Exercise 2.6 — Question 7

This question involves solving a system of three non-homogeneous linear equations using the matrix inversion method and/or Cramer's Rule.

Key Concepts

Matrix Inversion Method

For a system $AX=B$ where $A$ is a $3\times 3$ coefficient matrix:

$X=A^{-1}B$

Steps:

1. Write the system in matrix form $AX=B$.
2. Find $|A|$ (the determinant of $A$). If $|A|\ne 0$, the system has a unique solution.
3. Find the cofactor matrix of $A$, then the adjoint: $\text{adj}(A)=(\text{cofactor matrix}{)}^T$.
4. Compute $A^{-1}=\frac{1}{|A|}\cdot \text{adj}(A)$.
5. Multiply: $X=A^{-1}B$ to get the solution.

Cramer's Rule

For the system $AX=B$ with $|A|\ne 0$:

$x=\frac{|A_x|}{|A|},y=\frac{|A_y|}{|A|},z=\frac{|A_z|}{|A|}$

where $A_x$, $A_y$, $A_z$ are obtained by replacing the 1st, 2nd, and 3rd columns of $A$ with the column matrix $B$, respectively.

Evaluating a 3×3 Determinant (Cofactor Expansion)

For matrix $A=\left(\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right)$, expanding along Row 1:

$|A|=a_1(b_2{c}_3-b_3{c}_2)-b_1(a_2{c}_3-a_3{c}_2)+c_1(a_2{b}_3-a_3{b}_2)$

Finding the Inverse Using Row Operations

Form the augmented matrix $[A|I]$ and apply elementary row operations until the left side becomes $I$. The right side then becomes $A^{-1}$.

Worked Example

Solve the system using the matrix inversion method:

$x+y+z=6$ $2x-y+z=3$ $x+2y-z=2$

Step 1: Write in matrix form $AX=B$:

$A=\left(\begin{array}{ccc}1& 1& 1\\ 2& -1& 1\\ 1& 2& -1\end{array}\right),X=\left(\begin{array}{c}x\\ y\\ z\end{array}\right),B=\left(\begin{array}{c}6\\ 3\\ 2\end{array}\right)$

Step 2: Find $|A|$:

$|A|=1[(-1)(-1)-(1)(2)]-1[(2)(-1)-(1)(1)]+1[(2)(2)-(-1)(1)]$ $=1[1-2]-1[-2-1]+1[4+1]$ $=-1+3+5=7$

Since $|A|=7\ne 0$, a unique solution exists.

Step 3: Find cofactors and adjoint, then $A^{-1}=\frac{1}{7}\text{adj}(A)$.

Step 4: Compute $X=A^{-1}B$ to obtain $x$, $y$, $z$.