a + 5 & 7 + a + b \\ -2 + c + d & 2d + 4 \end{bmatrix}\\\\\)
⟹ \( \begin{bmatrix} 3a & 3b \\ 3c & 3d \end{bmatrix} = \begin{bmatrix} a + 5 & a + b + 7 \\ c + d – 2 & 2d + 4 \end{bmatrix}\)
By comparing the corresponding elements of the above two matrices, we will get:
3a = a + 5
⟹ 2a = 5
⟹ a = \(\frac{ 5 }{ 2 }\)
3b = a + b + 7
⟹ 3b = \(\frac{ 5 }{ 2 }\) + b + 7
⟹ 2b = \(\frac{ 5 }{ 2 }\) + 7
⟹ 2b = \(\frac{ 5 + 14 }{ 2 }\)
⟹ 2b = \(\frac{ 19 }{ 2 }\)
⟹ b = \(\frac{ 19 }{ 4 }\)
3d = 2d + 4
⟹ d = 4
3c = c + d – 2
⟹ 2c = 4 – 2
⟹ 2c = 2
⟹ c = 1
Hence, a = \(\frac{ 5 }{ 2 }\), b = \(\frac{ 19 }{ 4 }\), c = 1 and d = 4.
Q-13:If F(a) = \(\begin{bmatrix} cos a & -sin a & 0 \\ sin a & cos a & 0 \\ 0 & 0 & 1 \end{bmatrix}\), prove that F(a) F(b) = F( a + b ).
Solution: Based on the formula Matrices
F(a) = \(\begin{bmatrix} cos a & -sin a & 0 \\ sin a & cos a & 0 \\ 0 & 0 & 1 \end{bmatrix}\), F(b) = \(\begin{bmatrix} cos b & -sin b & 0 \\ sin b & cos b & 0 \\ 0 & 0 & 1 \end{bmatrix}\)
F(a + b) = \(\begin{bmatrix} cos \left ( a + b \right ) & -sin \left ( a + b \right ) & 0 \\ sin \left ( a + b \right ) & cos \left ( a + b \right ) & 0 \\ 0 & 0 & 1 \end{bmatrix}\)
F(a) F(b) = \(\begin{bmatrix} cos a & -sin a & 0 \\ sin a & cos a & 0 \\ 0 & 0 & 1 \end{bmatrix}\)\(\begin{bmatrix} cos b & -sin b & 0 \\ sin b & cos b & 0 \\ 0 & 0 & 1 \end{bmatrix}\)
= \(\begin{bmatrix} \left ( cos a \;cos b – sin a \;sin b + 0 \right ) & \left ( -cos a\; sin b – sin a \;cos b + 0 \right ) & 0 \\ \left ( sin a \;cos b + cos a \;sin b + 0 \right ) & -sin a \;sin b + cos a\; cos b + 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}\)
= \(\begin{bmatrix} cos \left ( a + b \right ) & -sin \left ( a + b \right ) & 0 \\ sin \left ( a + b \right ) & cos \left ( a + b \right ) & 0 \\ 0 & 0 & 1 \end{bmatrix}\) = F(a + b)
Hence, proved.
Q-14: Prove that:
(i) \(\begin{bmatrix} 6 & -2 \\ 7 & 8 \end{bmatrix}\begin{bmatrix} 3 & 2 \\ 4 & 5 \end{bmatrix}\neq \begin{bmatrix} 3 & 2 \\ 4 & 5 \end{bmatrix}\begin{bmatrix} 6 & -2 \\ 7 & 8 \end{bmatrix}\).
(ii) \(\begin{bmatrix} 2 & 3 & 4 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}\begin{bmatrix} -2 & 1 & 0 \\ 0 & -2 & 1 \\ 3 & 4 & 5 \end{bmatrix}\neq \begin{bmatrix} -2 & 1 & 0 \\ 0 & -2 & 1 \\ 3 & 4 & 5 \end{bmatrix}\begin{bmatrix} 2 & 3 & 4 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}\)
Solution:
(i) \(\begin{bmatrix} 6 & -2 \\ 7 & 8 \end{bmatrix}\begin{bmatrix} 3 & 2 \\ 4 & 5 \end{bmatrix}\)
= \(\\\begin{bmatrix} 6( 3 ) – 2( 4 ) & 6( 2 ) – 2( 5 ) \\ 7( 3 ) – 8( 4 ) & 7( 2 ) – 8( 5 ) \end{bmatrix}\)
= \(\\\begin{bmatrix} 18 – 8 & 12 – 10 \\ 21 – 24 & 14 – 40 \end{bmatrix}\)
= \(\\\begin{bmatrix} 10 & 2 \\ -3 & -26 \end{bmatrix}\)
\(\\\begin{bmatrix} 3 & 2 \\ 4 & 5 \end{bmatrix}\begin{bmatrix} 6 & -2 \\ 7 & 8 \end{bmatrix}\)= \(\\\begin{bmatrix} 3( 6 ) + 2( 7 ) & 3( -2 ) + 2( 8 ) \\ 4( 6 ) + 5( 7 ) & 4( -2 ) + 5( 8 ) \end{bmatrix}\)
= \(\\\begin{bmatrix} 18 + 14 & -6 + 16 \\ 24 + 35 & -8 + 40 \end{bmatrix}\)
= \(\\\begin{bmatrix} 32 & 10 \\ 59 & 32 \end{bmatrix}\)
Therefore, \(\begin{bmatrix} 6 & -2 \\ 7 & 8 \end{bmatrix}\begin{bmatrix} 3 & 2 \\ 4 & 5 \end{bmatrix}\) ≠ \(\begin{bmatrix} 3 & 2 \\ 4 & 5 \end{bmatrix}\begin{bmatrix} 6 & -2 \\ 7 & 8 \end{bmatrix}\)
Hence, proved.
