2.6 Q-4

Exercise 2.6 — Question 4

This question involves solving a system of 3 non-homogeneous linear equations in 3 unknowns. The standard methods covered are the Matrix Inversion Method and Cramer's Rule.

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Key Concepts

Consistent and Inconsistent Systems

A system $AX=B$ is:

• Consistent if it has at least one solution (unique or infinitely many).
• Inconsistent if it has no solution.

For a $3\times 3$ system:

• If $|A|\ne 0$, the system has a unique solution (consistent).
• If $|A|=0$, the system may be inconsistent or have infinitely many solutions.

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Matrix Inversion Method

Given the system $AX=B$, if $|A|\ne 0$, then:

$X=A^{-1}B$

Steps:

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

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Cramer's Rule

For the system $AX=B$ where $X=\left(\begin{array}{c}x\\ y\\ z\end{array}\right)$ and $|A|\ne 0$:

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

where:

• $A_x$ = matrix $A$ with the first column replaced by $B$
• $A_y$ = matrix $A$ with the second column replaced by $B$
• $A_z$ = matrix $A$ with the third column replaced by $B$

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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:

$|A|=a_1\left|\begin{array}{cc}{b}_2& c_2\\ b_3& c_3\end{array}\right|-b_1\left|\begin{array}{cc}{a}_2& c_2\\ a_3& c_3\end{array}\right|+c_1\left|\begin{array}{cc}{a}_2& b_2\\ a_3& b_3\end{array}\right|$

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Worked Example

Solve the system using Cramer's Rule:

$x+2y+z=8$ $2x+y+z=9$ $x-y+2z=3$

Step 1: Write coefficient matrix and constant matrix:

$A=\left(\begin{array}{ccc}1& 2& 1\\ 2& 1& 1\\ 1& -1& 2\end{array}\right),B=\left(\begin{array}{c}8\\ 9\\ 3\end{array}\right)$

Step 2: Find $|A|$:

$|A|=1(1\cdot 2-1\cdot (-1))-2(2\cdot 2-1\cdot 1)+1(2\cdot (-1)-1\cdot 1)$ $=1(2+1)-2(4-1)+1(-2-1)=3-6-3=-6$

Since $|A|=-6\ne 0$, the system is consistent with a unique solution.

Step 3: Find $|A_x|$, $|A_y|$, $|A_z|$:

$A_x=\left(\begin{array}{ccc}8& 2& 1\\ 9& 1& 1\\ 3& -1& 2\end{array}\right)$

$|A_x|=8(2+1)-2(18-3)+1(-9-3)=24-30-12=-18$

$A_y=\left(\begin{array}{ccc}1& 8& 1\\ 2& 9& 1\\ 1& 3& 2\end{array}\right)$

$|A_y|=1(18-3)-8(4-1)+1(6-9)=15-24-3=-12$

$A_z=\left(\begin{array}{ccc}1& 2& 8\\ 2& 1& 9\\ 1& -1& 3\end{array}\right)$

$|A_z|=1(3+9)-2(6-9)+8(-2-1)=12+6-24=-6$

Step 4: Apply Cramer's Rule:

$x=\frac{-18}{-6}=3,y=\frac{-12}{-6}=2,z=\frac{-6}{-6}=1$

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