Solution of Simultaneous Linear Equations — Maths STD 12 Science — Question
Gujarat BoardEnglish MediumSTD 12 ScienceMathsSolution of Simultaneous Linear Equations5 Marks
Question
Solve the follwing system of equations by matrix method:
$5x + 2y = 3$
$3x + 2y = 5$
✓
Answer
The above system can be written in matrix form as: $\begin{bmatrix}5&2\\ 3&2\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}3\\ 2\end{bmatrix}$ Or $\text{AX = B}$ Where, $\text{A}=\begin{bmatrix}5&2\\ 3&2\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix},\text{B}=\begin{bmatrix}3\\ 5\end{bmatrix}$Now, $\text{|A|}=10-6=4\neq0$
So, the above system has a unique solution, given by $\text{X}=\text{A}^{-1}\text{B}$ Let $C_{ij}$ be the co factor of $a_{ij}$ in A, then$\text{C}_{11} = 2,\text{C}_{12} = -3$
$\text{C}_{21} = -2,\text{C}_{22} = 5$
Also, $\text{Adj A}=\begin{bmatrix}2&-3\\ -2&5\end{bmatrix}^\text{T}=\begin{bmatrix}2&-2\\ -3&5\end{bmatrix}$ $\therefore\text{A}^{-1}=\frac{1}{|\text{A}|}\text{adj A}=\frac{1}{4}\begin{bmatrix}2&-2\\ -3&5\end{bmatrix}$ Now, $X = A^{-1}B =\frac{1}{4}\begin{bmatrix}2&-2\\ -3&5\end{bmatrix}\begin{bmatrix}3\\ 5\end{bmatrix}$ $=\frac{1}{4}\begin{bmatrix}-4\\ 16\end{bmatrix}$ $\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}-1\\ 4\end{bmatrix}$ Hence, x = -1 y = 4
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