UPSC NDA II 2024 Mathematics Paper

Concept:For any square matrix $A$, the product $A(\text{adj }A)$ equals $|A| I$, where $|A|$ is the determinant and $I$ is the identity matrix of the same order.Explanation:Given matrix $A = \begin{bmatrix}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{bmatrix}$.First, compute the determinant $|A|$:$|A| = 3\bigl((-3)(1) - (4)(-1)\bigr) - 2\bigl((-3)(1) - (4)(0)\bigr) + 0$$= 3(-3+4) - 2(-3+0) = 3(1) - 2(-3) = 3 + 6 = 9$.Wait, recalc carefully: correct is $3[(-3)(1) - (4)(-1)] - 2[(-3)(1) - (4)(0)] + 0[(-3)(-1) - (4)(-3)]$Actually step: $|A| = 3[(-3)(1) - (4)(-1)] - 2[(-3)(1) - (4)(0)] + 0[(-3)(-1) - (4)(-3)]$$= 3[-3+4] - 2[-3-0] + 0 = 3(1) - 2(-3) = 3+6=9$.But original solution says $|A|=1$, so there is a discrepancy. Let me re-evaluate: Using proper expansion: $A = \begin{bmatrix}3&-3&4\\2&-3&4\\0&-1&1\end{bmatrix}$Determinant along first row: $3\cdot\det\begin{bmatrix}-3&4\\-1&1\end{bmatrix} - (-3)\cdot\det\begin{bmatrix}2&4\\0&1\end{bmatrix} + 4\cdot\det\begin{bmatrix}2&-3\\0&-1\end{bmatrix}$$= 3[(-3)(1) - (4)(-1)] + 3[(2)(1) - (4)(0)] + 4[(2)(-1) - (-3)(0)]$$= 3[-3+4] + 3[2-0] + 4[-2+0] = 3(1) + 3(2) + 4(-2) = 3+6-8=1$.Ah, the original correct expansion yields 1. My earlier mistake: the cofactor signs: the second term should be $-(-3)=+3$, and third term $+4$. So correct: $3(1) + 3(2) + 4(-2) = 3+6-8=1$. So $|A| = 1$.Thus $A(\text{adj }A) = |A| I = 1 \cdot I = I$, where $I$ is the $3\times 3$ identity matrix $\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$.Answer:$\rm \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$ (Option D).