# Using matrix method, solve the system of equations

Question:

Using matrix method, solve the system of equations

3+ 2– 2= 3, + 2+ 3= 6, 2– = 2.

Solution:

Given system of equations are:

3+ 2– 2= 3

+ 2+ 3= 6 and

2– = 2

Or,

AX = B

So, $\left[\begin{array}{ccc}3 & 2 & -2 \\ 1 & 2 & 3 \\ 2 & -1 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}3 \\ 6 \\ 2\end{array}\right]$

Hence, $\quad X=A^{-1} B$

Now, for $\mathrm{A}^{-1}$ the co-factors are

$A_{11}=5, A_{12}=5, A_{13}=-5$

$A_{21}=0, A_{22}=7, A_{23}=7$

$A_{31}=10, A_{32}=+11$ and $A_{33}=4$

So, $\operatorname{adj} A=\left[\begin{array}{ccc}5 & 5 & -5 \\ 0 & 7 & 7 \\ 10 & -11 & 4\end{array}\right]^{T}=\left[\begin{array}{ccc}5 & 0 & 10 \\ 5 & 7 & -11 \\ -5 & 7 & 4\end{array}\right]$

$|A|=3(5)+2(5)+(-2)(-5)=35$

Thus,  $A^{-1}=\frac{\operatorname{adj} A}{|A|}=\frac{1}{35}\left[\begin{array}{ccc}5 & 0 & 10 \\ 5 & 7 & -11 \\ -5 & 7 & 4\end{array}\right]$

Now, $X=A^{-1} B$

$\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{35}\left[\begin{array}{ccc}5 & 0 & 10 \\ 5 & 7 & -11 \\ -5 & 7 & 4\end{array}\right]\left[\begin{array}{l}3 \\ 6 \\ 2\end{array}\right]=\frac{1}{35}\left[\begin{array}{c}15+20 \\ 15+42-22 \\ -15+42+8\end{array}\right]=\frac{1}{35}\left[\begin{array}{l}35 \\ 35 \\ 35\end{array}\right]=\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$

Therefore, $x=1, y=1$ and $z=1$

So,

$\left[\begin{array}{ccc}3 & 2 & -2 \\ 1 & 2 & 3 \\ 2 & -1 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}3 \\ 6 \\ 2\end{array}\right]$

Hence, $\quad X=A^{-1} B$

Now, for $\mathrm{A}^{-1}$ the co-factors are

$A_{11}=5, A_{12}=5, A_{13}=-5$

$A_{21}=0, A_{22}=7, A_{23}=7$

$A_{31}=10, A_{32}=411$ and $A_{33}=4$

So, $\operatorname{adj} A=\left[\begin{array}{ccc}5 & 5 & -5 \\ 0 & 7 & 7 \\ 10 & -11 & 4\end{array}\right]^{T}=\left[\begin{array}{ccc}5 & 0 & 10 \\ 5 & 7 & -11 \\ -5 & 7 & 4\end{array}\right]$

$|A|=3(5)+2(5)+(-2)(-5)=35$

Thus,

$A^{-1}=\frac{\operatorname{adj} A}{|A|}=\frac{1}{35}\left[\begin{array}{ccc}5 & 0 & 10 \\ 5 & 7 & -11 \\ -5 & 7 & 4\end{array}\right]$

Now, $X=A^{-1} B$

$\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{35}\left[\begin{array}{ccc}5 & 0 & 10 \\ 5 & 7 & -11 \\ -5 & 7 & 4\end{array}\right]\left[\begin{array}{l}3 \\ 6 \\ 2\end{array}\right]=\frac{1}{35}\left[\begin{array}{c}15+20 \\ 15+42-22 \\ -15+42+8\end{array}\right]=\frac{1}{35}\left[\begin{array}{l}35 \\ 35 \\ 35\end{array}\right]=\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$

Therefore, $x=1, y=1$ and $z=1$