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

Gujarat BoardEnglish MediumSTD 12 ScienceMathsSolution of Simultaneous Linear Equations4 Marks
Question
Solve the following system of equations by matrix method:
3x + y = 7
5x + 3y = 12
✓

Answer

The given system of equations can be written in matrix form as follows:
$\begin{bmatrix}3&1\\ 5&3\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}7\\ 12\end{bmatrix}$
AX = B
Here,
$\text{A}=\begin{bmatrix}3&1\\ 5&3\end{bmatrix},\text{x}=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}\text{ and }\text{B}=\begin{bmatrix}7\\ 12\end{bmatrix}$
Now,
$\text{|A|}=\begin{vmatrix}3&1\\ 5&3\end{vmatrix}$
$=9-5$
$=4\neq0$
So, the given system has a unique solution given by X = A-1 B.
Let Cij be the co-factors of the elements aij in A = [aij]. then,
$\text{C}_{11}=-{(-1)}^{1+1}{(3)}=3,\text{C}_{12}={(-1)}^{1+2}{(5)}=-5$
$\text{C}_{21}={(-1)}^{2+1}{(1)}=-1,\text{C}_{22}=-{(-1)}^{2+2}{(3)}=3$
$\text{adj A}=\begin{bmatrix}3&-5\\ -1&3\end{bmatrix}^\text{T}$
$=\begin{bmatrix}3&-1\\-5&3\end{bmatrix}$
$\text{A}^{-1}=\frac{1}{|\text{A}|}\text{adj A}$
$=\frac{1}{4}\begin{bmatrix}3&-1\\ -5&3\end{bmatrix}$
X = A-1B
$=\frac{1}{4}\begin{bmatrix}3&-1\\ -5&3\end{bmatrix}\begin{bmatrix}7\\ 12\end{bmatrix}$
$=\frac{1}{4}\begin{bmatrix}21-12\\ -35+36\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}\frac{9}{4}\\ \frac{1}{4}\end{bmatrix}$
$\therefore\ \text{x}=\frac{9}{4}\text{ and }\text{y}=\frac{1}{4}$

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