Solve the system of equations by using the inverse of the coefficient matrix of the equivalent matrix equation.

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The solution for the given system of equations x + 8y = -37, 4x + 8y = -52 is $\left[\begin{array}{ccc}&5&\\\\&4&\end{array}\right]$

Given,

System of equations as,

x + 8y = -37

4x + 8y = -52

We have to solve this by using the inverse of coefficient matrix of the equivalent matrix equation.

That is,

$A=\left[\begin{array}{ccc}a&&b\\\\c&&d\end{array}\right]$

$A^{-1} =\frac{1}{ad -bc} \left[\begin{array}{ccc}d&&-b\\\\-c&&a\end{array}\right]$

Now we can solve the equations.

Here we have,

x + 8y = -37

4x + 8y = -52

Now in matrix form,

$\left[\begin{array}{ccc}1&&8\\\\4&&8\end{array}\right]$ $\left[\begin{array}{ccc}&x&\\\\&y&\end{array}\right]$ $=\left[\begin{array}{ccc}&-37&\\\\&-52&\end{array}\right]$

A X B

We know that,

$A^{-1} =\frac{1}{ad -bc} \left[\begin{array}{ccc}d&&-b\\\\-c&&a\end{array}\right]$

Therefore,

$A^{-1} = \frac{1}{(1X8)-(4X8)} \left[\begin{array}{ccc}8&&-8\\\\-4&&1\end{array}\right]$

$=\frac{1}{8-32} \left[\begin{array}{ccc}8&&-8\\\\-4&&1\end{array}\right]$

$=\frac{1}{-24} \left[\begin{array}{ccc}8&&-8\\\\-4&&1\end{array}\right]$

$=\left[\begin{array}{ccc}\frac{-8}{24} &&\frac{8}{24} \\\\\frac{4}{24} &&\frac{-1}{24} \end{array}\right]$

Then,

$\left[\begin{array}{ccc}&x&\\\\&y&\end{array}\right] =$ $=\left[\begin{array}{ccc}\frac{-8}{24} &&\frac{8}{24} \\\\\frac{4}{24} &&\frac{-1}{24} \end{array}\right]$ $\left[\begin{array}{ccc}&\frac{37}{52} &\\\\\end{array}\right]$