Question

Solve by matrix method the system of equations: \(x-y+2z=7,\ 3x+4y-5z=-5,\ 2x-y+3z=12\).

Hint:

Form A, B; find A⁻¹ via cofactors and adjugate, then X = A⁻¹B (or check via Cramer's rule).

Answer 1

Step 1: Using Cramer's rule instead:
With \(\det A=4\) (as found above), compute \(\det A_x,\det A_y,\det A_z\) by replacing the respective column with \(B=(7,-5,12)\).

Step 2: Computing det(A_x):
\(A_x=\begin{bmatrix}7&-1&2\\-5&4&-5\\12&-1&3\end{bmatrix}\): \(\det A_x=7(12-5)-(-1)(-15+60)+2(5-48)=7(7)+1(45)+2(-43)=49+45-86=8\).

Step 3: Computing det(A_y):
\(A_y=\begin{bmatrix}1&7&2\\3&-5&-5\\2&12&3\end{bmatrix}\): \(\det A_y=1(-15+60)-7(9+10)+2(36+10)=45-133+92=4\).

Step 4: Computing det(A_z):
\(A_z=\begin{bmatrix}1&-1&7\\3&4&-5\\2&-1&12\end{bmatrix}\): \(\det A_z=1(48-5)+1(36+10)+7(-3-8)=43+46-77=12\).

Step 5: Applying Cramer's rule:
\(x=\dfrac{\det A_x}{\det A}=\dfrac84=2\), \(y=\dfrac{\det A_y}{\det A}=\dfrac44=1\), \(z=\dfrac{\det A_z}{\det A}=\dfrac{12}{4}=3\). Check: \(2-1+6=7\)✓, \(6+4-15=-5\)✓, \(4-1+9=12\)✓.

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