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

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsSolution of Simultaneous Linear Equations5 Marks
Question
If $\text{A}=\begin{bmatrix}1&-2&0\\ 2&1&3\\ 0&-2&1\end{bmatrix}$, find $A^{-1}.$ Using $A^{-1},$ solve the system of linear equations:
$x - 2y = 10, 2x + y + 3z = 8, -2y + z = 7$
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Answer

Here, $\text{A}=\begin{bmatrix}1&-2&0\\ 2&1&3\\ 0&-2&1\end{bmatrix}$ $\text{|A|}=1{(1+6)}+2{(2-0)}+0{(-4-0)}$ $=7+4+0$ $=11$
Let $C_{ij}$ be the co-factors of the elements $a_{ij}$ in $A = [a_{ij}]$,
Then, $\text{C}_{11}={(-1)}^{1+1}\begin{vmatrix}1&3\\ -2&1\end{vmatrix}=7,\\ \text{C}_{12}={(-1)}^{1+2}\begin{vmatrix}2&3\\ 0&1\end{vmatrix}=-2,\\ \text{C}_{13}={(-1)}^{1+3}\begin{vmatrix}2&1\\ 0&-2\end{vmatrix}=-4$
$\text{C}_{21}={(-1)}^{2+1}\begin{vmatrix}-2&0\\ -2&1\end{vmatrix}=2,\\ \text{C}_{22}={(-1)}^{2+2}\begin{vmatrix}1&0\\ 0&1\end{vmatrix}=2,\\ \text{C}_{23}={(-1)}^{2+3}\begin{vmatrix}1&-2\\ 0&-2\end{vmatrix}=2$ $\text{C}_{31}={(-1)}^{3+1}\begin{vmatrix}-2&0\\ 1&3\end{vmatrix}=-6,\\ \text{C}_{32}={(-1)}^{3+2}\begin{vmatrix}1&0\\ 2&3\end{vmatrix}=-3,\\ \text{C}_{23}={(-1)}^{3+3}\begin{vmatrix}1&-2\\ 2&1\end{vmatrix}=5$
$\therefore\text{adj A}=\begin{bmatrix}7&-2&-4\\ 2&1&2\\ -6&-3&5\end{bmatrix}^\text{T}$
$=\begin{bmatrix}7&2&-6\\ -2&1&-3\\ -4&2&5\end{bmatrix}$ $\Rightarrow\text{A}^{-1}=\frac{1}{\text{|A|}}\text{adj A}$
$=\frac{1}{11}\begin{bmatrix}7&2&-6\\ -2&1&-3\\ -4&2&5\end{bmatrix}$ or,
$\text{AX = B}$ where, $\text{A}=\begin{bmatrix}1&-2&0\\ 2&1&3\\ 0&-2&1\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}10\\ 8\\ 7\end{bmatrix}$
Now, $\therefore\text{X}=\text{A}^{-1}\text{B}$
$\Rightarrow\text{X}=\frac{1}{11}\begin{bmatrix}7&2&-6\\ -2&1&-3\\ -4&2&5\end{bmatrix}\begin{bmatrix}10\\ 8\\ 7\end{bmatrix}$
$\Rightarrow\text{X}=\frac{1}{11}\begin{bmatrix}70+16-42\ \\ -20+8-21\\ -40+16+35\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{11}\begin{bmatrix}44\\ -33\\ 11\end{bmatrix}$ $\therefore$ $x = 4, y = -3$ and $z = 1$

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