Exercise 2.6 — Question 5

This exercise covers solving a 3×3 non-homogeneous system $AX=B$ using the matrix inversion method and Cramer's Rule.

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Key Concepts

Matrix Inversion Method

For the system $AX=B$, if $|A|\ne 0$ (i.e., $A$ is non-singular), the unique solution is:

$X=A^{-1}B$

Steps:

1. Write the system in matrix form $AX=B$.
2. Compute $|A|$ using cofactor expansion.
3. Find the cofactor matrix, then the adjoint: $\text{adj}(A)=(\text{cofactor matrix}{)}^T$.
4. Compute $A^{-1}=\frac{1}{|A|}\text{adj}(A)$.
5. Multiply: $X=A^{-1}B$.

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Cramer's Rule

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

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

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

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Evaluating a 3×3 Determinant by Cofactor Expansion

Expanding along Row 1:

$|A|=a_{11}{C}_{11}+a_{12}{C}_{12}+a_{13}{C}_{13}$

where the cofactor $C_{ij}=(-1{)}^{i+j}{M}_{ij}$ and $M_{ij}$ is the minor (determinant of the 2×2 submatrix obtained by deleting row $i$ and column $j$).

The sign pattern for a 3×3 matrix is: $\left(\begin{array}{ccc}+& -& +\\ -& +& -\\ +& -& +\end{array}\right)$

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Worked Example

Solve the system using Cramer's Rule:

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

Step 1 — Coefficient matrix and $|A|$:

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

Expanding along Row 1:

$|A|=1\cdot \left|\begin{array}{cc}-1& 1\\ 2& -1\end{array}\right|-1\cdot \left|\begin{array}{cc}2& 1\\ 1& -1\end{array}\right|+1\cdot \left|\begin{array}{cc}2& -1\\ 1& 2\end{array}\right|$

$=1(1-2)-1(-2-1)+1(4+1)=-1+3+5=7$

Step 2 — Find $|A_x|$ (replace column 1 with $B=[6,3,2{]}^T$):

$A_x=\left(\begin{array}{ccc}6& 1& 1\\ 3& -1& 1\\ 2& 2& -1\end{array}\right),|A_x|=6(1-2)-1(-3-2)+1(6+2)=-6+5+8=7$

Step 3 — Find $|A_y|$ (replace column 2 with $B$):

$A_y=\left(\begin{array}{ccc}1& 6& 1\\ 2& 3& 1\\ 1& 2& -1\end{array}\right),|A_y|=1(-3-2)-6(-2-1)+1(4-3)=-5+18+1=14$

Step 4 — Find $|A_z|$ (replace column 3 with $B$):

$A_z=\left(\begin{array}{ccc}1& 1& 6\\ 2& -1& 3\\ 1& 2& 2\end{array}\right),|A_z|=1(-2-6)-1(4-3)+6(4+1)=-8-1+30=21$

Step 5 — Solution:

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