Q1415: If A is a $3\times3$ invertible matrix, show that for any sc...

Class CBSE Class 12 Mathematics Matrices and Determinants Q #1415

COMPETENCY BASED

ANALYZE

5 Marks 2025 AISSCE(Board Exam) LA

If A is a $3\times3$ invertible matrix, show that for any scalar $k\ne0$, $(kA)^{-1}=\frac{1}{k}A^{-1}$. Hence calculate $(3A)^{-1}$, where $A=\begin{bmatrix}2&-1&1\\ -1&2&-1\\ 1&-1&2\end{bmatrix}$.

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Detailed Solution

Step 1: Proving $(kA)^{-1}=\frac{1}{k}A^{-1}$

To prove $(kA)^{-1}=\frac{1}{k}A^{-1}$, we need to show that $(kA)(\frac{1}{k}A^{-1}) = I$ and $(\frac{1}{k}A^{-1})(kA) = I$, where $I$ is the identity matrix.

Step 2: Verifying $(kA)(\frac{1}{k}A^{-1}) = I$

$(kA)(\frac{1}{k}A^{-1}) = k(\frac{1}{k})AA^{-1} = 1 \cdot I = I$

Step 3: Verifying $(\frac{1}{k}A^{-1})(kA) = I$

$(\frac{1}{k}A^{-1})(kA) = k(\frac{1}{k})A^{-1}A = 1 \cdot I = I$

Step 4: Finding the determinant of A

Given $A=\begin{bmatrix}2&-1&1\\ -1&2&-1\\ 1&-1&2\end{bmatrix}$, we calculate the determinant of $A$ as follows: $$|A| = 2(4-1) - (-1)(-2+1) + 1(1-2) = 2(3) + 1(-1) + 1(-1) = 6 - 1 - 1 = 4$$

Step 5: Finding the Adjoint of A

The matrix of cofactors is: $C = \begin{bmatrix}3&1&-1\\ 1&3&1\\ -1&1&3\end{bmatrix}$ The adjoint of $A$ is the transpose of the cofactor matrix: $adj(A) = C^T = \begin{bmatrix}3&1&-1\\ 1&3&1\\ -1&1&3\end{bmatrix}$

Step 6: Finding the Inverse of A

The inverse of $A$ is given by $A^{-1} = \frac{1}{|A|}adj(A)$. $$A^{-1} = \frac{1}{4}\begin{bmatrix}3&1&-1\\ 1&3&1\\ -1&1&3\end{bmatrix} = \begin{bmatrix}\frac{3}{4}&\frac{1}{4}&-\frac{1}{4}\\ \frac{1}{4}&\frac{3}{4}&\frac{1}{4}\\ -\frac{1}{4}&\frac{1}{4}&\frac{3}{4}\end{bmatrix}$$

Step 7: Calculating $(3A)^{-1}$

Using the property $(kA)^{-1} = \frac{1}{k}A^{-1}$, we have $(3A)^{-1} = \frac{1}{3}A^{-1}$. $$(3A)^{-1} = \frac{1}{3} \cdot \frac{1}{4}\begin{bmatrix}3&1&-1\\ 1&3&1\\ -1&1&3\end{bmatrix} = \frac{1}{12}\begin{bmatrix}3&1&-1\\ 1&3&1\\ -1&1&3\end{bmatrix} = \begin{bmatrix}\frac{1}{4}&\frac{1}{12}&-\frac{1}{12}\\ \frac{1}{12}&\frac{1}{4}&\frac{1}{12}\\ -\frac{1}{12}&\frac{1}{12}&\frac{1}{4}\end{bmatrix}$$

Final Answer: $\frac{1}{12}\begin{bmatrix}3&1&-1\\ 1&3&1\\ -1&1&3\end{bmatrix}$<\/span>

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Pedagogical Audit

Bloom's Analysis: This is an ANALYZE question because the student needs to break down the problem into smaller parts, apply the properties of inverse matrices, and calculate the inverse of a given matrix.

Knowledge Dimension: PROCEDURAL

Justification: The question requires the student to apply a specific algorithm or method to find the inverse of a matrix and use the properties of scalar multiplication with matrices.<\/span>

Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. The question assesses the student's ability to apply the concepts of matrices and determinants to solve a problem involving the inverse of a matrix.