NCERT Solutions
Class 12 - Mathematics Part-1 - Chapter 4: DETERMINANTS
Exercise 4.4

Question. 1

Find adjoint of the matrix

\[ \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \]

Answer:

\(\operatorname{adj}A = \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}\)

Question. 2

Find adjoint of the matrix

\[ \begin{bmatrix} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{bmatrix} \]

Answer:

\(\operatorname{adj}A = \begin{bmatrix} 3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5 \end{bmatrix}\)

Question. 3

Verify that \(A(\operatorname{adj}A) = (\operatorname{adj}A)A = |A|I\) for

\[ A = \begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix} \]

Answer:

Question. 4

Verify that \(A(\operatorname{adj}A) = (\operatorname{adj}A)A = |A|I\) for

\[ A = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix} \]

Answer:

Question. 5

Find the inverse of the matrix (if it exists)

\[ \begin{bmatrix} 2 & -2 \\ 4 & 3 \end{bmatrix} \]

Answer:

\(A^{-1} = \frac{1}{14}\begin{bmatrix} 3 & 2 \\ -4 & 2 \end{bmatrix}\)

Question. 6

Find the inverse of the matrix (if it exists)

\[ \begin{bmatrix} -1 & 5 \\ -3 & 2 \end{bmatrix} \]

Answer:

\(A^{-1} = \frac{1}{13}\begin{bmatrix} 2 & -5 \\ 3 & -1 \end{bmatrix}\)

Question. 7

Find the inverse of the matrix (if it exists)

\[ \begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{bmatrix} \]

Answer:

\(A^{-1} = \frac{1}{10}\begin{bmatrix} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{bmatrix}\)

Question. 8

Find the inverse of the matrix (if it exists)

\[ \begin{bmatrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{bmatrix} \]

Answer:

\(A^{-1} = -\frac{1}{3}\begin{bmatrix} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{bmatrix}\)

Question. 9

Find the inverse of the matrix (if it exists)

\[ \begin{bmatrix} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{bmatrix} \]

Answer:

\(A^{-1} = -\frac{1}{3}\begin{bmatrix} -1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6 \end{bmatrix}\)

Question. 10

Find the inverse of the matrix (if it exists)

\[ \begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix} \]

Answer:

\(A^{-1} = \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}\)

Question. 11

Find the inverse of the matrix (if it exists)

\[ \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha \end{bmatrix} \]

Answer:

\(A^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha \end{bmatrix}\)

Question. 12

Let \(A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}\) and \(B = \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix}\). Verify that \((AB)^{-1} = B^{-1}A^{-1}\).

Answer:

Question. 13

If \(A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\), show that \(A^2 - 5A + 7I = O\). Hence, find \(A^{-1}\).

Answer:

\(A^{-1} = \frac{1}{7}\begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}\)

Question. 14

For the matrix \(A = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}\), find the numbers \(a\) and \(b\) such that \(A^2 + aA + bI = O\).

Answer:

\(a = -4,\; b = 1\)

Question. 15

For the matrix

\[ A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 2 & -1 & 3 \end{bmatrix} \]

show that \(A^3 - 6A^2 + 5A + 11I = O\). Hence, find \(A^{-1}\).

Answer:

\(A^{-1} = \frac{1}{11}\begin{bmatrix} -3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1 \end{bmatrix}\)

Question. 16

If

\[ A = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix} \]

verify that \(A^3 - 6A^2 + 9A - 4I = O\) and hence find \(A^{-1}\).

Answer:

\(A^{-1} = \frac{1}{4}\begin{bmatrix} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{bmatrix}\)

Question. 17

Let \(A\) be a nonsingular square matrix of order \(3 \times 3\). Then \(|\operatorname{adj}A|\) is equal to

Answer:

1

Question. 18

If \(A\) is an invertible matrix of order 2, then \(\det(A^{-1})\) is equal to

Answer:

1
NCERT Solutions Class 12 – Mathematics Part-1 – Chapter 4: DETERMINANTS – Exercise 4.4 | Detailed Answers