Class 12 RD Sharma Solutions Chapter 7 Adjoint and Inverse of a Matrix Exercise 7.1 | Set 1 (original) (raw)
Last Updated : 23 Jul, 2025
In matrix algebra, the concepts of the adjoint and inverse of a matrix are crucial for solving systems of linear equations finding the determinants, and more. The adjoint of a matrix is a matrix obtained by transposing the cofactor matrix of the original matrix while the inverse of a matrix is a matrix that when multiplied with the original matrix yields the identity matrix. Understanding these concepts allows for the efficient manipulation and solution of linear algebraic problems.
Adjoint and Inverse of a Matrix
Adjoint of a Matrix: The adjoint of a matrix A is the transpose of its cofactor matrix. If A is a square matrix, the cofactor matrix is formed by the computing the cofactor of each element in A. The adjoint is denoted by adj(A).
The inverse of a Matrix: The inverse of a matrix A is denoted by A−1 and is defined as the matrix that satisfies the equation A⋅A−1=I where I is the identity matrix. For a matrix A the inverse can be computed using the formula:
A^{-1} = \frac{1}{\det(A)} \cdot \text{adj}(A)
where det(A) is the determinant of A and adj(A) is the adjoint of A. The matrix must be square and its determinant should be non-zero for the inverse to exist.
Question 1. Find the adjoint of the following matrices:
Verify that (adj A)A = |A|I = A(adj A) for the above matrices:
(i) \begin{bmatrix}-3&5\\2&4\end{bmatrix}
**Solution:
Here, A = \begin{bmatrix}-3&5\\2&4\end{bmatrix}
Cofactors of A are:
C11 = 4 C12 = -2
C21 = -5 C22 = -3
adj A = \begin{bmatrix}C_{11}&C_{12}\\C_{21}&C_{22}\end{bmatrix}^T
(adj A) = \begin{bmatrix}4&-2\\-5&-3\end{bmatrix}^T
= \begin{bmatrix}4&-5\\-2&-3\end{bmatrix}
To Prove, (adj A)A = |A|I = A(adj A)
(adj A)A = \begin{bmatrix}4&-5\\-2&-3\end{bmatrix}\begin{bmatrix}-3&5\\2&4\end{bmatrix}=\begin{bmatrix}-22&0\\0&-22\end{bmatrix}
|A|I = \begin{bmatrix}-3&5\\2&4\end{bmatrix}\begin{bmatrix}1&0\\0&1\end{bmatrix} =(-22)\begin{bmatrix}1&0\\0&1\end{bmatrix} = \begin{bmatrix}-22&0\\0&-22\end{bmatrix}
A(adj A) = \begin{bmatrix}-3&5\\2&4\end{bmatrix}\begin{bmatrix}4&-5\\-2&-3\end{bmatrix}=\begin{bmatrix}-22&0\\0&-22\end{bmatrix}
Therefore, (adj A)A = |A|I = A(adj A)
Hence Proved
(ii) \begin{bmatrix}a&b\\c&d\end{bmatrix}
**Solution:
Here, A = \begin{bmatrix}a&b\\c&d\end{bmatrix}
Cofactors of A are:
C11 = d C12 = -c
C21 = -b C22 = a
(adj A) = \begin{bmatrix}d&-c\\-b&-a\end{bmatrix}^T
= \begin{bmatrix}d&-b\\-c&a\end{bmatrix}
To Prove, (adj A)A = |A|I = A(adj A)
(adj A)A = \begin{bmatrix}d&-b\\-c&a\end{bmatrix}\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix}ad-bc&bd-bd\\-ac+ac&-bc+ad\end{bmatrix}=\begin{bmatrix}ad-bc&0\\0&ad-bc\end{bmatrix}
|A|I =\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}1&0\\0&1\end{bmatrix} =(ad-bc)\begin{bmatrix}1&0\\0&1\end{bmatrix} =\begin{bmatrix}ad-bc&0\\0&ad-bc\end{bmatrix}