(ii) \(\\\begin{bmatrix} 2 & 3 & 4 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}\begin{bmatrix} -2 & 1 & 0 \\ 0 & -2 & 1 \\ 3 & 4 & 5 \end{bmatrix}\)
= \(\\\begin{bmatrix} 2( -2 ) + 3( 0 ) + 4( 3 ) & 2( 1 ) + 3( -2 ) + 4( 4 ) & 2( 0 ) + 3( 1 ) + 4( 5 ) \\ 0( -2 ) + 1( 0 ) + 0( 3 ) & 0( 1 ) + 1( -2 ) + 0( 4 ) & 0( 0 ) + 1( 1 ) + 0( 5 ) \\ 1( -2 ) + 1( 0 ) + 0( 3 ) & 1( 1 ) + 1( -2 ) + 0( 4 ) & 1( 0 ) + 1( 1 ) + 0( 5 ) \end{bmatrix}\)
= \(\\\begin{bmatrix} -4 + 0 + 12 & 2 – 6 + 16 & 0 + 3 + 20 \\ 0 + 0 + 0 & 0 – 2 + 0 & 0 + 1 + 0 \\ -2 + 0 + 0 & 1 – 2 + 0 & 0 + 1 + 0 \end{bmatrix}\)
= \(\\\begin{bmatrix} 8 & 12 & 23 \\ 0 & – 2 & 1 \\ -2 & -1 & 1 \end{bmatrix}\)
\(\\\begin{bmatrix} -2 & 1 & 0 \\ 0 & -2 & 1 \\ 3 & 4 & 5 \end{bmatrix}\begin{bmatrix} 2 & 3 & 4 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}\)= \(\\\begin{bmatrix} -2( 2 ) + 1( 0 ) + 0( 1 ) & -2( 3 ) + 1( 1 ) + 0( 1 ) & -2( 4 ) + 1( 0 ) + 0( 0 ) \\ 0( 2 ) – 2( 0 ) + 1( 1 ) & 0( 3 ) – 2( 1 ) + 1( 1 ) & 0( 4 ) – 2( 0 ) + 1( 0 ) \\ 3( 2 ) + 4( 0 ) + 5( 1 ) & 3( 3 ) + 4( 1 ) + 5( 1 ) & 3( 4 ) + 4( 0 ) + 5( 0 ) \end{bmatrix}\)
= \(\\\begin{bmatrix} -4 + 0 + 0 & -6 + 1 + 0 & -8 + 0 + 0 \\ 0 + 0 + 1 & 0 – 2 + 1 & 0 – 0 + 0 \\ 6 + 0 + 5 & 9 + 4 + 5 & 12 + 0 + 0 \end{bmatrix}\)
= \(\\\begin{bmatrix} -4 & -5 & -8 \\ 1 & – 1 & 0 \\ 11 & 18 & 12 \end{bmatrix}\)
Therefore, \(\begin{bmatrix} 2 & 3 & 4 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}\begin{bmatrix} -2 & 1 & 0 \\ 0 & -2 & 1 \\ 3 & 4 & 5 \end{bmatrix}\) ≠ \(\begin{bmatrix} -2 & 1 & 0 \\ 0 & -2 & 1 \\ 3 & 4 & 5 \end{bmatrix}\begin{bmatrix} 2 & 3 & 4 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}\)
Hence, proved Based on the formula Matrices.
Q-15: What will be the value of A2 – 5A + 6I, if A = \(\begin{bmatrix} 3 & 0 & 1 \\ 3 & 2 & 4 \\ 1 & -1 & 0 \end{bmatrix}\)?
