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MATRICES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMATRICES5 Marks
Question
Using matrices, solve the following system of equations:$\text{x + y + z} = 3, \text{x} - 2{\text{y}} + 3{z}=2\ \text{and} \ 2 \text{x - y + z} = 2 $
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Answer

The system of equations can be written as:AX = B, where $\text{A}= \begin{bmatrix} 1 & 1 & 1 \\ 1 & -2 & 3 \\ 2 & -1 & 1 \end{bmatrix} , X = \begin{bmatrix} x\\ y\\ z\end{bmatrix} , B = \begin{bmatrix} 3\\ 2\\ 2\end{bmatrix} $
OR
$X = A^{-1} B$
$|A| = 1 (- 2 + 3) -1 (1 - 6) + 1(-1+4) = 1+5+3 = 9$
$\text{A}_{11}=+\begin{vmatrix}-2&3\\-1 & 1 \end{vmatrix}=-2+3=1$
$\text{A}_{12}=-\begin{vmatrix}1&3\\2 & 1 \end{vmatrix}=-(1-6)=5$
$\text{A}_{13}=+\begin{vmatrix}1&-2\\2 & -1 \end{vmatrix}=-1+4=3$
$\text{A}_{21}=-\begin{vmatrix}1&1\\-1 & 1 \end{vmatrix}=-(1+1)=-2$
$\text{A}_{22}=+\begin{vmatrix}1&1\\2 &1 \end{vmatrix}=1-2=-1$
$\text{A}_{23}=-\begin{vmatrix}1&1\\2 & -1 \end{vmatrix}=-(-1-2)=3$
$\text{A}_{31}=+\begin{vmatrix}1&1\\-2 &3 \end{vmatrix}=3+2=5$
$\text{A}_{32}=-\begin{vmatrix}1&1\\1 & 3 \end{vmatrix}=-(3-1)=-2$
$\text{A}_{33}=+\begin{vmatrix}1&1\\1 & -2 \end{vmatrix}=-2-1=-3$
$Adj. A = \begin{bmatrix} 1 & -2 & 5 \\ 5 & -1 & -2 \\ 3 & 3 & -3 \end{bmatrix} $
$\therefore A^-1 = \frac{1}{9} \begin{bmatrix} 1 & -2 & 5 \\ 5 & -1 & -2 \\ 3 & 3 & -3 \end{bmatrix} $
$\therefore \begin{bmatrix} x \\ y \\z \end{bmatrix} =\frac{1}{9} \begin{bmatrix} 1 &-2 & 5 \\ 5& -1 & -2 \\ 3 & 3 & -3 \end{bmatrix} \begin{bmatrix} 3 \\ 2 \\2 \end{bmatrix} = \frac{1}{9} \begin{bmatrix} 9 \\ 9 \\9 \end{bmatrix}$
$\begin{bmatrix} 1 \\ 1 \\1 \end{bmatrix} \Rightarrow x = y = z = 1$

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