Q.Solve the following equations by using matrices :
$x - y + 2z = 7$
$3x + 4y - 5z = -5$
$2x - y + 3z = 12$

Set $AX=B$, find $A^{-1}=\frac{1}{4}\,\text{adj}\,A$, then $X=A^{-1}B=(2,1,3)^{T}$.

Write the equations in matrix form $AX=B$:

$A=\begin{pmatrix} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{pmatrix},\quad X=\begin{pmatrix} x \\ y \\ z \end{pmatrix},\quad B=\begin{pmatrix} 7 \\ -5 \\ 12 \end{pmatrix}.$

Step 1 — determinant of $A$:

$|A|=1(4\cdot3-(-5)(-1))-(-1)(3\cdot3-(-5)\cdot2)+2(3\cdot(-1)-4\cdot2)$

$=1(12-5)+1(9+10)+2(-3-8)=7+19-22=4 eq0,$

so $A^{-1}$ exists and the system has a unique solution.

Step 2 — cofactors:

$C_{11}=7,\ C_{12}=-19,\ C_{13}=-11,$

$C_{21}=1,\ C_{22}=-1,\ C_{23}=-1,$

$C_{31}=-3,\ C_{32}=11,\ C_{33}=7.$

Step 3 — adjoint (transpose of cofactor matrix):

$\text{adj}\,A=\begin{pmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{pmatrix}.$

Step 4 — inverse: $A^{-1}=\dfrac{1}{|A|}\text{adj}\,A=\dfrac14\begin{pmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{pmatrix}.$

Step 5 — solve $X=A^{-1}B$: …