Q81 / 150 · MHT CET 2024 (11 May Shift 1) · Mathematics
Let $A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right]$ and $B=\left[\begin{array}{l}4 \\ 0 \\ 2\end{array}\right]$ such that $\mathrm{AX}=\mathrm{B}$, then $\mathrm{X}=$
- A$\left[\begin{array}{c}-1 \\ 2 \\ 1\end{array}\right]$
- Correct answerB$\left[\begin{array}{c}2 \\ -1 \\ 1\end{array}\right]$
- C$\left[\begin{array}{c}-1 \\ 1 \\ 2\end{array}\right]$
- D$\left[\begin{array}{c}-2 \\ 1 \\ -1\end{array}\right]$
Solution
$\begin{aligned}
& \mathrm{AX}=\mathrm{B} \\
& \Rightarrow\left[\begin{array}{ccc}
1 & -1 & 1 \\
2 & 1 & -3 \\
1 & 1 & 1
\end{array}\right]\left[\begin{array}{l}
x_1 \\
x_2 \\
x_3
\end{array}\right]=\left[\begin{array}{l}
4 \\
0 \\
2
\end{array}\right]
\end{aligned}$
Applying $\mathrm{R}_2 \rightarrow \mathrm{R}_2-2 \mathrm{R}_1$ and $\mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_1$, $\left[\begin{array}{ccc} 1 & -1 & 1 \\ 0 & 3 & -5 \\ 0 & 2 & 0 \end{array}\right]\left[\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right]=\left[\begin{array}{c} 4 \\ -8 \\ -2 \end{array}\right]$ $\begin{array}{ll} \therefore \quad & x_1-x_2+x_3=4 ...(i)\\ & 3 x_2-5 x_3=-8 ...(ii)\\ \therefore & 2 x_2=-2 \Rightarrow x_2=-1 \end{array}$
From (ii), $3(-1)-5 x_3=-8 \Rightarrow x_3=1$
From (i), $\begin{aligned} & x_1+1+1=4 \Rightarrow x_1=2 \\ \therefore \quad & {\left[\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right]=\left[\begin{array}{c} 2 \\ -1 \\ 1 \end{array}\right] } \end{aligned}$
Applying $\mathrm{R}_2 \rightarrow \mathrm{R}_2-2 \mathrm{R}_1$ and $\mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_1$, $\left[\begin{array}{ccc} 1 & -1 & 1 \\ 0 & 3 & -5 \\ 0 & 2 & 0 \end{array}\right]\left[\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right]=\left[\begin{array}{c} 4 \\ -8 \\ -2 \end{array}\right]$ $\begin{array}{ll} \therefore \quad & x_1-x_2+x_3=4 ...(i)\\ & 3 x_2-5 x_3=-8 ...(ii)\\ \therefore & 2 x_2=-2 \Rightarrow x_2=-1 \end{array}$
From (ii), $3(-1)-5 x_3=-8 \Rightarrow x_3=1$
From (i), $\begin{aligned} & x_1+1+1=4 \Rightarrow x_1=2 \\ \therefore \quad & {\left[\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right]=\left[\begin{array}{c} 2 \\ -1 \\ 1 \end{array}\right] } \end{aligned}$
Hence, the correct answer is option: 'B'.