**Question 1: **Solve : 2x + 2y = 5 and 12x + 5y = 2

** Solution: **

The augmented matrix for the given system of equations is:

$\begin{bmatrix}

2 &2 :5\\

12 &5 :2

\end{bmatrix}$

Divide the first row by 2.

$\begin{bmatrix}

1 &1 :\frac{5}{2}\\

12 &5 :2

\end{bmatrix}$

Multiply the first row by 12.

$\begin{bmatrix}

12 &12 :30\\

12 &5 :2

\end{bmatrix}$

Subtract the 1st row from the 2nd row and restore the first row.

$\begin{bmatrix}

1 &1 :\frac{5}{2}\\

0 &-7 :-28

\end{bmatrix}$

Divide the second row by -7.

$\begin{bmatrix}

1 &1 :\frac{5}{2}\\

0 &1 :4

\end{bmatrix}$

Consider subtracting second row from the first row.

$\begin{bmatrix}

1 &0 :\frac{-3}{2}\\

0 &1 :4

\end{bmatrix}$

Therefore the given matrix has been reduced to an identity matrix. The solution is $x_{1}$ = $\frac{-3}{2}$ and $x_{2}$ = 4.

**Question 2: **Solve x + 5y - z = -11

3z = 12

2x + 4y - 2z = 8

** Solution: **

The augmented matrix for the given system of equations is:

$\begin{bmatrix}

1 &5&-1 :-11\\

0 &0&3 :12\\

2&4&-2 :8

\end{bmatrix}$

Interchange rows 2 and 3

$\begin{bmatrix}

1 &5 &-1:-11\\

2 &4&-2 :8\\

0 &0& 3: 12

\end{bmatrix}$

Multiply row 3 by $\frac{1}{3}$

$\begin{bmatrix}

1 &5 &-1:-11\\

2&4&-2 :8\\

0 &0& 1: 4

\end{bmatrix}$

Multiply row 2 by $\frac{-1}{2}$

$\begin{bmatrix}

1 &5 &-1:-11\\

-1 &-2&1 :-4\\

0 &0& 1: 4

\end{bmatrix}$

Add row 1 and row 2, place the result in second row.

$\begin{bmatrix}

1 &5 &-1:-11\\

0 &3&0 :-15\\

0 &0& 1: 4

\end{bmatrix}$

Multiply second row by $\frac{1}{3}$

$\begin{bmatrix}

1 &5 &-1:-11\\

0 &1&0 :-5\\

0 &0& 1: 4

\end{bmatrix}$

Multiply row 2 by -5 and add it to row 1. Replace row 1 with the result

$\begin{bmatrix}

1 &0 &-1:14\\

0 &1&0 :-5\\

0 &0& 1: 4

\end{bmatrix}$

Add row 3 to row 1 and row 1 with the result

$\begin{bmatrix}

1 &0 &0:18\\

0 &1&0 :-5\\

0 &0& 0: 4

\end{bmatrix}$

Therefore the given matrix has been reduced to an identity matrix. The solution is x =18, y = -5 and z = 4.