A(adj A) =\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}=\begin{bmatrix}ad-bc&0\\0&ad-bc\end{bmatrix}
Therefore, (adj A)A = |A|I = A(adj A)
Hence Proved
(iii) \begin{bmatrix}cos α&sin α\\sin α&cosα\end{bmatrix}
**Solution:
Here, A = \begin{bmatrix}cos α&sin α\\sin α&cosα\end{bmatrix}
Cofactors of A are:
C11 = cos α C12 = -sin α
C21 = -sin α C22 = cos α
(adj A) =\begin{bmatrix}cos α&-sin α\\-sin α&cosα\end{bmatrix}^T
=\begin{bmatrix}cos α&-sin α\\-sin α&cosα\end{bmatrix}
To Prove, (adj A)A = |A|I = A(adj A)
(adj A)A = \begin{bmatrix}cos α&-sin α\\-sin α&cosα\end{bmatrix}\begin{bmatrix}cos α&sin α\\sin α&cosα\end{bmatrix}
=\begin{bmatrix}cos^2α-sin^2α&cosαsinα-sinαcosα\\-cosαsinα+sinαcosα&-sin^2α+cos^2α\end{bmatrix}
=\begin{bmatrix}cos 2α&0\\0&cos2α\end{bmatrix}
|A|I = \begin{bmatrix}cos α&sin α\\sin α&cosα\end{bmatrix}\begin{bmatrix}1&0\\0&1\end{bmatrix}
=(cos^2α-sin^2α)\begin{bmatrix}1&0\\0&1\end{bmatrix}
=\begin{bmatrix}cos^2α-sin^2α&0\\0&cos^2α-sin^2α\end{bmatrix}
=\begin{bmatrix}cos2α&0\\0&cos2α\end{bmatrix}
A(adj A) = \begin{bmatrix}cos α&sin α\\sin α&cosα\end{bmatrix}\begin{bmatrix}cos α&-sin α\\-sin α&cosα\end{bmatrix}
=\begin{bmatrix}cos^2α-sin^2α&-cosαsinα+sinαcosα\\sinαcosα-cosαsinα&-sin^2α+cos^2α\end{bmatrix}
=\begin{bmatrix}cos 2α&0\\0&cos2α\end{bmatrix}
Therefore, (adj A)A = |A|I = A(adj A)
Hence Proved
(iv) \begin{bmatrix}1&tan α/2\\-tan α/2&1\end{bmatrix}
**Solution:
Here, A = \begin{bmatrix}1&tan α/2\\-tan α/2&1\end{bmatrix}
Cofactors of A are:
C11 = 1 C12 = -(-tan α/2) = tan α/2
C21 = -tan α/2 C22 = 1
adj A = \begin{bmatrix}1&tan α/2\\-tan α/2&1\end{bmatrix}^T
=\begin{bmatrix}1&-tan α/2\\tan α/2&1\end{bmatrix}
To Prove, (adj A)A = |A|I = A(adj A)
|A| = \begin{bmatrix}1&tan α/2\\-tan α/2&1\end{bmatrix}
= 1 + tan2 α/2
= sec2α/2
(adj)A = \begin{bmatrix}1&-tan α/2\\tan α/2&1\end{bmatrix}\begin{bmatrix}1&tan α/2\\-tan α/2&1\end{bmatrix}
=\begin{bmatrix}1+tan^2 α/2&tan α/2-tan α/2\\tan α/2-tan α/2&tan^2 α/2+1\end{bmatrix}
=\begin{bmatrix}sec^2 α/2&0\\0&sec^2 α/2\end{bmatrix}
|A|I = (sec2α/2)\begin{bmatrix}1&0\\0&1\end{bmatrix}
=\begin{bmatrix}sec^2 α/2&0\\0&sec^2 α/2\end{bmatrix}
A(adj A) = \begin{bmatrix}1&tan α/2\\-tan α/2&1\end{bmatrix}\begin{bmatrix}1&-tan α/2\\tan α/2&1\end{bmatrix}
=\begin{bmatrix}1+tan^2 α/2&-tan α/2+tan α/2\\-tan α/2+tan α/2&tan^2 α/2+1\end{bmatrix}
=\begin{bmatrix}sec^2 α/2&0\\0&sec^2 α/2\end{bmatrix}
Therefore, (adj A)A = |A|I = A(adj A)
Hence Proved
Question 2. Compute the adjoint of each of the following matrices:
Verify that (adj A)A = |A|I = A(adj A) for the above matrices:
(i) \begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}
**Solution:
Here, A = \begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}
Cofactors of A are
C11 = \begin{bmatrix}1&2\\2&1\end{bmatrix} = -3
C21 = \begin{bmatrix}2&2\\2&1\end{bmatrix} = 2
C31 = \begin{bmatrix}2&2\\1&2\end{bmatrix} = 2
C12 = \begin{bmatrix}2&2\\2&1\end{bmatrix} = 2
C22 = \begin{bmatrix}1&2\\2&1\end{bmatrix} =-3