Solution:
A2 = A × A
A2 = \(\\\begin{bmatrix} 3 & 0 & 1 \\ 3 & 2 & 4 \\ 1 & -1 & 0 \end{bmatrix}\) × \(\begin{bmatrix} 3 & 0 & 1 \\ 3 & 2 & 4 \\ 1 & -1 & 0 \end{bmatrix}\)
= \(\\\begin{bmatrix} 3( 3 ) + 3( 3 ) + 1( 1 ) & 3( 0 ) + 3( 2 ) + 1( -1 ) & 3( 1 ) + 3( 4 ) + 1( 0 ) \\ 3( 3 ) + 2( 3 ) + 4( 1 ) & 3( 0 ) + 2( 2 ) + 4( -1 ) & 3( 1 ) + 2( 4 ) + 4( 0 ) \\ 1( 3 ) – 1( 3 ) + 0( 1 ) & 1( 0 ) – 1( 2 ) + 0( -1 ) & 1( 1 ) – 1( 4 ) + 0( 0 ) \end{bmatrix}\)
= \(\\\begin{bmatrix} 9 + 9 + 1 & 0 + 6 – 1 & 3 + 12 + 0 \\ 9 + 6 + 4 & 0 + 4 – 4 & 3 + 8 + 0 \\ 3 – 3 + 0 & 0 – 2 + 0 & 1 – 4 + 0 \end{bmatrix}\)
= \(\\\begin{bmatrix} 19 & 5 & 15 \\ 19 & 0 & 11 \\ 0 & – 2 & -3 \end{bmatrix}\)
5A = 5\(\begin{bmatrix} 3 & 0 & 1 \\ 3 & 2 & 4 \\ 1 & -1 & 0 \end{bmatrix}\)
= \(\\\begin{bmatrix} 15 & 0 & 5 \\ 15 & 10 & 20 \\ 5 & -5 & 0 \end{bmatrix}\)
6I = 6 \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\)
= \(\\\begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix}\)
A2 – 5A + 6I = \(\begin{bmatrix} 19 & 5 & 15 \\ 19 & 0 & 11 \\ 0 & – 2 & -3 \end{bmatrix}\) – \(\begin{bmatrix} 15 & 0 & 5 \\ 15 & 10 & 20 \\ 5 & -5 & 0 \end{bmatrix}\) + \(\begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix}\)
= \(\\\begin{bmatrix} 19 + 15 + 6 & 5 + 0 + 0 & 15 + 5 + 0 \\ 19 + 15 + 0 & 0 + 10 + 6 & 11 + 20 + 0 \\ 0 + 5 + 0 & -2 -5 + 0 & -3 + 0 + 6 \end{bmatrix}\)
= \(\\\begin{bmatrix} 40 & 5 & 20 \\ 34 & 16 & 31 \\ 5 & -7 & 3 \end{bmatrix}\)
Hence,
A2 – 5A + 6I = \(\begin{bmatrix} 40 & 5 & 20 \\ 34 & 16 & 31 \\ 5 & -7 & 3 \end{bmatrix}\)
Q-16: If A = \(\begin{bmatrix} 0 & -tan\frac{ \beta }{ 2 } \\ tan\frac{ \beta }{ 2 } & 0 \end{bmatrix}\) and the identity matrix of order 2 be represented by I. Prove that:
I + A = ( I – A ) \(\begin{bmatrix} cos \beta & -sin \beta \\ sin \beta & cos \beta \end{bmatrix}\)
Solution: Based on the formula Matrices
LHS = I + A = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\) + \(\begin{bmatrix} 0 & -tan\frac{ \beta }{ 2 } \\ tan\frac{ \beta }{ 2 } & 0 \end{bmatrix}\)
= \(\\\begin{bmatrix} 1 & -tan\frac{ \alpha }{ 2 } \\ tan\frac{ \alpha }{ 2 } & 1 \end{bmatrix}\) ……… (i)
RHS = ( I – A ) \(\begin{bmatrix} cos \beta & -sin \beta \\ sin \beta & cos \beta \end{bmatrix}\)
= \(\\\left ( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} – \begin{bmatrix} 0 & -tan\frac{ \beta }{ 2 } \\ tan\frac{ \beta }{ 2 } & 0 \end{bmatrix} \right ) \begin{bmatrix} cos \beta & – sin\beta \\ sin\beta & cos\beta \end{bmatrix}\)
= \(\\\begin{bmatrix} 1 & tan\frac{ \beta }{ 2 } \\ -tan\frac{ \beta }{ 2 } & 1 \end{bmatrix} \;\; \begin{bmatrix} cos \beta & – sin\beta \\ sin\beta & cos\beta \end{bmatrix}\)
= \(\\\begin{bmatrix} cos \beta + sin \beta tan\frac{ \beta }{ 2 } & -sin \beta + cos \beta tan\frac{ \beta }{ 2 } \\ -cos \beta tan\frac{ \beta }{ 2 } + sin \beta tan\frac{ \beta }{ 2 } & sin \beta + cos \beta \end{bmatrix}\) ………..(ii)
= \(\\\begin{bmatrix} 1 – 2sin^{ 2 }\frac{ \beta }{ 2 } + 2sin\frac{ \beta }{ 2 }cos\frac{ \beta }{ 2 }tan\frac{ \beta }{ 2 } & -2sin\frac{ \beta }{ 2 }cos\frac{ \beta }{ 2 } + \left ( 2cos^{ 2 }\frac{ \beta }{ 2 } – 1 \right )tan\frac{ \beta }{ 2 } \\ -\left ( 2cos^{ 2 }\frac{ \beta }{ 2 } – 1 \right )tan\frac{ \beta }{ 2 } + 2sin\frac{ \beta }{ 2 }cos\frac{ \beta }{ 2 } & 2sin\frac{ \beta }{ 2 }cos\frac{ \beta }{ 2 }tan\frac{ \beta }{ 2 } + 1 – 2sin^{ 2 }\frac{ \beta }{ 2 } \end{bmatrix}\)
= \(\\\begin{bmatrix} 1 – 2sin^{ 2 }\frac{ \beta }{ 2 } + 2sin^{ 2 }\frac{ \beta }{ 2 } & -2sin\frac{ \beta }{ 2 }cos\frac{ \beta }{ 2 } + 2sin\frac{ \beta }{ 2 }cos\frac{ \beta } – tan\frac{ \beta }{ 2 } \\ -2sin\frac{ \beta }{ 2 }cos\frac{ \beta }{ 2 } + 2sin\frac{ \beta }{ 2 }cos\frac{ \beta } + tan\frac{ \beta }{ 2 } & 2sin^{ 2 }\frac{ \beta }{ 2 } + 1 – 2sin^{ 2 }\frac{ \beta }{ 2 } \end{bmatrix}\)
= \(\\\begin{bmatrix} 1 & -tan\frac{ \alpha }{ 2 } \\ tan\frac{ \alpha }{ 2 } & 1 \end{bmatrix}\)
= LHS
Hence, proved.
