2.3 Q-7

Exercise 2.3 — Question 7

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

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Matrix Inversion Method

For a system $AX=B$ where $A$ is a $3\times 3$ coefficient matrix, $X$ is the column vector of unknowns, and $B$ is the constant column vector:

$X=A^{-1}B\text{provided }|A|\ne 0$

Steps:

1. Write the system in matrix form $AX=B$.
2. Find $|A|$. If $|A|=0$, the matrix is singular and this method cannot be applied.
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.

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Cramer's Rule

For the system $AX=B$ with $|A|\ne 0$, the unique solution is:

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

where:

• $A_x$ = matrix $A$ with the 1st column replaced by $B$
• $A_y$ = matrix $A$ with the 2nd column replaced by $B$
• $A_z$ = matrix $A$ with the 3rd column replaced by $B$

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Worked Example (Q7)

Solve the following system using both methods:

$x+2y+z=8$ $2x-y+z=3$ $x+y-z=1$

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

$A=\left(\begin{array}{ccc}1& 2& 1\\ 2& -1& 1\\ 1& 1& -1\end{array}\right),X=\left(\begin{array}{c}x\\ y\\ z\end{array}\right),B=\left(\begin{array}{c}8\\ 3\\ 1\end{array}\right)$

Step 2: Find $|A|$

$|A|=1[(-1)(-1)-(1)(1)]-2[(2)(-1)-(1)(1)]+1[(2)(1)-(-1)(1)]$ $=1(1-1)-2(-2-1)+1(2+1)$ $=0+6+3=9\ne 0$

Since $|A|\ne 0$, the system has a unique solution.

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

$A_x=\left(\begin{array}{ccc}8& 2& 1\\ 3& -1& 1\\ 1& 1& -1\end{array}\right)$

$|A_x|=8(1-1)-2(-3-1)+1(3+1)=0+8+4=12$

$A_y=\left(\begin{array}{ccc}1& 8& 1\\ 2& 3& 1\\ 1& 1& -1\end{array}\right)$

$|A_y|=1(-3-1)-8(-2-1)+1(2-3)=-4+24-1=19$

$A_z=\left(\begin{array}{ccc}1& 2& 8\\ 2& -1& 3\\ 1& 1& 1\end{array}\right)$

$|A_z|=1(-1-3)-2(2-3)+8(2+1)=-4+2+24=22$

Step 4: Apply Cramer's Rule

$x=\frac{12}{9}=\frac{4}{3},y=\frac{19}{9},z=\frac{22}{9}$

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