C32 = \begin{bmatrix}1&2\\2&2\end{bmatrix} = 2
C13 = \begin{bmatrix}2&1\\2&2\end{bmatrix} = 2
C23 = \begin{bmatrix}1&2\\2&2\end{bmatrix} = 2
C33 = \begin{bmatrix}1&2\\2&1\end{bmatrix} = -3
adj A = \begin{bmatrix}C_{11}&C_{12}&C_{13}\\C_{21}&C_{22}&C_{23}\\C_{31}&C_{32}&C_{34}\end{bmatrix}^T
=\begin{bmatrix}-3&2&2\\2&-3&2\\2&2&-3\end{bmatrix}^T
=\begin{bmatrix}-3&2&2\\2&-3&2\\2&2&-3\end{bmatrix}
To Prove, (adj A)A = |A|I = A(adj A)
|A| = -3 + 4 + 4 = 5
(adj A)A = \begin{bmatrix}-3&2&2\\2&-3&2\\2&2&-3\end{bmatrix}\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}=\begin{bmatrix}5&0&0\\0&5&0\\0&0&5\end{bmatrix}
|A|I= (5)\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix} = \begin{bmatrix}5&0&0\\0&5&0\\0&0&5\end{bmatrix}
A(adj A) = \begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}\begin{bmatrix}-3&2&2\\2&-3&2\\2&2&-3\end{bmatrix}=\begin{bmatrix}5&0&0\\0&5&0\\0&0&5\end{bmatrix}
Therefore, (adj A)A = |A|I = A(adj A)
Hence Proved
(ii) \begin{bmatrix}1&2&5\\2&3&1\\-1&1&1\end{bmatrix}
**Solution:
Here, A = \begin{bmatrix}1&2&5\\2&3&1\\-1&1&1\end{bmatrix}
Cofactors of A are
C11 = \begin{bmatrix}3&1\\1&1\end{bmatrix} = 2
C12 = \begin{bmatrix}2&1\\-1&1\end{bmatrix} = -3
C13 = \begin{bmatrix}2&3\\-1&1\end{bmatrix} = 5
C21 = \begin{bmatrix}2&5\\1&1\end{bmatrix} = 3
C22 = \begin{bmatrix}1&5\\-1&1\end{bmatrix} = 6
C23 = \begin{bmatrix}1&2\\-1&1\end{bmatrix} = -3
C31 = \begin{bmatrix}2&5\\3&1\end{bmatrix} = -13
C32 = \begin{bmatrix}1&5\\2&1\end{bmatrix} = 9
C33 = \begin{bmatrix}1&2\\2&3\end{bmatrix} = -1
adj A = \begin{bmatrix}C_{11}&C_{12}&C_{13}\\C_{21}&C_{22}&C_{23}\\C_{31}&C_{32}&C_{34}\end{bmatrix}^T
= \begin{bmatrix}2&-3&5\\3&6&-3\\-13&9&-1\end{bmatrix}^T
=\begin{bmatrix}2&3&-13\\-3&6&9\\5&-3&-1\end{bmatrix}
To Prove, (adj A)A = |A|I = A(adj A)
|A| = 1(3 - 1) - 2(2 + 1) + 5(2 + 3)
= 2 - 6 + 25 = 21
(adj A)A = \begin{bmatrix}2&3&-13\\-3&6&9\\5&-3&-1\end{bmatrix}\begin{bmatrix}1&2&5\\2&3&1\\-1&1&1\end{bmatrix}=\begin{bmatrix}21&0&0\\0&21&0\\0&0&21\end{bmatrix}
|A|I = (21)\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}21&0&0\\0&21&0\\0&0&21\end{bmatrix}
A(adj A) = \begin{bmatrix}1&2&5\\2&3&1\\-1&1&1\end{bmatrix}\begin{bmatrix}2&3&-13\\-3&6&9\\5&-3&-1\end{bmatrix}=\begin{bmatrix}21&0&0\\0&21&0\\0&0&21\end{bmatrix}
Therefore, (adj A)A = |A|I = A(adj A)
Hence Proved
(iii) \begin{bmatrix}2&-1&3\\4&2&5\\0&4&-1\end{bmatrix}
**Solution:
Here, A = \begin{bmatrix}2&-1&3\\4&2&5\\0&4&-1\end{bmatrix}
Cofactors of A are
C11 = \begin{bmatrix}2&5\\4&-1\end{bmatrix} = -22
C12 = -\begin{bmatrix}4&5\\0&-1\end{bmatrix} = 4
C13 = \begin{bmatrix}4&2\\0&4\end{bmatrix} = 16
C21 = -\begin{bmatrix}-1&3\\4&-1\end{bmatrix} = 11
C22 = \begin{bmatrix}2&3\\0&-1\end{bmatrix} = -2
C23 = -\begin{bmatrix}2&-1\\0&4\end{bmatrix} = -8
C31 = \begin{bmatrix}-1&3\\2&5\end{bmatrix} = -11
C32 = -\begin{bmatrix}2&3\\4&5\end{bmatrix} = 2
C33 = \begin{bmatrix}2&-1\\4&2\end{bmatrix} = 8
