Question
Solve the following system of equations by matrix method:
3x + 7y = 4
x + 2y = -1
3x + 7y = 4
x + 2y = -1
$\begin{bmatrix}3&7\\ 1&2\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}4\\ -1\end{bmatrix}$
Or AX = BWhere $\text{A}=\begin{bmatrix}3&7\\ 1&2\end{bmatrix},\text{x}=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}\text{ and }\text{B}=\begin{bmatrix}4\\ -1\end{bmatrix}$
Now,$\text{|A|}=-1\neq0$
So, the above system has a unique solution, given by X = A-1 B Now, let Cij be the co-factors of aij in A $\text{C}_{11} = 2,\text{C}_{12} = -1$$\text{C}_{21} = -7,\text{C}_{22} = 3$
$\text{Adj A}=\begin{bmatrix}2&-1\\ -7&3\end{bmatrix}^\text{T}=\begin{bmatrix}2&-7\\ -1&3\end{bmatrix}$
$\therefore\text{A}^{-1}=\frac{1}{\text{|A|}}.\text{adj A}=\frac{1}{(-1)}\begin{bmatrix}2&-7\\ -1&3\end{bmatrix}$
Now, X = A-1 B$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\frac{1}{-1}\begin{bmatrix}2&-7\\ -1&3\end{bmatrix}\begin{bmatrix}4\\ -1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\frac{1}{-1}\begin{bmatrix}15\\ -7\end{bmatrix}=\begin{bmatrix}-15\\ 7\end{bmatrix}$
Hence, x = -15 y = 7Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\frac{\text{x}-2}{3}=\frac{\text{y}+3}{-2},\text{z}=5$ and $\frac{\text{x}+1}{1}=\frac{2\text{y}-3}{3}=\frac{\text{z}-5}{2}$