CBSE 2025, Set 3

Question

If $A$ is a $3	imes3$ invertible matrix, show that for any scalar $k
e0$, $(kA)^{-1}=\frac{1}{k}A^{-1}$. Hence calculate $(3A)^{-1}$, where $A=\begin{pmatrix}2&-1&1\\ -1&2&-1\\ 1&-1&2\end{pmatrix}$.

Edvaya · Accepted Answer

The correct answer is Subjective. **Step 1:** Prove the property $(kA)^{-1} = \frac{1}{k}A^{-1}$. Let $B = kA$. For any invertible matrix $B$, we know that $B B^{-1} = I$, where $I$ is the identity matrix. Substitute $B = kA$: $$(kA)(kA)^{-1} = I$$ Since $k$ is a non-zero scalar, we can factor it outside the matrix product: $$k (A(kA)^{-1}) = I$$ Multiply both sides from the left by $A^{-1}$: $$A^{-1} [k (A(kA)^{-1})] = A^{-1} I$$ $$k (A^{-1}A) (kA)^{-1} = A^{-1}$$ $$k I (kA)^{-1} = A^{-1} \implies k(kA)^{-1} = A^{-1}$$ Dividing both sides by $k$ (since $k 
e 0$): $$(kA)^{-1} = \frac{1}{k}A^{-1}$$ **Hence Proved.** **Step 2:** Find the determinant of matrix $A$. $$A = \begin{pmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{pmatrix}$$ Expanding along the first row: $$|A| = 2(4 - 1) - (-1)(-2 - (-1)) + 1(1 - 2)$$ $$|A| = 2(3) + 1(-1) + 1(-1)$$ $$|A| = 6 - 1 - 1 = 4$$ Since $|A| = 4 
e 0$, $A$ is invertible. **Step 3:** Find the cofactors of $A$. $C_{11} = (4 - 1) = 3$ $C_{12} = -(-2 + 1) = 1$ $C_{13} = (1 - 2) = -1$ $C_{21} = -(-2 + 1) = 1$ $C_{22} = (4 - 1) = 3$ $C_{23} = -(-2 + 1) = 1$ $C_{31} = (1 - 2) = -1$ $C_{32} = -(-2 + 1) = 1$ $C_{33} = (4 - 1) = 3$ The cofactor matrix is $\begin{pmatrix} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{pmatrix}$. **Step 4:** Find $A^{-1}$. The adjoint matrix $adj(A)$ is the transpose of the cofactor matrix: $$adj(A) = \begin{pmatrix} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{pmatrix}^T = \begin{pmatrix} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{pmatrix}$$ $$A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{4} \begin{pmatrix} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{pmatrix}$$ **Step 5:** Calculate $(3A)^{-1}$. Using the property proved in Step 1 where $k = 3$: $$(3A)^{-1} = \frac{1}{3} A^{-1}$$ $$(3A)^{-1} = \frac{1}{3} \left[ \frac{1}{4} \begin{pmatrix} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{pmatrix} ight]$$ $$(3A)^{-1} = \frac{1}{12} \begin{pmatrix} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{pmatrix}$$

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