adj A = \begin{bmatrix}-22&4&16\\11&-2&-8\\-11&2&8\end{bmatrix}^T
= \begin{bmatrix}-22&11&-11\\4&-2&2\\16&-8&8\end{bmatrix}
To Prove, (adj A)A = |A|I = A(adj A)
|A| = 2(-2 - 20) + 1(-4 - 0) + 3(16 - 0)
= -44 - 4 + 48 = 0
(adj A)A = \begin{bmatrix}-22&11&-11\\4&-2&2\\16&-8&8\end{bmatrix}\begin{bmatrix}2&-1&3\\4&2&5\\0&4&-1\end{bmatrix}=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}
|A|I = (0)\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}
A(adj A) = \begin{bmatrix}2&-1&3\\4&2&5\\0&4&-1\end{bmatrix}\begin{bmatrix}-22&11&-11\\4&-2&2\\16&-8&8\end{bmatrix}=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}
Therefore, (adj A)A = |A|I = A(adj A)
Hence Proved
(iv) \begin{bmatrix}2&0&-1\\5&1&0\\1&1&3\end{bmatrix}
**Solution:
Here, A = \begin{bmatrix}2&0&-1\\5&1&0\\1&1&3\end{bmatrix}
Cofactors of the A are
C11 = \begin{bmatrix}1&0\\1&3\end{bmatrix} = 3
C12 = -\begin{bmatrix}5&0\\1&3\end{bmatrix} = -15
C13 = \begin{bmatrix}5&1\\1&1\end{bmatrix} = 4
C21 = \begin{bmatrix}0&-1\\1&3\end{bmatrix} = -1
C22 = \begin{bmatrix}2&-1\\1&3\end{bmatrix} = 7
C23 = \begin{bmatrix}2&0\\1&1\end{bmatrix} = -2
C31 = \begin{bmatrix}0&-1\\1&0\end{bmatrix} = 1
C32 = \begin{bmatrix}2&-1\\5&0\end{bmatrix} = -5
C33 = \begin{bmatrix}2&0\\5&1\end{bmatrix} = 2
adj A = \begin{bmatrix}3&-15&4\\-1&7&-2\\1&-5&2\end{bmatrix}^T
=\begin{bmatrix}3&-1&1\\-15&7&-5\\4&-2&2\end{bmatrix}
To Prove, (adj A)A = |A|I = A(adj A)
|A| = 2(3 - 0) - 0(15 - 0) - 1(5 - 1)
= 6 - 4 = 2
(adj A)A = \begin{bmatrix}3&-1&1\\-15&7&-5\\4&-2&2\end{bmatrix}\begin{bmatrix}2&0&-1\\5&1&0\\1&1&3\end{bmatrix}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}
|A|I = (2)\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}
A (adj A) = \begin{bmatrix}2&0&-1\\5&1&0\\1&1&3\end{bmatrix}\begin{bmatrix}3&-1&1\\-15&7&-5\\4&-2&2\end{bmatrix}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}
Therefore, (adj A)A = |A|I = A(adj A)
Hence Proved
Question 3. For the matrix A = \begin{bmatrix}1&-1&1\\2&3&0\\18&2&10\end{bmatrix} , show that A(adj A) = O.
Solution:
Cofactor of A are,
C11 = 30 C12 = -20 C13 = -50
C21 = 12 C22 = -8 C23 = -20
C31 = -3 C32 = 2 C33 = 5
adj A = \begin{bmatrix}30&-20&-50\\12&-8&-20\\-3&2&5\end{bmatrix}^T
= \begin{bmatrix}30&12&-3\\-20&-8&2\\-50&-20&5\end{bmatrix}
A(adj A) = \begin{bmatrix}1&-1&1\\2&3&0\\18&2&10\end{bmatrix} \begin{bmatrix}30&12&-3\\-20&-8&2\\-50&-20&5\end{bmatrix}
= \begin{bmatrix}1&-1&1\\2&3&0\\18&2&10\end{bmatrix}(0)
= 0
Hence Proved
Question 4. If A = \begin{bmatrix}-4&-3&-3\\1&0&1\\4&4&3\end{bmatrix} , show that adj A = A.
**Solution:
Here, A = \begin{bmatrix}-4&-3&-3\\1&0&1\\4&4&3\end{bmatrix}
Cofactor of A are,
C11 = -4 C12 = 1 C13 = 4
C21 = -3 C22 = 0 C23 = 4
C31 = 4 C32 = 4 C33 = 3
adj A = \begin{bmatrix}-4&1&4\\-3&0&4\\-3&1&3\end{bmatrix}^T
= \begin{bmatrix}-4&-3&-3\\1&0&1\\4&4&3\end{bmatrix}
Therefore, adj A = A
Question 5. If A = \begin{bmatrix}-1&-2&-2\\2&1&-2\\2&-2&1\end{bmatrix} , show that adj A = 3AT.
