Question

Solve the system of linear equations by Matrix Method: \(2x-3y+5z=11\), \(3x+2y-4z=-5\), \(x+y-2z=-3\).

Hint:

Write as AX=B, compute det(A), then use Cramer's rule or A⁻¹B.

Answer 1

Step 1: Build $A^{-1}$ explicitly using the adjoint method, instead of Cramer's rule column-swapping:
With $A=\begin{bmatrix}2&-3&5\\3&2&-4\\1&1&-2\end{bmatrix}$, first confirm $\det(A)=-1$ (same computation as before).

Step 2: Find every cofactor of $A$ to build the adjoint:
$C_{11}=2(-2)-(-4)(1)=0$, $C_{12}=-[3(-2)-(-4)(1)]=2$, $C_{13}=3(1)-2(1)=1$, $C_{21}=-[(-3)(-2)-5(1)]=-1$, $C_{22}=2(-2)-5(1)=-9$, $C_{23}=-[2(1)-(-3)(1)]=-5$, $C_{31}=(-3)(-4)-5(2)=2$, $C_{32}=-[2(-4)-5(3)]=23$, $C_{33}=2(2)-(-3)(3)=13$.

Step 3: Transpose the cofactor matrix to get the adjoint, then divide by the determinant:
$\text{adj}(A)=\begin{bmatrix}0&-1&2\\2&-9&23\\1&-5&13\end{bmatrix}$, so $A^{-1}=\dfrac{1}{-1}\begin{bmatrix}0&-1&2\\2&-9&23\\1&-5&13\end{bmatrix}=\begin{bmatrix}0&1&-2\\-2&9&-23\\-1&5&-13\end{bmatrix}$.

Step 4: Multiply $A^{-1}B$ to get $X$:
$x=0(11)+1(-5)+(-2)(-3)=-5+6=1$. $y=-2(11)+9(-5)+(-23)(-3)=-22-45+69=2$. $z=-1(11)+5(-5)+(-13)(-3)=-11-25+39=3$.

Final Answer:
$x=1$, $y=2$, $z=3$, matching Cramer's rule exactly. \[ \boxed{x=1,\ y=2,\ z=3} \]