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Solution of Simultaneous Linear Equations — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsSolution of Simultaneous Linear Equations1 Mark
MCQ
Let $\text{X}=\begin{bmatrix}\text{x}_1\\\text{x}_2\\\text{x}_3\end{bmatrix},\text{A}=\begin{bmatrix}1&-1&2\\2&0&1\\3&2&1\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}3\\1\\4\end{bmatrix}$. If $AX = B,$ then $X$ is equal to:
  • ✓
    $\begin{bmatrix}1\\2\\3\end{bmatrix}$
  • B
    $\begin{bmatrix}-1\\-2\\-3\end{bmatrix}$
  • C
    $\begin{bmatrix}-1\\2\\3\end{bmatrix}$
  • D
    $\begin{bmatrix}0\\2\\1\end{bmatrix}$
✓

Answer

Correct option: A.
$\begin{bmatrix}1\\2\\3\end{bmatrix}$
Here,
$\text{A}=\begin{bmatrix}1&-1&2\\2&0&1\\3&2&1\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}_1\\\text{x}_2\\\text{x}_3\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}3\\1\\4\end{bmatrix}$
$|\text{A}|=1(0-2)+1(2-3)+2(4-0)$
$=-2-1+8$
$=5$
Let $C_{ij}$ be the co$-$factors of the elements $a_{ij}$ in $A = [a_{ij}]$. Then,
$\text{C}_{11}=(-1)^{1+1}\begin{vmatrix}0&1\\2&1\end{vmatrix}=-2,$ $\text{C}_{12}=(-1)^{1+2}\begin{vmatrix}2&1\\3&1\end{vmatrix}=1,$ $\text{C}_{13}=(-1)^{1+3}\begin{vmatrix}2&0\\3&2\end{vmatrix}=4$
$\text{C}_{21}=(-1)^{2+1}\begin{vmatrix}-1&2\\2&1\end{vmatrix}=5,$ $\text{C}_{22}=(-1)^{2+2}\begin{vmatrix}1&2\\3&1\end{vmatrix}=-5,$ $\text{C}_{23}=(-1)^{2+3}\begin{vmatrix}1&-1\\3&2\end{vmatrix}=-5$
$\text{C}_{31}=(-1)^{3+1}\begin{vmatrix}-1&2\\0&1\end{vmatrix}=-1,$ $\text{C}_{32}=(-1)^{3+2}\begin{vmatrix}1&2\\2&1\end{vmatrix}=3,$ $\text{C}_{33}=(-1)^{3+3}\begin{vmatrix}1&-1\\2&0\end{vmatrix}=2$
$\text{adj }\text{A}=\begin{bmatrix}-2&5&-1\\5&-5&-5\\-1&2&3\end{bmatrix}^\text{T}$
$=\begin{bmatrix}-2&5&-1\\1&-5&3\\4&-5&2\end{bmatrix}$
$\Rightarrow\text{A}^{-1}=\frac{1}{|\text{A}|}\text{adj }\text{A}$
$=\frac{1}{5}\begin{bmatrix}-2&5&-1\\1&-5&3\\4&-5&2\end{bmatrix}$
$\therefore\ \text{X}=\text{A}^{-1}\text{B}$
$\Rightarrow\text{X}=\frac{1}{5}\begin{bmatrix}-2&5&-1\\1&-5&3\\4&-5&2\end{bmatrix}\begin{bmatrix}3\\1\\4\end{bmatrix}$
$\Rightarrow\text{X}=\frac{1}{5}\begin{bmatrix}-6+5-4\\3-5+12\\12-5+8\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}_1\\\text{x}_2\\\text{x}_3\end{bmatrix}=\frac{1}{5}\begin{bmatrix}-5\\10\\15\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}_1\\\text{x}_2\\\text{x}_3\end{bmatrix}=\begin{bmatrix}-1\\2\\3\end{bmatrix}$

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