JEE Main 2026 — 22 January, Evening Shift — Question 9

Given $\text{adj}(A) = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$. We know $A^{-1} = \frac{\text{adj}(A)}{|A|}$. First, find $|\text{adj}(A)| = |A|^{n-1}$ where $n=3$, so $|\text{adj}(A)| = |A|^2$. Compute $|\text{adj}(A)| = 4(0\cdot3 - 5\cdot(-2)) - 2((-5)\cdot3 - 5\cdot1) + 2((-5)\cdot(-2) - 0\cdot1)$. $= 4(0+10) - 2(-15-5) + 2(10-0) = 40 - 2(-20) + 20 = 40 + 40 + 20 = 100$. Thus $|A|^2 = 100 \Rightarrow |A| = \pm 10$. Now $X = A^{-1}B = \frac{\text{adj}(A)}{|A|} B = \pm \frac{1}{10} \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix} \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$. Compute product: $\begin{bmatrix} 4\cdot4 + 2\cdot0 + 2\cdot2 \\ -5\cdot4 + 0\cdot0 + 5\cdot2 \\ 1\cdot4 + (-2)\cdot0 + 3\cdot2 \end{bmatrix} = \begin{bmatrix} 16+0+4 \\ -20+0+10 \\ 4+0+6 \end{bmatrix} = \begin{bmatrix} 20 \\ -10 \\ 10 \end{bmatrix}$. Thus $X = \pm \frac{1}{10} \begin{bmatrix} 20 \\ -10 \\ 10 \end{bmatrix} = \pm \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}$. Hence $x+y+z = \pm (2-1+1) = \pm 2$,

so $|x+y+z| = 2$.