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Matrics — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsMatrics2 Marks
MCQ
Let $X =\left[\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right], A =\left[\begin{array}{ccc}1 & -1 & 2 \\ 2 & 0 & 1 \\ 3 & 2 & 1\end{array}\right]$ and $B =\left[\begin{array}{l}3 \\ 1 \\ 4\end{array}\right]$. If $AX = B$, then $X =$
  • A
    $\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$
  • ✓
    $\left[\begin{array}{c}-1 \\ 2 \\ 3\end{array}\right]$
  • C
    $\left[\begin{array}{c}-1 \\ -2 \\ 3\end{array}\right]$
  • D
    $\left[\begin{array}{l}-1 \\ -2 \\ -3\end{array}\right]$
✓

Answer

Correct option: B.
$\left[\begin{array}{c}-1 \\ 2 \\ 3\end{array}\right]$
(B) $A X=B$
$\left[\begin{array}{ccc}1 & -1 & 2 \\ 2 & 0 & 1 \\ 3 & 2 & 1\end{array}\right]\left[\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right]=\left[\begin{array}{l}3 \\ 1 \\ 4\end{array}\right]$
Applying $R _1 \rightarrow 2 R _1+ R _3$,
$\left[\begin{array}{lll}5 & 0 & 5 \\ 2 & 0 & 1 \\ 3 & 2 & 1\end{array}\right]\left[\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right]=\left[\begin{array}{c}10 \\ 1 \\ 4\end{array}\right]$
Applying $R _1 \rightarrow R _1-5 R _2$,
$\left[\begin{array}{ccc}-5 & 0 & 0 \\ 2 & 0 & 1 \\ 3 & 2 & 1\end{array}\right]\left[\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right]-\left[\begin{array}{l}5 \\ 1 \\ 4\end{array}\right]$
$\begin{array}{ll}\therefore & -5 x_1=5 \Rightarrow x_1=-1 \\ & 2 x_1+x_3=1 \Rightarrow x_3=3\end{array}$
$3 x_1+2 x_2+x_3=4 \Rightarrow x_2=2$
$\therefore \quad X =\left[\begin{array}{c}-1 \\ 2 \\ 3\end{array}\right]$

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