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Q10:
Find the inverse of the matrix (if it exists): $\begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}$
Solution :
Step 1: Establishing the Matrix and Checking for Invertibility
Given the matrix $A = \begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}$.
An inverse $A^{-1}$ exists if and only if $\det(A) \neq 0$ [Per the definition of non-singular matrices].
Expanding along the first column:
$\det(A) = 1 \begin{vmatrix} 2 & -3 \\ -2 & 4 \end{vmatrix} - 0 \begin{vmatrix} -1 & 2 \\ -2 & 4 \end{vmatrix} + 3 \begin{vmatrix} -1 & 2 \\ 2 & -3 \end{vmatrix}$$\det(A) = 1(8 - 6) - 0 + 3(3 - 4)$
$\det(A) = 2 + 3(-1) = 2 - 3 = -1$
Since $\det(A) = -1 \neq 0$, the matrix $A$ is non-singular and the inverse $A^{-1}$ exists.
Step 2: Calculating the Matrix of Cofactors
The cofactor $C_{ij}$ is given by $(-1)^{i+j} M_{ij}$, where $M_{ij}$ is the minor of element $a_{ij}$.
- $C_{11} = + \begin{vmatrix} 2 & -3 \\ -2 & 4 \end{vmatrix} = 8 - 6 = 2$
- $C_{12} = - \begin{vmatrix} 0 & -3 \\ 3 & 4 \end{vmatrix} = -(0 - (-9)) = -9$
- $C_{13} = + \begin{vmatrix} 0 & 2 \\ 3 & -2 \end{vmatrix} = 0 - 6 = -6$
- $C_{21} = - \begin{vmatrix} -1 & 2 \\ -2 & 4 \end{vmatrix} = -(-4 - (-4)) = 0$
- $C_{22} = + \begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} = 4 - 6 = -2$
- $C_{23} = - \begin{vmatrix} 1 & -1 \\ 3 & -2 \end{vmatrix} = -(-2 - (-3)) = -1$
- $C_{31} = + \begin{vmatrix} -1 & 2 \\ 2 & -3 \end{vmatrix} = 3 - 4 = -1$
- $C_{32} = - \begin{vmatrix} 1 & 2 \\ 0 & -3 \end{vmatrix} = -(-3 - 0) = 3$
- $C_{33} = + \begin{vmatrix} 1 & -1 \\ 0 & 2 \end{vmatrix} = 2 - 0 = 2$
Step 3: Finding the Adjoint Matrix $adj(A)$
The adjoint matrix is the transpose of the cofactor matrix [Per the properties of the Adjoint of a matrix]:
$adj(A) = \begin{bmatrix} C_{11} & C_{21} & C_{31} \\ C_{12} & C_{22} & C_{32} \\ C_{13} & C_{23} & C_{33} \end{bmatrix} = \begin{bmatrix} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2 \end{bmatrix}$Step 4: Calculating $A^{-1}$
Using the formula $A^{-1} = \frac{1}{\det(A)} adj(A)$:
$A^{-1} = \frac{1}{-1} \begin{bmatrix} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2 \end{bmatrix}$$A^{-1} = \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}$
Final Solution: $A^{-1} = \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}$
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): $\begin{bmatrix} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{bmatrix}$
- Q11: Find the inverse of the matrix (if it exists): $\begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha \end{bmatrix}$
- 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
Chapters in CBSE - Class 12 Mathematics
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