Q-17: There is the fund of Rs 30,000 from a trust which should be invested in any of the two different types of bonds. The first bond will pay 5% interest every year, and the second bond will pay 7% interest every year. Using the principle of matrix multiplication, determine how Rs 30,000 will be divided between the two different types of bonds, if the trust fund will obtain an annual total interest of:
(a) Rs 2,000 (b) Rs 1,800
Solution: Based on the formula Matrices
(a) Lets assume, in the first bond, Rs. q is invested.
The sum of the total money thus invested in the second bond is Rs. ( 30000 – q ).
As per the data given to us, the first bond needs to pay 5% interest every year and the second bond needs to pay 7% interest every year.
Hence, in order to have an annual interest of Rs. 2000, we have
S.I for 1 year = \(\frac{ Principle \times Rate \times 1 }{ 100 }\)
⟹ \(\begin{bmatrix} q & \left ( 30000 – q \right ) \end{bmatrix}\begin{bmatrix} \frac{ 5 }{ 100 } \\ \frac{ 7 }{ 100 } \end{bmatrix} = 2000\)
⟹ \(\frac{ 5q }{ 100 } + \frac{ 7 \left ( 30000 – q \right ) }{ 100 } = 2000\)
⟹ 5q + 7( 30000 – q ) = 200000
⟹ 5q + 210000 – 7q = 200000
⟹ 210000 – 200000 = 7q – 5q
⟹ 2q = 10000
⟹ q = 5000
To have an annual interest of Rs. 2000, the trusty providing fund must invest Rs. 5000 in the first bond and remaining Rs. 25000 in the second bond.
(b) Let have assumption that, in the first bond, Rs. q is invested.
The sum of the total money thus invested in the second bond is Rs. ( 30000 – q ).
As per the data given , the first bond needs to pay 5% interest every year and the second bond needs to pay 7% interest every year.
Hence, in order to have an annual interest of Rs. 1800, we have
S.I for 1 year = \(\frac{ Principle \times Rate \times 1 }{ 100 }\)
⟹ \(\begin{bmatrix} q & \left ( 30000 – q \right ) \end{bmatrix}\begin{bmatrix} \frac{ 5 }{ 100 } \\ \frac{ 7 }{ 100 } \end{bmatrix} = 1800\)
⟹ \(\frac{ 5q }{ 100 } + \frac{ 7 \left ( 30000 – q \right ) }{ 100 } = 1800\)
⟹ 5q + 7( 30000 – q ) = 180000
⟹ 5q + 210000 – 7q = 180000
⟹ 210000 – 180000 = 7q – 5q
⟹ 2q = 30000
⟹ q = 15000
In order to have an annual interest of Rs. 1800, the trusty providing fund must invest Rs. 15000 in the first bond and remaining Rs. 15000 in the second bond.
Q-18: A particular school has the bookshop of 8 dozen of the physics books, 10 dozen of the economics books, 10 dozen of the chemistry books. Their selling prices are Rs 60, Rs 40 and Rs 80 each, respectively. What will be the total amount of the bookshop which will be received by selling all of the books using the matrix algebra?
Solution:
The bookshop of 8 dozen of the physics books, 10 dozen of the economics books, 10 dozen of the chemistry books.
The selling prices for the physics, chemistry and the economics book are Rs. 60, Rs. 40 and Rs. 80, respectively.
The total amount of the bookshop received by selling all of the books by using the matrix algebra will be represented in the form given below:
\(12\begin{bmatrix} 8 & 10 & 10 \end{bmatrix}\begin{bmatrix} 60 \\ 40 \\ 80 \end{bmatrix}\)= 12 [8 × 60 + 10 × 40 + 10 × 80]
= 12 [480 + 400 + 800]
= 12 [1680]
= 20160
Hence we proved, the bookshop will get Rs. 20160 by the selling of all of the books.
Q-19: Let us consider P, Q, R, S and T be the matrices with order 2 × m, 3 × n, 2 × o, m× 3 and o× n, respectively. Then, the restriction on m, n and o such that, TQ + SQ is defined as:
Solution: Based on the formula Matrices
Matrices T and Q are having the orders o × n and 3 × n, respectively.
Thus, hence formed matrix TQ will be defined if n = 3. Consequently, TQ matrix will be have the order o × n. Matrices S and Q are having the orders of m × 3 and 3 × n, respectively.
