Q52 / 150 · MHT CET 2020 (15 Oct Shift 1) · Mathematics
If $\mathrm{AX}=\mathrm{B}$, where $\mathrm{A}=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & -1 & 0 \\ 3 & 3 & -4\end{array}\right], \mathrm{B}=\left[\begin{array}{l}1 \\ 1 \\ 2\end{array}\right]$ and $\mathrm{X}=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$, then $x+y+z=$
- A$2$
- Correct answerB$3$
- C$6$
- D$1$
Solution
Given $A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & -1 & 0 \\ 3 & 3 & -4\end{array}\right]$ and $B=\left[\begin{array}{l}1 \\ 1 \\ 2\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $A X=B$ $\therefore \mathrm{AA}^{-1} \mathrm{X}=\mathrm{A}^{-1} \mathrm{~B} \Rightarrow \mathrm{IX}=\mathrm{A}^{-1} \mathrm{~B} \Rightarrow \mathrm{X}=\mathrm{A}^{-1} \mathrm{~B}$
Now $|\mathrm{A}|=4+(-8)+9=5$
$$
\begin{aligned}
\therefore A^{-1} &=\left[\begin{array}{lll}
4 & -1 & 1 \\
8 & -7 & 2 \\
9 & -6 & 1
\end{array}\right] \times \frac{1}{5} \\
\therefore\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right] &=\frac{1}{5}\left[\begin{array}{ccc}
4 & -1 & 1 \\
8 & -7 & 2 \\
9 & -6 & 1
\end{array}\right]\left[\begin{array}{l}
1 \\
1 \\
2
\end{array}\right] \\
&=\frac{1}{5}\left[\begin{array}{ll}
8-1+2 \\
8-7+4 \\
9-6+2
\end{array}\right]=\frac{1}{5}\left[\begin{array}{l}
5 \\
5 \\
5
\end{array}\right] \\
y &=\left[\begin{array}{l}
1 \\
1 \\
1
\end{array}\right]
\end{aligned}
$$
On comparing both side, we get
$$
x=y=z=1 \Rightarrow x+y+z=1+1+1=3
$$
Hence, the correct answer is option: 'B'.