**Solution:
Here, A = \begin{bmatrix}-1&-2&-2\\2&1&-2\\2&-2&1\end{bmatrix}
Cofactor of A are,
C11 = -3 C12 = -6 C13 = -6
C21 = 6 C22 = 3 C23 = -6
C31 = 6 C32 = -6 C33 = 3
adj A = \begin{bmatrix}-3&-6&-6\\6&3&-6\\6&-6&3\end{bmatrix}^T
= \begin{bmatrix}-3&6&6\\-6&3&-6\\-6&-6&3\end{bmatrix}
AT= \begin{bmatrix}-1&2&2\\-2&1&-2\\-2&-2&1\end{bmatrix}
Now, 3AT = 3 \begin{bmatrix}-1&2&2\\-2&1&-2\\-2&-2&1\end{bmatrix} = \begin{bmatrix}-3&6&6\\-6&3&-6\\-6&-6&3\end{bmatrix}
adj A = 3.AT
Hence Proved
Question 6. Find A(adj A) for the matrix A =\begin{bmatrix}1&-2&3\\0&2&-1\\-4&5&2\end{bmatrix} .
**Solution:
Here, A = \begin{bmatrix}1&-2&3\\0&2&-1\\-4&5&2\end{bmatrix}
Cofactor of A are,
C11 = 9 C12 = 4 C13 = 8
C21 = 19 C22 = 14 C23 = 3
C31 = -4 C32 = 1 C33 = 2
adj A = \begin{bmatrix}C_{11}&C_{12}&C_{13}\\C_{21}&C_{22}&C_{23}\\C_{31}&C_{32}&C_{34}\end{bmatrix}^T
= \begin{bmatrix}9&4&8\\19&14&3\\-4&1&2\end{bmatrix}^T
= \begin{bmatrix}9&19&-4\\4&14&1\\8&3&2\end{bmatrix}
A(adj A) = \begin{bmatrix}1&-2&3\\0&2&-1\\-4&5&2\end{bmatrix} \begin{bmatrix}9&19&-4\\4&14&1\\8&3&2\end{bmatrix}
= \begin{bmatrix}25&0&0\\0&25&0\\0&0&25\end{bmatrix}
= 25\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}
= 25I3
Question 7. Find the inverse of each of the following matrices:
(i) \begin{bmatrix}cos θ&sinθ\\-sin θ&cos θ\end{bmatrix}
**Solution:
Here, A = \begin{bmatrix}cos θ&sinθ\\-sin θ&cos θ\end{bmatrix}
|A| = cos2θ + sin2θ = 1
Hence, inverse of A exist
Cofactors of A are,
Cofactor of A are,
C11 = cos θ C12 = sin θ
C21 = -sin θ C22 = cos θ
adj A = \begin{bmatrix}C_{11}&C_{12}\\C_{21}&C_{22}\end{bmatrix}^T
= \begin{bmatrix}cos θ&sinθ\\-sin θ&cos θ\end{bmatrix}^T
= \begin{bmatrix}cos θ&-sinθ\\sin θ&cos θ\end{bmatrix}
A-1 = 1/|A|. adj A
=1/1. \begin{bmatrix}cos θ&-sinθ\\sin θ&cos θ\end{bmatrix}
(ii) \begin{bmatrix}0&1\\1&0\end{bmatrix}
**Solution:
Here, A = \begin{bmatrix}0&1\\1&0\end{bmatrix}
|A| = -1
Hence, inverse of A exist
Cofactor of A are,
C11 = 0 C12 = -1
C21 = -1 C22 = 0
adj A = \begin{bmatrix}C_{11}&C_{12}\\C_{21}&C_{22}\end{bmatrix}^T
= \begin{bmatrix}0&-1\\-1&0\end{bmatrix}^T
= \begin{bmatrix}0&-1\\-1&0\end{bmatrix}
A-1 = 1/|A|. adj A
= \frac{1}{-1}\begin{bmatrix}0&-1\\-1&0\end{bmatrix}
= \begin{bmatrix}0&1\\1&0\end{bmatrix}
(iii) \begin{bmatrix}a&b\\c&\frac{(a+bc)}{a}\end{bmatrix}
**Solution:
Here, A = \begin{bmatrix}a&b\\c&\frac{(a+bc)}{a}\end{bmatrix}
|A| = a(1 + bc)/a - bc = 1 + bc - bc = 1
Hence, inverse of A exists.
