Solve the following system of equations by matrix method: (i) 3x + 4y + 2z = 8 2y – 3z = 3 x – 2y + 6z = -2 (ii) 2x + y + z = 2 x + 3y – z = 5 3x + y

Solve the following system of equations by matrix method:
(i) 3x + 4y + 2z = 8
2y – 3z = 3
x – 2y + 6z = -2
(ii) 2x + y + z = 2
x + 3y – z = 5
3x + y – 2z = 6

Solve the following system of equations by matrix method:
(i) 3x + 4y + 2z = 8
2y – 3z = 3
x – 2y + 6z = -2
(ii) 2x + y + z = 2
x + 3y – z = 5
3x + y – 2z = 6

Solution:

(i) Given that $3 x+4 y+2 z=8$
$\begin{array}{l} 2 y-3 z=3 \\ x-2 y+6 z=-2 \end{array}$
We can write the given system in matrix form as:
$\left[\begin{array}{rrr} 2 & 1 & 1 \\ 1 & 3 & -1 \\ 3 & 1 & -2 \end{array}\right]\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{Z} \end{array}\right]=\left[\begin{array}{l} 2 \\ 5 \\ 6 \end{array}\right] \text { Or } \mathrm{A} \mathrm{X}=\mathrm{B}$
$A=\left[\begin{array}{ccc}2 & 1 & 1 \\ 1 & 3 & -1 \\ 3 & 1 & -2\end{array}\right], X=\left[\begin{array}{l}\mathrm{x} \\ \mathrm{y} \\ \mathrm{z}\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{l}2 \\ 5 \\ 6\end{array}\right]$
Now, $|\mathrm{A}|=2^{\left|\begin{array}{ll}3 & -1 \\ 1 & -2\end{array}\right|-1\left|\begin{array}{ll}1 & -1 \\ 3 & -2\end{array}\right|+1\left|\begin{array}{ll}1 & 3 \\ 3 & 1\end{array}\right|}$
$\begin{array}{l} =2(-6+1)-1(-2+3)+1(1-9) \\ =-10-1-8 \\ =-19 \end{array}$
Therefore, the above system of equations has a unique solution, given by
$X=A^{-1} B$
Cofactors of A are
$\begin{array}{l} C_{11}=(-1)^{1+1}-6+1=-5 \\ C_{21}=(-1)^{2+1}(24+4)=-28 \\ C_{31}=(-1)^{3+1}-1-3=-4 \\ C_{12}=(-1)^{1+2}-2+3=-1 \\ C_{22}=(-1)^{2+1}-4-3=-7 \\ C_{32}=(-1)^{3+1}-2-1=3 \\ C_{13}=(-1)^{1+2} 1-9=-8 \\ C_{23}=(-1)^{2+1} 2-3=-1 \\ C_{33}=(-1)^{3+1} 6-1=5 \end{array}$
$\begin{array}{l} \text { Adj } A=\left[\begin{array}{ccc} -5 & -1 & -8 \\ 3 & -7 & 1 \\ -4 & 3 & 5 \end{array}\right]^{T} \\ =\left[\begin{array}{ccc} -5 & 3 & -4 \\ -1 & -7 & 3 \\ -8 & 1 & 5 \end{array}\right] \end{array}$
$\mathrm{A}^{-1}=\stackrel{1}{|\mathrm{~A}|} \operatorname{adj} \mathrm{A}$
Now, $X=A^{-1} B=\frac{1}{-19}\left[\begin{array}{ccc}-5 & 3 & -4 \\ -1 & -7 & 3 \\ -8 & 1 & 5\end{array}\right]\left[\begin{array}{l}2 \\ 5 \\ 6\end{array}\right]$
$\begin{array}{l} X=^{\frac{1}{-19}}\left[\begin{array}{c} -10+15-24 \\ -2-35+18 \\ -16+5+30 \end{array}\right] \\ X=^{\frac{1}{-19}}\left[\begin{array}{c} -19 \\ -19 \\ 19 \end{array}\right] \\ X=\left[\begin{array}{c} 1 \\ 1 \\ -1 \end{array}\right] \end{array}$
As a result, $X=1, Y=1$ and $Z=-1$

(ii) Given that $2 x+y+z=2$
$\begin{array}{l} x+3 y-z=5 \\ 3 x+y-2 z=6 \end{array}$
We can write the given system in matrix form as:
$\left[\begin{array}{ccc} 2 & 1 & 1 \\ 1 & 3 & -1 \\ 3 & 1 & -2 \end{array}\right]\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right]=\left[\begin{array}{l} 2 \\ 5 \\ 6 \end{array}\right]_{\text {Or } A \mathrm{X}=\mathrm{B}}$
$A=\left[\begin{array}{ccc}2 & 1 & 1 \\ 1 & 3 & -1 \\ 3 & 1 & -2\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{l}2 \\ 5 \\ 6\end{array}\right]$
Now, $|A|=2\left|\begin{array}{ll}3 & -1 \\ 1 & -2\end{array}\right|-1\left|\begin{array}{ll}1 & -1 \\ 3 & -2\end{array}\right|+1\left|\begin{array}{ll}1 & 3 \\ 3 & 1\end{array}\right|$
$\begin{array}{l} =2(-6+1)-1(-2+3)+1(1-9) \\ =-10-1-8 \\ =-19 \end{array}$
Therefore, the above system of equations has a unique solution, given by
$X=A^{-1} B$
Cofactors of $A$ are
$\begin{array}{l} C_{11}=(-1)^{1+1}-6+1=-5 \\ C_{21}=(-1)^{2+1}(24+4)=-28 \\ C_{31}=(-1)^{3+1}-1-3=-4 \\ C_{12}=(-1)^{1+2}-2+3=-1 \\ C_{22}=(-1)^{2+1}-4-3=-7 \\ C_{32}=(-1)^{3+1}-2-1=3 \\ C_{13}=(-1)^{1+2} 1-9=-8 \\ C_{23}=(-1)^{2+1} 2-3=-1 \\ C_{33}=(-1)^{3+1} 6-1=5 \end{array}$
$\begin{array}{l} \text { Adj } A=\left[\begin{array}{ccc} -5 & -1 & -8 \\ 3 & -7 & 1 \\ -4 & 3 & 5 \end{array}\right]^{T} \\ =\left[\begin{array}{ccc} -5 & 3 & -4 \\ -1 & -7 & 3 \\ -8 & 1 & 5 \end{array}\right] \\ A^{-1}=\frac{1}{|A|} a d j A \end{array}$
Now, $X=A^{-1} B=\frac{1}{-19}\left[\begin{array}{ccc}-5 & 3 & -4 \\ -1 & -7 & 3 \\ -8 & 1 & 5\end{array}\right]\left[\begin{array}{l}2 \\ 5 \\ 6\end{array}\right]$
$\begin{array}{l} X=\frac{1}{-19}\left[\begin{array}{c} -10+15-24 \\ -2-35+18 \\ -16+5+30 \end{array}\right] \\ X=\frac{1}{-19}\left[\begin{array}{c} -19 \\ -19 \\ 19 \end{array}\right] \end{array}$
$\mathrm{X}=\left[\begin{array}{c} 1 \\ 1 \\ -1 \end{array}\right]$
As a result, $X=1, Y=1$ and $Z=-1$