As the number of columns in the matrix S is equal to the number of rows in the matrix Q, so, the matrix SQ hence is well-defined and will be of the order m × n. Matrices PY and WY will be added at the situation, when their orders will be the same.
Since, the matrix TQ has the order o × n and the matrix SQ has the order m × n. Hence, we will have o = m.
Therefore, n = 3 and o = m are the restrictions on m, n and o, such that TQ + SQ will be defined.
Q-20: Let us consider P, Q, R, S and T be the matrices with order 2 × m, 3 × n, 2 × o, m× 3 and o× n, respectively. If m = o, then find the order for the matrix 7P – 5R.
(a). o × 2
(b). 2 × m
(c). m × 3
(d). o × n
Solution:
The matrix P has the order 2 × m.
Thus, the matrix 7P will also have the same order. The matrix R is of the order 2 × o, i.e., 2 × m [As m = o]
Thus, the matrix 5R will also have the same order.
Now Based on the formula Matrices, both of the matrices 7P and 5R are of the order 2 × m.
Thus, the matrix 7P – 5R is defined and will have the order 2 × m.
Q-1: What is the transpose of each of the following matrices?
(i) \(\begin{bmatrix} 6 \\ \frac{ 1 }{ 2 }\\ -2 \end{bmatrix}\)
(ii) \(\begin{bmatrix} 2 & -2 \\ 3 & 4 \end{bmatrix}\)
Solution:
(i) Let M = \(\begin{bmatrix} 6 \\ \frac{ 1 }{ 2 }\\ -2 \end{bmatrix}\)
Then,
MT = \(\begin{bmatrix} 6 & \frac{ 1 }{ 2 } & -2 \end{bmatrix}\)
(ii) Let M = \(\begin{bmatrix} 2 & -2 \\ 3 & 4 \end{bmatrix}\)
Then,
MT = \(\begin{bmatrix} 2 & 3 \\ -2 & 4 \end{bmatrix}\) Based on the formula Matrices
Q-2: If A = \(\begin{bmatrix} -2 & 3 & 4 \\ 6 & 7 & 10 \\ -3 & 2 & 2 \end{bmatrix}\) and B = \(\begin{bmatrix} -5 & 2 & -6 \\ 2 & 3 & 1 \\ 2 & 4 & 2 \end{bmatrix}\), then verify that
(A + B)’ = A’ + B’
Solution:
A = \(\begin{bmatrix} -2 & 3 & 4 \\ 6 & 7 & 10 \\ -3 & 2 & 2 \end{bmatrix}\), B = \(\begin{bmatrix} -5 & 2 & -6 \\ 2 & 3 & 1 \\ 2 & 4 & 2 \end{bmatrix}\)
A + B = \(\begin{bmatrix} -2 & 3 & 4 \\ 6 & 7 & 10 \\ -3 & 2 & 2 \end{bmatrix}\) + \(\begin{bmatrix} -5 & 2 & -6 \\ 2 & 3 & 1 \\ 2 & 4 & 2 \end{bmatrix}\)
= \(\begin{bmatrix} -2 – 5 & 3 + 2 & 4 – 6 \\ 6 + 2 & 7 + 3 & 10 + 1 \\ -3 + 2 & 2 + 4 & 2 + 2 \end{bmatrix}\)
= \(\begin{bmatrix} -7 & 5 & -2 \\ 8 & 10 & 10 \\ -1 & 6 & 4 \end{bmatrix}\)
Thus,
(A + B)’ = \(\begin{bmatrix} -7 & 8 & -1 \\ 5 & 10 & 6 \\ -2 & 10 & 4 \end{bmatrix}\)
A’ + B’ = \(\begin{bmatrix} -2 & 6 & -3 \\ 3 & 8 & 2 \\ 4 & 10 & 2 \end{bmatrix}\) + \(\begin{bmatrix} -5 & 2 & 2 \\ 2 & 3 & 4 \\ -6 & 1 & 2 \end{bmatrix}\)
= \(\begin{bmatrix} -2 – 5 & 6 + 2 & -3 + 2 \\ 3 + 2 & 8 + 3 & 2 + 4 \\ 4 – 6 & 9 + 1 & 2 + 2 \end{bmatrix}\)
= \(\begin{bmatrix} -7 & 8 & -1 \\ 5 & 11 & 6 \\ -2 & 10 & 4 \end{bmatrix}\)
Thus, (A + B)’ = A’ + B’
Hence, proved Based on the formula Matrices.
Q-3: If A’ = \(\begin{bmatrix} -3 & 4 \\ 2 & 3 \end{bmatrix}\) and B = \(\begin{bmatrix} -2 & 1 \\ 2 & 3 \end{bmatrix}\), now find (A + 2B)’.
