2.6 Q-2

Exercise 2.6 — Question 2

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

1. Matrix Inversion Method
2. 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: $|A|\ne 0$ (i.e., $A$ must be non-singular/invertible)

Steps:

1. Write the system in matrix form $AX=B$
2. Compute $|A|$ (determinant of $A$)
3. Find $\text{adj}(A)$ (adjoint = transpose of cofactor matrix)
4. Compute $A^{-1}=\frac{1}{|A|}\cdot \text{adj}(A)$
5. Solution: $X=A^{-1}B$

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Cramer's Rule

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

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

Where:

• $A_x$ = 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

Solve the system using both methods: $x+2y+z=8$ $2x-y+z=3$ $x+y-2z=-3$

Step 1 — Matrix Form

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

Step 2 — Compute $|A|$

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

$=1(2-1)-2(-4-1)+1(2+1)=1+10+3=14$

Since $|A|=14\ne 0$, both methods apply.

Method 1 — Matrix Inversion

Cofactor matrix $C$:

$C_{11}=+\left|\begin{array}{cc}-1& 1\\ 1& -2\end{array}\right|=1,C_{12}=-\left|\begin{array}{cc}2& 1\\ 1& -2\end{array}\right|=5,C_{13}=+\left|\begin{array}{cc}2& -1\\ 1& 1\end{array}\right|=3$

$C_{21}=-\left|\begin{array}{cc}2& 1\\ 1& -2\end{array}\right|=5,C_{22}=+\left|\begin{array}{cc}1& 1\\ 1& -2\end{array}\right|=-3,C_{23}=-\left|\begin{array}{cc}1& 2\\ 1& 1\end{array}\right|=1$

$C_{31}=+\left|\begin{array}{cc}2& 1\\ -1& 1\end{array}\right|=3,C_{32}=-\left|\begin{array}{cc}1& 1\\ 2& 1\end{array}\right|=1,C_{33}=+\left|\begin{array}{cc}1& 2\\ 2& -1\end{array}\right|=-5$

$\text{adj}(A)=C^T=\left(\begin{array}{ccc}1& 5& 3\\ 5& -3& 1\\ 3& 1& -5\end{array}\right)$

$A^{-1}=\frac{1}{14}\left(\begin{array}{ccc}1& 5& 3\\ 5& -3& 1\\ 3& 1& -5\end{array}\right)$

$X=A^{-1}B=\frac{1}{14}\left(\begin{array}{c}1(8)+5(3)+3(-3)\\ 5(8)+(-3)(3)+1(-3)\\ 3(8)+1(3)+(-5)(-3)\end{array}\right)=\frac{1}{14}\left(\begin{array}{c}14\\ 28\\ 42\end{array}\right)=\left(\begin{array}{c}1\\ 2\\ 3\end{array}\right)$

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

Method 2 — Cramer's Rule

$|A_x|=\left|\begin{array}{ccc}8& 2& 1\\ 3& -1& 1\\ -3& 1& -2\end{array}\right|=8(2-1)-2(-6+3)+1(3-3)=8+6+0=14$

$|A_y|=\left|\begin{array}{ccc}1& 8& 1\\ 2& 3& 1\\ 1& -3& -2\end{array}\right|=1(-6+3)-8(-4-1)+1(-6-3)=-3+40-9=28$

$|A_z|=\left|\begin{array}{ccc}1& 2& 8\\ 2& -1& 3\\ 1& 1& -3\end{array}\right|=1(3-3)-2(-6-3)+8(2+1)=0+18+24=42$

$x=\frac{14}{14}=1,y=\frac{28}{14}=2,z=\frac{42}{14}=3$

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

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Summary