Question

How do you solve a matrix with variables?

Answer 1

Linear equations in two or three variables can be solved using matrix method or by Row reduction or Gaussian elimination method.

Step 1. Solve the system of equation given below using matrices:

$7x+5y=3$

$3x-2y=22$

Write the matrix on the left as the product of coefficients and variables that is

$\left[\begin{array}{c}7x+5y\\ 3x-2y\end{array}\right]=\left[\begin{array}{c}3\\ 22\end{array}\right]$

Now,

$$\begin{gathered} \left[\begin{array}{cc}7& 5\\ 3& -2\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}3\\ 22\end{array}\right] \\ \uparrow \uparrow \hspace{0.17em}\hspace{0.17em}\uparrow \hspace{0.17em} \end{gathered}$$ …(1)

coefficient variable constant

matrix matrix matrix

Step 2. Find coefficient matrix inverse.

The inverse of $\left[\begin{array}{cc}7& 5\\ 3& -2\end{array}\right]$ is

$\begin{array}{rcl}& =& \frac{1}{7(-2)-3\times 5}\left[\begin{array}{cc}-2& -5\\ -3& 7\end{array}\right]\end{array}$ $\left[\mathbf{\because }{\mathit{A}}^{\mathbf{-}\mathbf{1}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\left|A\right|}\mathbf{.}\mathit{A}\mathit{d}\mathit{j}\mathbf{}\mathit{A}\right]$

$\begin{array}{rcl}& =& \frac{-1}{29}\left[\begin{array}{cc}-2& -5\\ -3& 7\end{array}\right]\\ & =& \left[\begin{array}{cc}\frac{2}{29}& \frac{5}{29}\\ \frac{3}{29}& \frac{-7}{29}\end{array}\right]\end{array}$

Step 3. Multiply equation (1) by inverse matrix, we get

$\begin{array}{rcl}\left[\begin{array}{cc}\frac{2}{29}& \frac{5}{29}\\ \frac{3}{29}& \frac{-7}{29}\end{array}\right]\left[\begin{array}{cc}7& 5\\ 3& -2\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]& =& \left[\begin{array}{cc}\frac{2}{29}& \frac{5}{29}\\ \frac{3}{29}& \frac{-7}{29}\end{array}\right]\left[\begin{array}{c}3\\ 22\end{array}\right]\\ \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]& =& \left[\begin{array}{c}4\\ -5\end{array}\right]\end{array}$

$\begin{array}{rcl}\left[\begin{array}{c}x\\ y\end{array}\right]& =& \left[\begin{array}{c}4\\ -5\end{array}\right]\end{array}$

The solution is $\mathbf{(}\mathbf{4}\mathbf{,}\mathbf{}\mathbf{-}\mathbf{5}\mathbf{)}$.

Hence this way we can solve a matrix with variables.