If X=[ array l x y z array ] is a solution of the system of equations A X=B , where adj A=[ array ccc 4 & 2 & 2 -5 & 0 & 5 1 & -2 & 3 array ] and B=[ array l 4 0 2 array ] , then |x+y+z| is equal to :

Correct answer: A. 2 X = (1)/(|A|)adj(A)· B . |adj(A)| = 4(10)-2(-20)+2(10) = 100 , so |A|^2=100 |A|=±10 . adj(A)· B = pmatrix 16+0+4 -20+0+10 4+0+6 pmatrix = pmatrix 20 -10 10 pmatrix x+y+z = (20-10+10)/(|A|)=(20)/(±10)=±2 |x+y+z|=2

JEE Main 202622 JanuaryEvening Shift

Exam: JEE Main
Year: 2026
Date: 22 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Matrices

JEE Main 2026 · 22 January · Evening Shift · Mathematics · Matrices

If X=[ array l x y z array ] is a solution of the system of...

X=\left[

\begin{array}{l}x \\ y \\ z\end{array}

\right]

is a solution of the system of equations

A X=B

, where

\operatorname{adj} A=\left[

\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}

\right]

and

\mathrm{B}=\left[

\begin{array}{l}4 \\ 0 \\ 2\end{array}

\right]

, then

|x+y+z|

is equal to :

Options

A2
B $\frac{3}{2}$
C1
D3

Correct answer

A. 2

Step-by-step solution

X = \frac{1}{|A|}\text{adj}(A)\cdot B

|\text{adj}(A)| = 4(10)-2(-20)+2(10) = 100

, so

|A|^2=100 \Rightarrow |A|=\pm10

\text{adj}(A)\cdot B = \begin{pmatrix}16+0+4\\-20+0+10\\4+0+6\end{pmatrix}=\begin{pmatrix}20\\-10\\10\end{pmatrix}

x+y+z = \frac{20-10+10}{|A|}=\frac{20}{\pm10}=\pm2

|x+y+z|=2

Practice Matrices on Quantrex Academy →

More from Matrices