Matrices – Adjoint and Inverse – JEE Main 22 Jan 2026 Shift 2

Solution:

We are given the system $AX = B$, so $X = A^{-1}B$.
We know that $A^{-1} = \frac{\text{adj } A}{|A|}$.
Let $C = \text{adj } A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$.
We need to find $|A|$. We use the property $|\text{adj } A| = |A|^{n-1} = |A|^2$ (for a $3 \times 3$ matrix).

Calculate $|\text{adj } A|$:
$$ |C| = 4(0 – (-10)) – 2(-15 – 5) + 2(10 – 0) $$
$$ |C| = 4(10) – 2(-20) + 2(10) = 40 + 40 + 20 = 100 $$
So, $|A|^2 = 100 \Rightarrow |A| = \pm 10$.

Now, find $X$:
$$ X = \frac{1}{|A|} (\text{adj } A) B $$
$$ (\text{adj } A) B = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix} \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix} = \begin{bmatrix} 16 + 0 + 4 \\ -20 + 0 + 10 \\ 4 + 0 + 6 \end{bmatrix} = \begin{bmatrix} 20 \\ -10 \\ 10 \end{bmatrix} $$

$$ X = \pm \frac{1}{10} \begin{bmatrix} 20 \\ -10 \\ 10 \end{bmatrix} = \pm \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix} $$
Thus, $x = \pm 2, y = \mp 1, z = \pm 1$.

We need to find $|x+y+z|$:
$$ |x+y+z| = |\pm 2 \mp 1 \pm 1| = |\pm (2 – 1 + 1)| = |\pm 2| = 2 $$

Ans. (4)