Exercise 2.6 — Question 6

Topic Overview

This question involves solving a 3×3 non-homogeneous system of linear equations $AX=B$ using:

1. Matrix Inversion Method
2. Cramer's Rule

It also requires understanding consistent and inconsistent systems and using row operations to find the inverse.

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Key Concepts

Consistent and Inconsistent Systems

A system $AX=B$ is:

• Consistent if $\text{rank}(A)=\text{rank}([A|B])$ — at least one solution exists.
  • Unique solution if rank equals the number of unknowns.
  • Infinitely many solutions if rank is less than the number of unknowns.
• Inconsistent if $\text{rank}(A)\ne \text{rank}([A|B])$ — no solution exists.

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Matrix Inversion Method

For the system $AX=B$:

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

Steps:

1. Write the system in matrix form $AX=B$.
2. Find $|A|$ (determinant of $A$).
3. If $|A|\ne 0$, find $A^{-1}$ using row operations on $[A|I]$.
4. Compute $X=A^{-1}B$.

Finding $A^{-1}$ by Row Operations (Gauss-Jordan):

Form $[A|I]$ and apply elementary row operations until the left side becomes $I$: $[A|I]\stackrel{\text{row operations}}{\to }[I|A^{-1}]$

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Cramer's Rule

For the system: $a_{1}x+b_{1}y+c_{1}z=d_1$ $a_{2}x+b_{2}y+c_{2}z=d_2$ $a_{3}x+b_{3}y+c_{3}z=d_3$

Let $D=|A|$. Then:

$x=\frac{D_x}{D},y=\frac{D_y}{D},z=\frac{D_z}{D}$

where:

• $D_x$ = determinant with the $x$-column replaced by $[d_1,d_2,d_3{]}^T$
• $D_y$ = determinant with the $y$-column replaced by $[d_1,d_2,d_3{]}^T$
• $D_z$ = determinant with the $z$-column replaced by $[d_1,d_2,d_3{]}^T$

Condition: Cramer's Rule applies only when $D=|A|\ne 0$.

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Evaluating a 3×3 Determinant by 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:

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

The cofactor 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 both methods: $x+y+z=6$ $2x-y+z=3$ $x+2y-z=2$

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

Step 1: Find $|A|$

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

Since $|A|=7\ne 0$, the system is consistent with a unique solution.

Step 2: Cramer's Rule

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

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

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

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

Solution: $x=1,y=2,z=3$

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