Question

If $\left[\begin{array}{c}x-y-z\\ -y+z\\ z\end{array}\right]=\left[\begin{array}{c}0\\ 5\\ 3\end{array}\right]$ then $x,y,z$ are

• A. $5,2,2$
• B. $1,–2,3$
• C. $0,–3,3$
• D. $11,8,3$
• E. $4,1,3$

Answer 1

The correct option is B

$1,–2,3$

The explanation for the correct answer.

Given: $\left[\begin{array}{c}x-y-z\\ -y+z\\ z\end{array}\right]=\left[\begin{array}{c}0\\ 5\\ 3\end{array}\right]$

On comparing both the matrix we get

$$\begin{gathered} x-y-z=0-\left(1\right) \\ -y+z=5-\left(2\right) \\ z=3-\left(3\right) \end{gathered}$$

Put the value of $z$ in $\left(2\right)$

$$\begin{gathered} \Rightarrow -y+3=5 \\ \Rightarrow -y=2 \\ \Rightarrow y=-2 \end{gathered}$$

Put the value of $y,z$in $\left(1\right)$

$$\begin{gathered} \Rightarrow x-\left(-2\right)-3=0 \\ \Rightarrow x=1 \end{gathered}$$

Therefore, $x=1,y=-2,z=3$

Hence, option (b) is correct.