Question
Solve the following system of equations by matrix method:
$5x + 7y + 2 = 0$
$5x + 7y + 2 = 0$
$4x + 6y + 3 = 0$
✓
Answer
The given system of equations can be written in matrix from as follws: $\begin{bmatrix}5&7\\ 4&6\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}-2\\ -3\end{bmatrix}$ $\text{AX = B}$Here,
$\text{A}=\begin{bmatrix}5&7\\ 4&6\end{bmatrix},\text{X=}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}$ And$\text{ B}=\begin{bmatrix}-2\\ -3\end{bmatrix}$ Now, $|\text{A}|=\begin{bmatrix}5&7\\ 4&6\end{bmatrix}\\ $ $=30-28$ $=2\neq0$ The given system has a unique solution given by $\text{X}=\text{A}^{-1 }\text{B.}$ Let $C_{ij}$ be the cofactors of the elements $a_{ij}$ in $A = [a_{ij}]$. Then, $\text{C}_{11}=(-1)^{1+1}(6)=6,\text{C}_{12}=(-1)^{1+2}(4)=-4$ $\text{C}_{21}=(-1)^{2+1}(7)=-7,\text{C}_{22}=(-1)^{2+2}(5)$ $\text{adj}\ \text{A}=\begin{bmatrix}6&-4\\ -7&5\end{bmatrix}=\begin{bmatrix}6&-7\\ -4&5\end{bmatrix}$ $\text{A}^{-1}=\frac{1}{\text{|A|}}\text{ adj}\text{ A}$
$\Rightarrow \text{A}^{-1}=\frac{1}{2}\begin{bmatrix}6&-7\\ -4&5\end{bmatrix}$ $\text{X}=\text{A}^{-1}\text{B}$ $=\frac{1}{2}\begin{bmatrix}6&-7\\ -4&5\end{bmatrix}\begin{bmatrix}-2\\-3\end{bmatrix} $ $=\frac{1}{2}\begin{bmatrix}-12+21\\ 8-15\end{bmatrix} $
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}\frac{9}{2}\\ \frac{-7}{2}\end{bmatrix}$ $\therefore\text{X}=\frac{9}{2}$ And $\text{ y }=\frac{-7}{2}$
$\text{A}=\begin{bmatrix}5&7\\ 4&6\end{bmatrix},\text{X=}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}$ And$\text{ B}=\begin{bmatrix}-2\\ -3\end{bmatrix}$ Now, $|\text{A}|=\begin{bmatrix}5&7\\ 4&6\end{bmatrix}\\ $ $=30-28$ $=2\neq0$ The given system has a unique solution given by $\text{X}=\text{A}^{-1 }\text{B.}$ Let $C_{ij}$ be the cofactors of the elements $a_{ij}$ in $A = [a_{ij}]$. Then, $\text{C}_{11}=(-1)^{1+1}(6)=6,\text{C}_{12}=(-1)^{1+2}(4)=-4$ $\text{C}_{21}=(-1)^{2+1}(7)=-7,\text{C}_{22}=(-1)^{2+2}(5)$ $\text{adj}\ \text{A}=\begin{bmatrix}6&-4\\ -7&5\end{bmatrix}=\begin{bmatrix}6&-7\\ -4&5\end{bmatrix}$ $\text{A}^{-1}=\frac{1}{\text{|A|}}\text{ adj}\text{ A}$
$\Rightarrow \text{A}^{-1}=\frac{1}{2}\begin{bmatrix}6&-7\\ -4&5\end{bmatrix}$ $\text{X}=\text{A}^{-1}\text{B}$ $=\frac{1}{2}\begin{bmatrix}6&-7\\ -4&5\end{bmatrix}\begin{bmatrix}-2\\-3\end{bmatrix} $ $=\frac{1}{2}\begin{bmatrix}-12+21\\ 8-15\end{bmatrix} $
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}\frac{9}{2}\\ \frac{-7}{2}\end{bmatrix}$ $\therefore\text{X}=\frac{9}{2}$ And $\text{ y }=\frac{-7}{2}$
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