Solution:

In a Right Triangle one angle is of $90^{\circ}$ and the other two angles are acute angles. Let the two acute angles be of $x$ and $y$ degrees.

According to the given conditions:

One acute angle $x$ is $12^{\circ}$ more than twice the other acute angle $y$.

$2x+12=y$

$2x-y=-12 \quad\quad\quad\quad\quad...~(1)$

In a Right Triangle the sum of the two acute angles is $90^{\circ}$.

$x+y=90 \quad\quad\quad\quad\quad...~(2)$

Method 1: Matrix Inversion Method

Write the linear equations $(1)$ and $(2)$ in matrix form.

$AX = B$

$\left[ \begin{matrix} 2 & -1 \\ 1 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = $\left[ \begin{matrix} -12 \\ 90 \\ \end{matrix} \right]$

According to matrix inversion method, solution $X$ can be found by the following formula:

$X=A^{-1}B$

$\left| \text{A} \right|=\left| \begin{matrix} 2 & -1 \\ 1 & 1 \\ \end{matrix} \right|=2\times 1-\left( -1 \right)\times 1=2+1=3~\ne ~0$

The coefficient matrix $A$ is non-singular therefore $A^{-1}$ exists.

$\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = ${{\text{A}}^{-1}}\left[ \begin{matrix} -12 \\ 90 \\ \end{matrix} \right]$

= $\frac{1}{\left| \text{A} \right|}~Adj~\text{A}\left[ \begin{matrix} -12 \\ 90 \\ \end{matrix} \right]$

= $\frac{1}{3}\left[ \begin{matrix} 1 & 1 \\ -1 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} -12 \\ 90 \\ \end{matrix} \right]$

= $\frac{1}{3}\left[ \begin{matrix} 1~\times \left( -12 \right)+1\times 90 \\ -1~\times \left( -12 \right)+2\times 90 \\ \end{matrix} \right]$

= $\frac{1}{3}\left[ \begin{matrix} -12+90 \\ 12+180 \\ \end{matrix} \right]$

= $\frac{1}{3}\left[ \begin{matrix} 78 \\ 192 \\ \end{matrix} \right]$

$\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]= \left[ \begin{matrix} 26 \\ 64 \\ \end{matrix} \right]$

Therefore $x=26$ and $y=64$

Hence the acute angles of the Right Triangle are $26^{\circ}$ and $64^{\circ}$

Method 2: Cramer’s Rule

Write the linear equations $(1)$ and $(2)$ in matrix form.

$AX = B$

$\left[ \begin{matrix} 2 & -1 \\ 1 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = $\left[ \begin{matrix} -12 \\ 90 \\ \end{matrix} \right]$

According to Cramer's Rule:

$x=\frac{\left| {{A}_{x}} \right|}{\left| A \right|}$

$y=\frac{\left| {{A}_{y}} \right|}{\left| A \right|}$

$\left| \text{A} \right|=\left| \begin{matrix} 2 & -1 \\ 1 & 1 \\ \end{matrix} \right|=2\times 1-\left( -1 \right)\times 1=2+1=3~\ne ~0$

The coefficient matrix $A$ is non-singular therefore solution exists.

For ${{A}_{x}}$, the $x$ column is replaced with the constant column $B$

${{A}_{x}}=\left[ \begin{matrix} -12 & -1 \\ 90 & 1 \\ \end{matrix} \right]$

$\left| {{A}_{x}} \right|=\left| \begin{matrix} -12 & -1 \\ 90 & 1 \\ \end{matrix} \right|=\left( -12 \right)\times 1-\left( -1 \right)\times 90=-12+90=78$

$x=\frac{\left| {{A}_{x}} \right|}{\left| A \right|}=\frac{78}{3}=26$

For ${{A}_{y}}$, the $y$ column is replaced with the constant column $B$

${{A}_{y}}=\left[ \begin{matrix} 2 & -12 \\ 1 & 90 \\ \end{matrix} \right]$

$\left| {{A}_{y}} \right|=\left| \begin{matrix} 2 & -12 \\ 1 & 90 \\ \end{matrix} \right|=90\times 2-\left( -12 \right)~\times 1=180+12=192$

$y=\frac{\left| {{A}_{y}} \right|}{\left| A \right|}=\frac{192}{3}=64$

Hence the acute angles of the Right Triangle are $26^{\circ}$ and $64^{\circ}$