JEE Mathematics 2026 Question with Solution

Given: $AX=B$, $adj\,A=\begin{bmatrix}4&2&2\\-5&0&5\\1&-2&3\end{bmatrix}$ and $B=\begin{bmatrix}4\\0\\2\end{bmatrix}$.

Find: $|x+y+z|$.

Using the inverse-adjoint relation,

$$X=A^{-1}B=\frac{1}{|A|}(adj\,A)B$$

Now compute

$$(adj\,A)B= \begin{bmatrix} 4&2&2\\ -5&0&5\\ 1&-2&3 \end{bmatrix} \begin{bmatrix} 4\\ 0\\ 2 \end{bmatrix}$$

So,

$$=\begin{bmatrix} 16+0+4\\ -20+0+10\\ 4+0+6 \end{bmatrix} =\begin{bmatrix} 20\\ -10\\ 10 \end{bmatrix}$$

Hence,

$$X=\frac{1}{|A|}\begin{bmatrix}20\\-10\\10\end{bmatrix}$$

Therefore,

$$x+y+z=\frac{1}{|A|}(20-10+10)=\frac{20}{|A|}$$

For a $3\times 3$ matrix,

$$|adj\,A|=|A|^2$$

From the given adjoint matrix,

$$|adj\,A|=100$$

So,

$$|A|=10$$

Thus,

$$|x+y+z|=\left|\frac{20}{10}\right|=2$$

Therefore, the correct option is D.

Given: The solution vector is $X=\begin{bmatrix}x\\y\\z\end{bmatrix}$ and $AX=B$.

Find: the value of $|x+y+z|$.

The hint suggests using

$$X=\frac{1}{|A|}(adj\,A)B$$

First multiply the given matrices entry-wise:

$$(adj\,A)B= \begin{bmatrix} 4&2&2\\ -5&0&5\\ 1&-2&3 \end{bmatrix} \begin{bmatrix} 4\\ 0\\ 2 \end{bmatrix}$$

The first component is

$$4\cdot 4+2\cdot 0+2\cdot 2=20$$

The second component is

$$-5\cdot 4+0\cdot 0+5\cdot 2=-10$$

The third component is

$$1\cdot 4+(-2)\cdot 0+3\cdot 2=10$$

So,

$$(adj\,A)B=\begin{bmatrix}20\\-10\\10\end{bmatrix}$$

Hence,

$$X=\frac{1}{|A|}\begin{bmatrix}20\\-10\\10\end{bmatrix}$$

This gives

$$x=\frac{20}{|A|},\quad y=\frac{-10}{|A|},\quad z=\frac{10}{|A|}$$

Therefore,

$$x+y+z=\frac{20-10+10}{|A|}=\frac{20}{|A|}$$

Now use the determinant identity for a $3\times 3$ matrix:

$$|adj\,A|=|A|^{3-1}=|A|^2$$

From the worked solution,

$$|adj\,A|=100$$

Therefore,

$$|A|=10$$

and so

$$|x+y+z|=\left|\frac{20}{10}\right|=2$$

Thus, the required value is $2$.