Cofactor of A are,
C11 = (1 + bc)/a C12 = -c
C21 = -b C22 = a
adj A =\begin{bmatrix}C_{11}&C_{12}\\C_{21}&C_{22}\end{bmatrix}^T
= \begin{bmatrix}\frac{(a+bc)}{a} &-c\\-b&a\end{bmatrix}^T
= \begin{bmatrix}\frac{(a+bc)}{a}&-b\\-c&a\end{bmatrix}
A-1 = 1/|A|. adj A
= 1/1 \begin{bmatrix}\frac{(a+bc)}{a}&-b\\-c&a\end{bmatrix}
= \begin{bmatrix}\frac{(a+bc)}{a}&-b\\-c&a\end{bmatrix}
(iv) \begin{bmatrix}2&5\\-3&1\end{bmatrix}
**Solution:
Here, A = \begin{bmatrix}2&5\\-3&1\end{bmatrix}
|A| = 2 + 15 = 17
Hence, inverse of A exists.
Cofactor of A are,
C11 = 1 C12 = 3
C21 = -5 C22 = 2
adj A = \begin{bmatrix}C_{11}&C_{12}\\C_{21}&C_{22}\end{bmatrix}^T
= \begin{bmatrix}1&3\\-5&2\end{bmatrix}^T
= \begin{bmatrix}1&-5\\3&2\end{bmatrix}
A-1 = 1/|A|. adj A
= \frac{1}{17}\begin{bmatrix}1&-5\\3&2\end{bmatrix}
= \begin{bmatrix}\frac{1}{17}&\frac{-5}{17}\\\frac{3}{17}&\frac{2}{17}\end{bmatrix}
Question 8. Find the inverse of each of the following matrices.
(i) \begin{bmatrix}1&2&3\\2&3&1\\3&1&2\end{bmatrix}
**Solution:
Here, A = \begin{bmatrix}1&2&3\\2&3&1\\3&1&2\end{bmatrix}
|A| = 1(6 - 1) - 2(4 - 3) + 3(2 - 9)
= 5 - 2 - 21 = -18
Therefore, inverse of A exists
Cofactors of A are:
C11 = 5 C12 = -1 C13 = -7
C21 = -1 C22 = -7 C23 = 5
C31 = -7 C32 = 5 C33 = -1
adj A = \begin{bmatrix}C_{11}&C_{12}&C_{13}\\C_{21}&C_{22}&C_{23}\\C_{31}&C_{32}&C_{34}\end{bmatrix}^T
= \begin{bmatrix}5&-1&-7\\-1&-7&5\\-7&5&-1\end{bmatrix}^T
= \begin{bmatrix}5&-1&-7\\-1&-7&5\\-7&5&-1\end{bmatrix}
A-1 = 1/|A|. adj A
Hence, A-1 = \frac{1}{-18}\begin{bmatrix}5&-1&-7\\-1&-7&5\\-7&5&-1\end{bmatrix}
=\begin{bmatrix}-5/18&1/18&7/18\\1/18&7/18&-5/18\\7/18&-5/18&1/18\end{bmatrix}
(ii) \begin{bmatrix}1&2&5\\1&-1&-1\\2&3&-1\end{bmatrix}
**Solution:
Here, A = \begin{bmatrix}1&2&5\\1&-1&-1\\2&3&-1\end{bmatrix}
|A| = 1(1 + 3) - 2(-1 + 2) + 5(3 + 2)
= 4 - 2 - 25 = 27
Therefore, inverse of A exists
Cofactors of A are:
C11 = 4 C12 = -1 C13 = 5
C21 = -17 C22 = -11 C23 = 1
C31 = 3 C32 = 6 C33 = -3
adj A = \begin{bmatrix}C_{11}&C_{12}&C_{13}\\C_{21}&C_{22}&C_{23}\\C_{31}&C_{32}&C_{34}\end{bmatrix}^T
= \begin{bmatrix}4&-1&5\\17&-11&1\\3&6&-3\end{bmatrix}^T
= \begin{bmatrix}4&17&3\\-1&-11&6\\5&1&-3\end{bmatrix}
A-1 = 1/|A|. adj A
Hence, A-1 = \frac{1}{27}\begin{bmatrix}4&17&3\\-1&-11&6\\5&1&-3\end{bmatrix}
= \begin{bmatrix}4/27&17/27&3/27\\-1/27&-11/27&6/27\\5/27&1/27&-3/27\end{bmatrix}
= \begin{bmatrix}4/27&17/27&1/9\\-1/27&-11/27&2/9\\5/27&1/27&-1/9\end{bmatrix}
(iii) \begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}
**Solution:
Here, A = \begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}
|A| = 2(4 - 1) - (-1)(-2 + 1) + 1(1 - 2)
= 6 - 1 - 1 = 4
Therefore, inverse of A exists
Cofactors of A are:
C11 = 3 C12 = 1 C13 = -1
C21 = 1 C22 = 3 C23 = 1
C31 = -1 C32 = 1 C33 = 3
