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: Key Formula or Approach:
Write the system as \(AX=B\) where \(A=\begin{bmatrix}2&-3&5\\3&2&-4\\1&1&-2\end{bmatrix}\), \(X=\begin{bmatrix}x\\y\\z\end{bmatrix}\), \(B=\begin{bmatrix}11\\-5\\-3\end{bmatrix}\). If \(\det(A)\ne0\), the unique solution is \(X=A^{-1}B\), which can be computed via Cramer's rule as \(x=\dfrac{\det(A_x)}{\det(A)}\), etc.

Step 2: Computing \(\det(A)\):
\[ \det(A) = 2[2(-2)-(-4)(1)] - (-3)[3(-2)-(-4)(1)] + 5[3(1)-2(1)] \]
\[ = 2(-4+4) + 3(-6+4) + 5(3-2) = 0 - 6 + 5 = -1 \]
Since \(\det(A)=-1\ne0\), a unique solution exists.

Step 3: Finding \(x\) using \(A_x\) (replace column 1 with \(B\)):
\[ \det(A_x) = \begin{vmatrix}11&-3&5\\-5&2&-4\\-3&1&-2\end{vmatrix} = 11(-4+4)+3(10-12)+5(-5+6) = 0-6+5=-1 \]
\[ x = \frac{\det(A_x)}{\det(A)} = \frac{-1}{-1} = 1 \]

Step 4: Finding \(y\) using \(A_y\) (replace column 2 with \(B\)):
\[ \det(A_y) = \begin{vmatrix}2&11&5\\3&-5&-4\\1&-3&-2\end{vmatrix} = 2(10-12)-11(-6+4)+5(-9+5) = -4+22-20=-2 \]
\[ y = \frac{\det(A_y)}{\det(A)} = \frac{-2}{-1} = 2 \]

Step 5: Finding \(z\) using \(A_z\) (replace column 3 with \(B\)):
\[ \det(A_z) = \begin{vmatrix}2&-3&11\\3&2&-5\\1&1&-3\end{vmatrix} = 2(-6+5)+3(-9+5)+11(3-2) = -2-12+11=-3 \]
\[ z = \frac{\det(A_z)}{\det(A)} = \frac{-3}{-1} = 3 \]

Step 6: Verifying:
\(2(1)-3(2)+5(3)=2-6+15=11\) ✓; \(3(1)+2(2)-4(3)=3+4-12=-5\) ✓; \(1+2-2(3)=1+2-6=-3\) ✓.

Final Answer:
\(x=1,\ y=2,\ z=3\). \[ \boxed{x=1,\ y=2,\ z=3} \]