Solution:
We know that:
A = \({ \left ( {A}’ \right ) }’\)
Thus,
\({ A }’ \) = \(\begin{bmatrix} -3 & 4 \\ 2 & 3 \end{bmatrix}\)
⟹ A = \(\begin{bmatrix} -3 & 2 \\ 4 & 3 \end{bmatrix}\)
Now,
A + 2B = \(\begin{bmatrix} -3 & 2 \\ 4 & 3 \end{bmatrix}\) + 2\(\begin{bmatrix} -2 & 1 \\ 2 & 3 \end{bmatrix}\)
= \(\begin{bmatrix} -3 & 2 \\ 4 & 3 \end{bmatrix}\) + \(\begin{bmatrix} -4 & 2 \\ 4 & 6 \end{bmatrix}\)
= \(\begin{bmatrix} -3 – 4 & 2 + 2 \\ 4 + 4 & 3 + 6 \end{bmatrix}\)
= \(\begin{bmatrix} -7 & 4 \\ 8 & 9 \end{bmatrix}\)
Therefore,\({ \left ( A + 2B \right ) }’\) = \(\begin{bmatrix} -7 & 8 \\ 4 & 9 \end{bmatrix}\) Based on the formula Matrices
Q-4: Prove that \({ \left ( AB \right ) }’ = { B }'{ A }’\) for the matrices A and B where,
A = \( \begin{bmatrix} 2 \\ -5 \\ 4 \end{bmatrix}\), B = \( \begin{bmatrix} -2 & 3 & 2 \end{bmatrix}\)
Solution:
AB = \( \begin{bmatrix} 2 \\ -5 \\ 4 \end{bmatrix}\)\( \begin{bmatrix} -2 & 3 & 2 \end{bmatrix}\)
= \(\begin{bmatrix} -4 & 6 & 4 \\ 10 & -15 & -10 \\ -8 & 12 & 8 \end{bmatrix}\)
Thus,
\({ \left ( AB \right ) }’\) = \(\begin{bmatrix} -4 & 10 & -8 \\ 6 & -15 & 12 \\ 4 & -10 & 8 \end{bmatrix}\)
Now,
\({ A }’\) = \(\begin{bmatrix} 2 & -5 & 4 \end{bmatrix}\)
\({ B }’\) = \( \begin{bmatrix} -2 \\ 3 \\ 2 \end{bmatrix}\)
Thus,
{ B }’ { A }’= \(\begin{bmatrix} -4 & 10 & -8 \\ 6 & -15 & 12 \\ 4 & -10 & 8 \end{bmatrix}\)
= \({ \left ( AB \right ) }’\)
Hence, proved Based on the formula Matrices.
Q-5: If M = \(\begin{bmatrix} cos \beta & sin \beta \\ -sin \beta & cos \beta \end{bmatrix}\), then prove that M’M = I.
Solution:
M = \(\begin{bmatrix} cos \beta & sin \beta \\ -sin \beta & cos \beta \end{bmatrix}\)
M’ = \(\begin{bmatrix} cos \beta & -sin \beta \\ sin \beta & cos \beta \end{bmatrix}\)
⟹ M’M = \(\begin{bmatrix} cos \beta & -sin \beta \\ sin \beta & cos \beta \end{bmatrix}\) \(\begin{bmatrix} cos \beta & sin \beta \\ -sin \beta & cos \beta \end{bmatrix}\)
= \(\begin{bmatrix} \left ( cos \beta \right )\left ( cos \beta \right ) + \left ( -sin \beta \right )\left ( -sin \beta \right ) & \left ( cos \beta \right )\left ( sin \beta \right ) + \left ( -sin \beta \right )\left ( cos \beta \right ) \\ \left ( sin \beta \right )\left ( cos \beta \right ) + \left ( cos \beta \right )\left ( -sin \beta \right ) & \left ( sin \beta \right )\left ( sin \beta \right ) + \left ( cos \beta \right )\left ( cos \beta \right ) \end{bmatrix}\)
= \(\begin{bmatrix} cos^{ 2 } \beta + sin^{ 2 } & cos \beta . sin \beta – sin \beta . cos \beta \\ sin \beta cos \beta – cos \beta . sin \beta & sin^{ 2 } \beta + cos^{ 2 } \beta \end{bmatrix}\)
= \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)
= I
Hence, proved Based on the formula Matrices.
Q-6: Prove that the matrix A = \(\begin{bmatrix} 2 & -2 & 6 \\ -2 & 3 & 2 \\ 6 & 2 & 4 \end{bmatrix}\) is a symmetric matrix.
Solution:
Given in the data that
\({ A }’\) = \(\begin{bmatrix} 2 & -2 & 6 \\ -2 & 3 & 2 \\ 6 & 2 & 4 \end{bmatrix}\) which is equal to A.
Hence, \({ A }’\) = A
Therefore, the given matrix A = \(\begin{bmatrix} 2 & -2 & 6 \\ -2 & 3 & 2 \\ 6 & 2 & 4 \end{bmatrix}\) is a symmetric matrix Based on the formula Matrices.
Q-7: Prove that ( A + \({ A }’\) ) is a symmetric matrix for the matrix A = \(\begin{bmatrix} 2 & 6 \\ 7 & 8 \end{bmatrix}\).
Solution: Based on the formula Matrices
\({ A }’\) = \(\begin{bmatrix} 2 & 7 \\ 6 & 8 \end{bmatrix}\)
A + \({ A }’\) = \(\begin{bmatrix} 2 & 6 \\ 7 & 8 \end{bmatrix}\) + \(\begin{bmatrix} 2 & 7 \\ 6 & 8 \end{bmatrix}\)
= \(\begin{bmatrix} 4 & 13 \\ 13 & 16 \end{bmatrix}\)
\({ \left ( A + { A }’ \right ) }’\) = \(\begin{bmatrix} 4 & 13 \\ 13 & 16 \end{bmatrix}\) = A + \({ A }’\)
Hence, (A + \({ A }’\) ) is a symmetric matrix.