adj A = \begin{bmatrix}C_{11}&C_{12}&C_{13}\\C_{21}&C_{22}&C_{23}\\C_{31}&C_{32}&C_{34}\end{bmatrix}^T
= \begin{bmatrix}3&1&-1\\1&3&1\\-1&1&3\end{bmatrix}^T
= \begin{bmatrix}3&1&-1\\1&3&1\\-1&1&3\end{bmatrix}
A-1 = 1/|A|. adj A
Hence, A-1 = \frac{1}{4}\begin{bmatrix}3&1&-1\\1&3&1\\-1&1&3\end{bmatrix}
= \begin{bmatrix}3/4&1/4&-1/4\\1/4&3/4&1/4\\-1/4&1/4&3/4\end{bmatrix}
(iv) \begin{bmatrix}2&0&-1\\5&1&0\\0&1&3\end{bmatrix}
**Solution:
Here, A = \begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}
|A| = 2(3 - 0) - 0 + 1(5)
= 6 - 5 = 1
Therefore, inverse of A exists
Cofactors of A are:
C11 = 3 C12 = -15 C13 = 5
C21 = -1 C22 = 6 C23 = -2
C31 = 1 C32 = -5 C33 = 2
adj A = \begin{bmatrix}C_{11}&C_{12}&C_{13}\\C_{21}&C_{22}&C_{23}\\C_{31}&C_{32}&C_{34}\end{bmatrix}^T
= \begin{bmatrix}3&-15&5\\-1&6&-2\\1&-5&2\end{bmatrix}^T
= \begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}
A-1 = 1/|A|. adj A
Hence, A-1 = \frac{1}{1}\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}
= \begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}
(v) \begin{bmatrix}0&1&-1\\4&-3&4\\3&-3&4\end{bmatrix}
**Solution:
Here, A = \begin{bmatrix}0&1&-1\\4&-3&4\\3&-3&4\end{bmatrix}
|A| = 0 - 1(16 - 12) - 1(-12 + 9)
= -4 + 3 = -1
Therefore, inverse of A exists
Cofactors of A are:
C11 = 0 C12 = -4 C13 = -3
C21 = -1 C22 = 3 C23 = 3
C31 = 1 C32 = -4 C33 = -4
adj A = \begin{bmatrix}C_{11}&C_{12}&C_{13}\\C_{21}&C_{22}&C_{23}\\C_{31}&C_{32}&C_{34}\end{bmatrix}^T
= \begin{bmatrix}0&-4&-3\\-1&3&3\\1&-4&-4\end{bmatrix}^T
= \begin{bmatrix}0&-1&1\\-4&3&-4\\-3&3&-4\end{bmatrix}
A-1 = 1/|A|. adj A
Hence, A-1 = \frac{1}{-1}\begin{bmatrix}0&-1&1\\-4&3&-4\\-3&3&-4\end{bmatrix}
= \begin{bmatrix}0&-1&1\\-4&3&-4\\-3&3&-4\end{bmatrix}
(vi) \begin{bmatrix}0&0&-1\\3&4&5\\-2&-4&-7\end{bmatrix}
**Solution:
Here, A = \begin{bmatrix}0&0&-1\\3&4&5\\-2&-4&-7\end{bmatrix}
|A| = 0 - 0 - 1(-12 + 8)
= -1(-4) = 4
Therefore, inverse of A exists
Cofactors of A are:
C11 = -8 C12 = 11 C13 = -4
C21 = 4 C22 = -2 C23 = 0
C31 = 4 C32 = -3 C33 = 0
adj A = \begin{bmatrix}C_{11}&C_{12}&C_{13}\\C_{21}&C_{22}&C_{23}\\C_{31}&C_{32}&C_{34}\end{bmatrix}^T
= \begin{bmatrix}-8&11&-4\\4&-2&0\\4&-3&0\end{bmatrix}^T
= \begin{bmatrix}-8&4&4\\11&-2&-3\\-4&0&-0\end{bmatrix}
A-1 = 1/|A|. adj A
Hence, A-1 = \frac{1}{4}\begin{bmatrix}-8&4&4\\11&-2&-3\\-4&0&-0\end{bmatrix}
= \begin{bmatrix}-2&1&1\\11/4&-1/2&-3/4\\-1&0&-0\end{bmatrix}
(vii) \begin{bmatrix}1&0&0\\0&cosα &sinα\\0&sinα&-cosα \end{bmatrix}
**Solution:
Here, A = \begin{bmatrix}1&0&0\\0&cosα &sinα\\0&sinα&-cosα \end{bmatrix}
|A| = -cos2α - sin2α
= -(cos2α + sin2α) = -1
Therefore, inverse of A exists
Cofactors of A are:
C11 = -1 C12 = 0 C13 = -0
C21 = 0 C22 = -cosα C23 = -sinα
C31 = 0 C32 = -sinα C33 = cosα
adj A = \begin{bmatrix}C_{11}&C_{12}&C_{13}\\C_{21}&C_{22}&C_{23}\\C_{31}&C_{32}&C_{34}\end{bmatrix}^T
