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Using matrices, solve the following system of equations
$2x+3y+3z=5;\text{ }x-2y+z=-4;\text{ }3x-y-2z=3.$

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Hint: In this, to find the solution of the system of equation we used the method of reduction which is as follows:
We start with converting the linear system of equation into matrix equation AX=b. Then we perform the suitable row transformation on the matrix A. Using row transformations on A reduce it to an upper triangular matrix or lower triangular matrix. The same operations are performed simultaneously on matrix b. After this step we rewrite the equation again in the form of the system of the linear equation. Now they are in such a form that they can easily be solved by the elimination method.


Complete step by step answer:
The given system of equation is
$\begin{align}
  & 2x+3y+3z=5;\text{ } \\
 & x-2y+z=-4;\text{ } \\
 & 3x-y-2z=3. \\
\end{align}$
The above system of equations can be written in the form AX=b as follows
$\left[ \begin{matrix}
   2 & 3 & 3 \\
   1 & -2 & 1 \\
   3 & -1 & -2 \\
\end{matrix} \right]\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\left[ \begin{matrix}
   5 \\
   -4 \\
   3 \\
\end{matrix} \right]$
Now, we will interchange the row 1 by row 2, we get
$\left[ \begin{matrix}
   1 & -2 & 1 \\
   2 & 3 & 3 \\
   3 & -1 & -2 \\
\end{matrix} \right]\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\left[ \begin{matrix}
   -4 \\
   5 \\
   3 \\
\end{matrix} \right]$
Now, we will multiply row 1 by 2 and then subtract by row 2, we get
  $\left[ \begin{matrix}
   1 & -2 & 1 \\
   0 & 7 & 1 \\
   3 & -1 & -2 \\
\end{matrix} \right]\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\left[ \begin{matrix}
   -4 \\
   13 \\
   3 \\
\end{matrix} \right]$
 Now, we will multiply row 1 by 3 and then subtract by row 3, we get
$\left[ \begin{matrix}
   1 & -2 & 1 \\
   0 & 7 & 1 \\
   0 & 5 & -5 \\
\end{matrix} \right]\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\left[ \begin{matrix}
   -4 \\
   13 \\
   15 \\
\end{matrix} \right]$
Now we will divide row 2 by 7, we get
$\left[ \begin{matrix}
   1 & -2 & 1 \\
   0 & 1 & \dfrac{1}{7} \\
   0 & 5 & -5 \\
\end{matrix} \right]\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\left[ \begin{matrix}
   -4 \\
   \dfrac{13}{7} \\
   15 \\
\end{matrix} \right]$
Now, we will multiply row 2 by 5 and then subtract by row 3, we get
$\left[ \begin{matrix}
   1 & -2 & 1 \\
   0 & 1 & \dfrac{1}{7} \\
   0 & 0 & -\dfrac{40}{7} \\
\end{matrix} \right]\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\left[ \begin{matrix}
   -4 \\
   \dfrac{13}{7} \\
   \dfrac{40}{7} \\
\end{matrix} \right]$
Now A is reduced to an upper triangular matrix, we rewrite the equations in the original form as
$\begin{align}
  & x-2y+z=-4...(1) \\
 & y+\dfrac{1}{7}z=\dfrac{13}{7}\text{ }...\text{(2)} \\
 & -\dfrac{40}{7}z=\dfrac{40}{7}\text{ }...\text{(3)} \\
\end{align}$
From equation (3)
 $z=-1$
Using $z=-1$ in the equation (2) we get
$y+\dfrac{1}{7}\left( -1 \right)=\dfrac{13}{7}$
$y-\dfrac{1}{7}=\dfrac{13}{7}$
By adding both sides by $\dfrac{1}{7}$, we get
$y=\dfrac{13}{7}+\dfrac{1}{7}=\dfrac{13+1}{7}$
$y=\dfrac{14}{7}=2$
$y=2$
Using $y=2$ and $z=-1$ in equation (1), we get
$x-2\left( 2 \right)-1=-4$
$x-5=-4$
By adding both sides by 5, we get
$x=-4+5=1$
$x=1$
Hence the values of x, y and z are 1, 2 and -1 respectively,
The required solution is $x=1$, $y=2$ and $z=-1$

Note:
In this method we cannot use the column transformation. Alternative method, the alternative method is known method of inversion. In this method we use the inverse of the matrix. Consider a system of equation which has matrix form AX=b then by finding inverse of matrix A and pre-multiplying to matrix for AX=b, we get
$\operatorname{X}={{A}^{-1}}b$
Which is the solution of system