Q-8: If M and N are the symmetric matrices with having same order, then MN – NM will be a
(a) Symmetric matrix
(b) Skew symmetric matrix.
(c) Identity matrix
(d) Zero matrix
Solution:
As per the data given, M and N are two symmetric matrices.
Thus, we have have got
\({ A }’ = A\) and \({ B }’ = B\) ……..(i)
Let us consider,
\({ \left( MN – NM \right ) }’ = { \left ( MN \right ) }’ – { \left ( NM \right ) }’ \;\;\;\;\;\;\;\; \left [ {\left ( M – N \right )}’ = { M }’ – { N }’ \right ]\)= \({ N }'{ M }’ – { M }'{ N }’ \;\;\;\;\;\;\;\; \left [ {\left ( MN \right )}’ = { N }'{ M }’ \right ]\)
= \(BA – MN \;\;\;\;\;\;\;\;\;\; \left [ by \; (i) \right ]\)
= – (MN – NM)
Thus, (MN – NM )’ = – (MN – NM)
Hence, (MN – NM) is a skew- symmetric matrix.
Therefore, the correct answer is (b).
Q-9: Consider M = \(\begin{bmatrix} cos \beta & – sin \beta \\ sin \beta & cos \beta \end{bmatrix}\).
Then M + M’ = I, if β is:
(a) \(\frac{ \pi }{ 3 }\)
(b) \(\frac{ \pi }{ 6 }\)
(c) \(\frac{ 3 \pi }{ 2 }\)
(d) \(\frac{ \pi }{ 2 }\)
Solution: Based on the formula Matrices
M = \(\begin{bmatrix} cos \beta & -sin \beta \\ sin \beta & cos \beta \end{bmatrix}\)
M’ = \(\begin{bmatrix} cos \beta & sin \beta \\ -sin \beta & cos \beta \end{bmatrix}\)
⟹ M + M’ = I
⟹ \(\begin{bmatrix} cos \beta & -sin \beta \\ sin \beta & cos \beta \end{bmatrix}\) + \(\begin{bmatrix} cos \beta & sin \beta \\ -sin \beta & cos \beta \end{bmatrix}\) = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)
⟹ \(\begin{bmatrix} 2 cos \beta & 0 \\ 0 & 2 cos \beta \end{bmatrix}\) = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)
Thus,
\( 2 cos \beta = 1 \)⟹ cos β = \(\frac{ 1 }{ 2 }\)
⟹ cos β = \(cos \frac{ \pi }{ 3 }\)
⟹ β = \(\frac{ \pi }{ 3 }\)
Therefore, (a) is the correct answer.
Q-1: What will be the inverse of the matrix \(\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}\), if any exists?
Solution:
Let, M = \(\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}\)
As we know, M = IM
⟹ \(\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}\) = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)M
⟹ \(\begin{bmatrix} 1 & -1 \\ 0 & 5 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix}\)M [ R2 → R2 – 2 R1 ]
⟹ \(\begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ \frac{ -2 }{ 5 } & \frac{ 1 }{ 5 } \end{bmatrix}\)M [ R2 → \(\frac{ 1 }{ 5 } \) R2]
⟹ \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} \frac{ 3 }{ 5 } & \frac{ 1 }{ 5 } \\ \frac{ -2 }{ 5 } & \frac{ 1 }{ 5 } \end{bmatrix}\)M [ R2 → R1 + R2 ]
Hence, M-1 = \(\begin{bmatrix} \frac{ 3 }{ 5 } & \frac{ 1 }{ 5 } \\ \frac{ -2 }{ 5 } & \frac{ 1 }{ 5 } \end{bmatrix}\)
Q-2: What will be the inverse of the matrix \(\begin{bmatrix} 2 & -6 \\ 1 & -2 \end{bmatrix}\), if any exists?
Solution:
Let, M = \(\begin{bmatrix} 2 & -6 \\ 1 & -3 \end{bmatrix}\)
As we know, M = IM
⟹ \(\begin{bmatrix} 2 & -6 \\ 1 & -3 \end{bmatrix}\) = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)M
⟹ \(\begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}\)M [ C2 → C2 + 3 C1 ]
⟹ \(\begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 3 \\ -1 & 1 \end{bmatrix}\)M [ C1 → C1 – C2]
⟹ \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} -1 & 3 \\ \frac{ -1 }{ 2 } & 1 \end{bmatrix}\)M [ C1 → \(\frac{ 1 }{ 2 }\) C1]
Hence, M-1 = \(\begin{bmatrix} \frac{ 3 }{ 5 } & \frac{ 1 }{ 5 } \\ \frac{ -2 }{ 5 } & \frac{ 1 }{ 5 } \end{bmatrix}\)
Q-3: What will be the inverse of the matrix \(\begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}\), if any exists?
