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Q11:
Find the inverse of the matrix (if it exists): $\begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha \end{bmatrix}$
Solution :
Step 1: Establishing the Matrix and Existence Condition
Let the given matrix be $A$:
$A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha \end{bmatrix}$The inverse $A^{-1}$ exists if and only if $|A| \neq 0$ [Per the property of Non-singular Matrices].
Step 2: Calculating the Determinant $|A|$
Expanding along the first row ($R_1$):
$|A| = 1 \cdot \begin{vmatrix} \cos\alpha & \sin\alpha \\ \sin\alpha & -\cos\alpha \end{vmatrix} - 0 \cdot (\dots) + 0 \cdot (\dots)$ $|A| = 1 \cdot [(\cos\alpha)(-\cos\alpha) - (\sin\alpha)(\sin\alpha)]$ $|A| = -(\cos^2\alpha + \sin^2\alpha)$Applying the Pythagorean identity $\cos^2\alpha + \sin^2\alpha = 1$ [Trigonometric Identity]:
$|A| = -1$Since $|A| \neq 0$, the inverse $A^{-1}$ exists and is given by $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.
Step 3: Calculating the Matrix of Cofactors $C_{ij}$
The cofactor $C_{ij}$ is defined as $(-1)^{i+j}M_{ij}$ where $M_{ij}$ is the minor of element $a_{ij}$.
- $C_{11} = +(\cos\alpha(-\cos\alpha) - \sin^2\alpha) = -1$
- $C_{12} = -(0 - 0) = 0$
- $C_{13} = +(0 - 0) = 0$
- $C_{21} = -(0 - 0) = 0$
- $C_{22} = +(1(-\cos\alpha) - 0) = -\cos\alpha$
- $C_{23} = -(1(\sin\alpha) - 0) = -\sin\alpha$
- $C_{31} = +(0 - 0) = 0$
- $C_{32} = -(1(\sin\alpha) - 0) = -\sin\alpha$
- $C_{33} = +(1(\cos\alpha) - 0) = \cos\alpha$
Step 4: Constructing the Adjoint Matrix $\text{adj}(A)$
The adjoint is the transpose of the cofactor matrix, $\text{adj}(A) = [C_{ij}]^T$:
$\text{adj}(A) = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -\cos\alpha & -\sin\alpha \\ 0 & -\sin\alpha & \cos\alpha \end{bmatrix}$Step 5: Final Computation of $A^{-1}$
Using $A^{-1} = \frac{1}{-1} \text{adj}(A) = -1 \cdot \text{adj}(A)$:
$A^{-1} = -1 \begin{bmatrix} -1 & 0 & 0 \\ 0 & -\cos\alpha & -\sin\alpha \\ 0 & -\sin\alpha & \cos\alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha \end{bmatrix}$Final Solution: The inverse of the matrix is $\begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha \end{bmatrix}$
More Questions from Class 12 Mathematics Determinants EXERCISE 4.4
- Q1: Find adjoint of the matrix $\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$
- Q2: Find adjoint of the matrix $\begin{bmatrix} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{bmatrix}$
- Q3: Verify $A (adj A) = (adj A) A = |A| I$ for the matrix $\begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix}$
- Q4: Verify $A (adj A) = (adj A) A = |A| I$ for the matrix $\begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix}$
- Q5: Find the inverse of the matrix (if it exists): $\begin{bmatrix} 2 & -2 \\ 4 & 3 \end{bmatrix}$
- Q6: Find the inverse of the matrix (if it exists): $\begin{bmatrix} -1 & 5 \\ -3 & 2 \end{bmatrix}$
- Q7: Find the inverse of the matrix (if it exists): $\begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{bmatrix}$
- Q8: Find the inverse of the matrix (if it exists): $\begin{bmatrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{bmatrix}$
- Q9: Find the inverse of the matrix (if it exists): $\begin{bmatrix} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{bmatrix}$
- Q10: Find the inverse of the matrix (if it exists): $\begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}$
- Q12: Let $A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}$ and $B = \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix}$. Verify that $(AB)^{-1} = B^{-1} A^{-1}$.
- Q13: If $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$, show that $A^2 – 5A + 7I = O$. Hence find $A^{-1}$.
- Q14: For the matrix $A = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}$, find the numbers $a$ and $b$ such that $A^2 + aA + bI = O$.
- Q15: For the matrix $A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}$ Show that $A^3– 6A^2 + 5A + 11 I = O$. Hence, find $A^{-1}$.
- Q16: 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 $A^{-1}$
- Q17: Let $A$ be a nonsingular square matrix of order $3 \times 3$. Then $|adj A|$ is equal to
- Q18: If $A$ is an invertible matrix of order 2, then det $(A^{-1})$ is equal to
CBSE Solutions for Class 12 Mathematics Determinants
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