2.6 Q-8

Exercise 2.6 — Question 8

Topic: Solving a 3×3 Non-Homogeneous System using Matrix Inversion Method and Cramer's Rule

SLO: M-11-A-19 — Solve a system of 3×3 non-homogeneous linear equations by using matrix inversion method and Cramer's Rule.

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Key Concepts

Matrix Inversion Method

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

$A=\left(\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right),X=\left(\begin{array}{c}x\\ y\\ z\end{array}\right),B=\left(\begin{array}{c}{d}_1\\ d_2\\ d_3\end{array}\right)$

Condition: $|A|\ne 0$ (A must be non-singular)

Solution: $X=A^{-1}B$, where $A^{-1}=\frac{1}{|A|}\text{adj}(A)$

Steps:

1. Write the system in matrix form $AX=B$.
2. Compute $|A|$. If $|A|=0$, this method cannot be used.
3. Find the cofactor matrix of $A$, then take its transpose to get $\text{adj}(A)$.
4. Compute $A^{-1}=\frac{1}{|A|}\text{adj}(A)$.
5. Multiply: $X=A^{-1}B$ to find $x$, $y$, $z$.

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Cramer's Rule

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

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

where:

• $A_x$ = $A$ with column 1 replaced by $B$
• $A_y$ = $A$ with column 2 replaced by $B$
• $A_z$ = $A$ with column 3 replaced by $B$

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Worked Example

Solve the system: $x+y+z=6$ $2x-y+z=3$ $x+2y-z=2$

Step 1: Matrix form

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

Step 2: Compute $|A|$

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

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

Step 3 (Cramer's Rule): Compute $|A_x|$, $|A_y|$, $|A_z|$

$A_x=\left(\begin{array}{ccc}6& 1& 1\\ 3& -1& 1\\ 2& 2& -1\end{array}\right)$

$|A_x|=6(1-2)-1(-3-2)+1(6+2)=-6+5+8=7$

$A_y=\left(\begin{array}{ccc}1& 6& 1\\ 2& 3& 1\\ 1& 2& -1\end{array}\right)$

$|A_y|=1(-3-2)-6(-2-1)+1(4-3)=-5+18+1=14$

$A_z=\left(\begin{array}{ccc}1& 1& 6\\ 2& -1& 3\\ 1& 2& 2\end{array}\right)$

$|A_z|=1(-2-6)-1(4-3)+6(4+1)=-8-1+30=21$

Step 4: Solution

$x=\frac{7}{7}=1,y=\frac{14}{7}=2,z=\frac{21}{7}=3$

$\boxed{x=1,y=2,z=3}$

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