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Q6:
Find the inverse of the matrix (if it exists): $\begin{bmatrix} -1 & 5 \\ -3 & 2 \end{bmatrix}$
Solution :
Step 1: Establishing the Theoretical Framework
To find the inverse of a square matrix $A$, denoted as $A^{-1}$, we must apply the formula:
$A^{-1} = \frac{1}{|A|} \text{adj}(A)$This operation is defined only if the determinant $|A| \neq 0$, in which case the matrix is termed non-singular and invertible [Per the Invertibility Criterion for Square Matrices].
Step 2: Calculating the Determinant
Given the matrix $A = \begin{bmatrix} -1 & 5 \\ -3 & 2 \end{bmatrix}$, we calculate the determinant $|A|$ using the expansion formula for a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, where $|A| = ad - bc$.
$|A| = (-1)(2) - (5)(-3)$ $|A| = -2 - (-15)$ $|A| = -2 + 15$ $|A| = 13$Since $|A| = 13 \neq 0$, the inverse exists [By the Non-Singularity Theorem].
Step 3: Determining the Adjoint Matrix
For a $2 \times 2$ matrix, the adjugate (adjoint) is found by interchanging the diagonal elements and changing the signs of the off-diagonal elements [Per the definition of the Adjoint of a $2 \times 2$ matrix].
Given $A = \begin{bmatrix} -1 & 5 \\ -3 & 2 \end{bmatrix}$:- Swap $a_{11}$ and $a_{22}$: $(-1 \leftrightarrow 2)$
- Negate $a_{12}$ and $a_{21}$: $(5 \rightarrow -5, -3 \rightarrow 3)$
Step 4: Constructing the Inverse Matrix
Substitute the results from Step 2 and Step 3 into the inverse formula:
$A^{-1} = \frac{1}{13} \begin{bmatrix} 2 & -5 \\ 3 & -1 \end{bmatrix}$ $A^{-1} = \begin{bmatrix} \frac{2}{13} & -\frac{5}{13} \\ \frac{3}{13} & -\frac{1}{13} \end{bmatrix}$Step 5: Visual Representation of Transformation
Final Solution: The inverse of the matrix is $\begin{bmatrix} \frac{2}{13} & -\frac{5}{13} \\ \frac{3}{13} & -\frac{1}{13} \end{bmatrix}$.
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): $\begin{bmatrix} 2 & -2 \\ 4 & 3 \end{bmatrix}$
- 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): $\begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}$
- 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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