Question

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

Hint:

Form \(AX=B\), compute \(A^{-1}\) via the adjugate, and evaluate \(X=A^{-1}B\).

Answer 1

Step 1: Understanding the Concept:
Write the system as \(AX=B\) and solve \(X=A^{-1}B\), where \(A\) is the coefficient matrix, \(X=(x,y,z)^T\), and \(B\) is the constants column.

Step 2: Setting up A, X, B:
\(A=\begin{bmatrix}2&3&3\\1&-2&1\\3&-1&-2\end{bmatrix}\), \(X=\begin{bmatrix}x\\y\\z\end{bmatrix}\), \(B=\begin{bmatrix}5\\-4\\3\end{bmatrix}\).

Step 3: Computing det(A):
Expanding along row 1: \(\det A=2\big((-2)(-2)-1(-1)\big)-3\big(1(-2)-1(3)\big)+3\big(1(-1)-(-2)(3)\big)\)\[=2(4+1)-3(-2-3)+3(-1+6)=10+15+15=40.\] (nonzero, so \(A^{-1}\) exists and the system has a unique solution)

Step 4: Computing the cofactors and A inverse:
Working out all cofactors gives \(\text{adj}(A)=\begin{bmatrix}5&3&9\\5&-13&1\\5&11&-7\end{bmatrix}\), so \(A^{-1}=\dfrac{1}{40}\begin{bmatrix}5&3&9\\5&-13&1\\5&11&-7\end{bmatrix}\).

Step 5: Computing X = A^{-1}B:
\(A^{-1}B=\dfrac{1}{40}\begin{bmatrix}5(5)+3(-4)+9(3)\\5(5)+(-13)(-4)+1(3)\\5(5)+11(-4)+(-7)(3)\end{bmatrix}=\dfrac{1}{40}\begin{bmatrix}25-12+27\\25+52+3\\25-44-21\end{bmatrix}=\dfrac{1}{40}\begin{bmatrix}40\\80\\-40\end{bmatrix}\).

Final Answer:
\(X=\begin{bmatrix}1\\2\\-1\end{bmatrix}\), so \(\boxed{x=1,\ y=2,\ z=-1}\).