Solution:
Let, M = \(\begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}\)
As we know Based on the formula Matrices, M = IM
⟹ \(\begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}\) = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)M
⟹ \(\begin{bmatrix} 1 & \frac{ -1 }{ 2 } \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} \frac{ 1 }{ 6 } & 0 \\ 0 & 1 \end{bmatrix}\)M [ R1 → \(\frac{ 1 }{ 6 }\) R1 ]
⟹ \(\begin{bmatrix} 1 & \frac{ -1 }{ 2 } \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} \frac{ 1 }{ 6 } & 0 \\ \frac{ 1 }{ 3 } & 1 \end{bmatrix}\)M [ R2 → R2 +2R1]
Here, in the above matrix in LHS side, there is only zero in the second row.
Hence, M-1 does not exist.
Q-4: What will be the inverse of the matrix \(\begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}\), if any exists?
Solution:
Let assume, M = \(\begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}\)
As we know, M = IM
⟹ \(\begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}\) = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)M
⟹ \(\begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\)M [ R1 → R1 + R2 ]
⟹ \(\begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}\)M [ R2 → R2 + R1]
⟹ \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}\)M [ R2 → R1 + R2]
Hence we solved, M-1 = \( \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}\)
Q-5: What will be the inverse of the matrix \(\begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix}\), if any exists?
Solution: Based on the formula Matrices
Let, M = \(\begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix}\)
As we know, M = IM
⟹ \(\begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix}\) = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)M
⟹ \(\begin{bmatrix} 0 & 0 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} 1 & – \frac{ 1 }{ 2 } \\ 0 & 1 \end{bmatrix}\)M [ R1 → R1 – \(\frac{ 1 }{ 6 }\) R2 ]
Here, in the above matrix in LHS side, there is only zero in the second row.
Hence we solved, M-1 does not exist.
Q-6: What will be the inverse of the matrix \(\begin{bmatrix} 3 & 10 \\ 2 & 7 \end{bmatrix}\), if any exists?
Solution: Based on the formula Matrices
Let, M = \(\begin{bmatrix} 3 & 10 \\ 2 & 7 \end{bmatrix}\)
As we know, M = IM
⟹ \(\begin{bmatrix} 3 & 10 \\ 2 & 7 \end{bmatrix}\) = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)M
⟹ \(\begin{bmatrix} 1 & 3 \\ 2 & 7 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}\)M [ R1 → R1 – R2 ]
⟹ \(\begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ -2 & 3 \end{bmatrix}\)M [ R2 → R2 – 2R1]
⟹ \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & -10 \\ -2 & 3 \end{bmatrix}\)M [ R1 → R1 – 3 R2]
Hence we have , M-1 = \( \begin{bmatrix} 7 & -10 \\ -2 & 3 \end{bmatrix}\)
Q-7: What will be the inverse of the matrix \(\begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}\), if any exists?
Solution:
Let, M = \(\begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}\)
As we know Based on the formula Matrices, M = IM
\(\begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}\) = \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\)M
⟹ \(\begin{bmatrix} 1 & 0 & \frac{ -1 }{ 2 } \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}\) = \(\begin{bmatrix} \frac{ 1 }{ 2 } & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\)M [ R1 → \(\frac{ 1 }{ 2 }\)R1 ]
⟹ \(\begin{bmatrix} 1 & 0 & \frac{ -1 }{ 2 } \\ 0 & 1 & \frac{ 5 }{ 2 } \\ 0 & 1 & 3 \end{bmatrix}\) = \(\begin{bmatrix} \frac{ 1 }{ 2 } & 0 & 0 \\ \frac{ -5 }{ 2 } & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\)M [ R2 → R2 – 5 R1]
⟹ \(\begin{bmatrix} 1 & 0 & \frac{ -1 }{ 2 } \\ 0 & 1 & \frac{ 5 }{ 2 } \\ 0 & 1 & \frac{ 1 }{ 2 } \end{bmatrix}\) = \(\begin{bmatrix} \frac{ 1 }{ 2 } & 0 & 0 \\ \frac{ -5 }{ 2 } & 1 & 0 \\ \frac{ 5 }{ 2 } & -1 & 1 \end{bmatrix}\)M [ R3 → R3 – R2]
⟹ \(\begin{bmatrix} 1 & 0 & \frac{ -1 }{ 2 } \\ 0 & 1 & \frac{ 5 }{ 2 } \\ 0 & 0 & 1 \end{bmatrix}\) = \(\begin{bmatrix} \frac{ 1 }{ 2 } & 0 & 0 \\ \frac{ -5 }{ 2 } & 1 & 0 \\ 5 & -2 & 2 \end{bmatrix}\)M [ R3 → 2R3]
⟹ \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\) = \(\begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix}\)M [ R1 → R1 + \(\frac{ 1 }{ 2 }\) R3, and R2 → R2 – \(\frac{ 5 }{ 2 }\) R3]
Hence we have, M-1 = \(\begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix}\)
Q-8: The matrices M and N can be inverse of each other if and only if
(a) MN = NM (b) MN = NM = 0
(c) MN = 0, NM = I (d) MN = NM = I
Solution:
As we know Based on the formula Matrices, if M is a square matrix of order a, and if there exists another square matrix N of the same order b, such that MN = NM = I, then N will be said to be the inverse of M. We can see that M is the inverse of N.
Therefore, matrices M and N will be inverses of each other only if MN = NM = I.