= \begin{bmatrix}-1&0&0\\0&- cosα &- sinα \\0&-sinα & cosα \end{bmatrix}^T
= \begin{bmatrix}-1&0&0\\0&- cosα &- sinα \\0&-sinα & cosα \end{bmatrix}
A-1 = 1/|A|. adj A
Hence, A-1 = \frac{1}{-1}\begin{bmatrix}-1&0&0\\0&- cosα &- sinα \\0&-sinα & cosα \end{bmatrix}
= \begin{bmatrix}1&0&0\\0&cosα &sinα \\0&sinα &-cosα \end{bmatrix}
**Question 9. (i) \begin{bmatrix}1&3&3\\1&4&3\\1&3&4\end{bmatrix}
**Solution:
Here, A = \begin{bmatrix}1&3&3\\1&4&3\\1&3&4\end{bmatrix}
|A| = 1(16 - 9) - 3(4 - 3) + 3(3 - 4)
= 7 - 3 - 3 = 1
Therefore, inverse of A exists
Cofactors of A are:
C11 = 7 C12 = -1 C13 = -1
C21 = -3 C22 = 1 C23 = 0
C31 = -3 C32 = 0 C33 = 1
adj A = \begin{bmatrix}C_{11}&C_{12}&C_{13}\\C_{21}&C_{22}&C_{23}\\C_{31}&C_{32}&C_{34}\end{bmatrix}^T
= \begin{bmatrix}7&-1&-1\\-3&1&0\\-3&0&1\end{bmatrix}^T
= \begin{bmatrix}7&-3&-3\\-1&1&0\\-1&0&1\end{bmatrix}
A-1 = 1/|A|. adj A
Hence, A-1 = 1/1\begin{bmatrix}7&-3&-3\\-1&1&0\\-1&0&1\end{bmatrix}
=\begin{bmatrix}7&-3&-3\\-1&1&0\\-1&0&1\end{bmatrix}
To verify A-1A = \begin{bmatrix}7&-3&-3\\-1&1&0\\-1&0&1\end{bmatrix} \begin{bmatrix}1&3&3\\1&4&3\\1&3&4\end{bmatrix}
= \begin{bmatrix}7-3-3&21-12-9&21-9-12\\-1+1&-3+4&-3+3\\-1+1&-3+3&-3+4\end{bmatrix}
= \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}
(ii) \begin{bmatrix}2&3&1\\3&4&1\\3&7&2\end{bmatrix}
**Solution:
Here, A = \begin{bmatrix}2&3&1\\3&4&1\\3&7&2\end{bmatrix}
|A| = 2(8 - 7) - 3(6 - 3) + 1(21 - 12)
= 2 - 3(3) + 1(9) = 2
Therefore, inverse of A exists
Cofactors of A are:
C11 = 1 C12 = -3 C13 = 9
C21 = 1 C22 = 1 C23 = -5
C31 = -1 C32 = 1 C33 = -1
adj A = \begin{bmatrix}C_{11}&C_{12}&C_{13}\\C_{21}&C_{22}&C_{23}\\C_{31}&C_{32}&C_{34}\end{bmatrix}^T
= \begin{bmatrix}1&-3&9\\1&1&-5\\-1&1&-1\end{bmatrix}^T
= \begin{bmatrix}1&1&-1\\-3&1&1\\9&-5&-1\end{bmatrix}
A-1 = 1/|A|. adj A
Hence, A-1 = \frac{1}{2}\begin{bmatrix}1&1&-1\\-3&1&1\\9&-5&-1\end{bmatrix}
To verify A-1A = \frac{1}{2}\begin{bmatrix}1&1&-1\\-3&1&1\\9&-5&-1\end{bmatrix} \begin{bmatrix}2&3&1\\3&4&1\\3&7&2\end{bmatrix}
= \frac{1}{2}\begin{bmatrix}2+3-3&3+4-7&1+1-2\\-6+3+3&-9+4+7&-3+1+2\\18-15-3&27-20-7&9-5-2\end{bmatrix}
= \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}
**Read More:
Summary
Chapter 7, Exercise 7.1 | Set 1 typically focuses on the concepts of adjoint and inverse matrices. The main topics covered usually include:
- Definition and calculation of cofactors and minors
- Construction of adjoint matrices
- Properties of adjoint matrices
- Definition and calculation of inverse matrices
- Properties of inverse matrices
- Relationship between adjoint and inverse matrices
- Application of adjoint and inverse matrices in solving linear equations
This exercise emphasizes understanding the fundamental concepts of adjoint and inverse matrices, their properties, and their applications in solving mathematical problems.