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Matrics — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsMatrics2 Marks
MCQ
Let $X =\left[\begin{array}{l}x \\ y \\ z\end{array}\right], D =\left[\begin{array}{c}3 \\ 5 \\ 11\end{array}\right]$ and $A =\left[\begin{array}{rrr}1 & -1 & -2 \\ 2 & 1 & 1 \\ 4 & -1 & -2\end{array}\right]$, if $X=A^{-1} D$, then $X$ is equal to
  • A
    $\left[\begin{array}{l}1 \\ 0 \\ 2\end{array}\right]$
  • ✓
    $\left[\begin{array}{c}\frac{8}{3} \\ \frac{-1}{3} \\ 0\end{array}\right]$
  • C
    $\left[\begin{array}{c}\frac{-8}{3} \\ 1 \\ 0\end{array}\right]$
  • D
    $\left[\begin{array}{c}\frac{8}{3} \\ \frac{1}{3} \\ -1\end{array}\right]$
✓

Answer

Correct option: B.
$\left[\begin{array}{c}\frac{8}{3} \\ \frac{-1}{3} \\ 0\end{array}\right]$
(B) $X = A ^{-1} D$
$\Rightarrow A X=D$
$\left[\begin{array}{ccc}1 & -1 & -2 \\ 2 & 1 & 1 \\ 4 & -1 & -2\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}3 \\ 5 \\ 11\end{array}\right]$
Applying $R _1 \rightarrow R _1+ R _2, R _3 \rightarrow R _3+ R _2$,
$\left[\begin{array}{ccc}3 & 0 & -1 \\ 2 & 1 & 1 \\ 6 & 0 & -1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}8 \\ 5 \\ 16\end{array}\right]$
Applying $R _3 \rightarrow R _3- R _1$,
$\left[\begin{array}{ccc}3 & 0 & -1 \\ 2 & 1 & 1 \\ 3 & 0 & 0\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}8 \\ 5 \\ 8\end{array}\right]$
$\begin{aligned} \therefore \quad & 3 x=8 \Rightarrow x=\frac{8}{3} \\ & 3 x-z=8 \Rightarrow z=0\end{aligned}$
$2 x+y+z=5 \Rightarrow y=\frac{-1}{3}$
$\therefore \quad X=\left[\begin{array}{c}\frac{8}{3} \\ \frac{-1}{3} \\ 0\end{array}\right]$

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