Exercise 2.6 — Question 3

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

Linked SLOs: M-11-A-19, M-11-A-16

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Problem

Solve the following system of linear equations using (i) the Matrix Inversion Method and (ii) Cramer's Rule:

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

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Method 1: Matrix Inversion Method

Write the system as $AX=B$:

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

Step 1: Find $|A|$

Expand along Row 1 using cofactors:

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

$=1(2-(-1))-2(4-1)+1(-2-1)$

$=1(3)-2(3)+1(-3)=3-6-3=-6$

Since $|A|=-6\ne 0$, the matrix is non-singular and $A^{-1}$ exists.

Step 2: Find the Cofactor Matrix

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

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

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

Step 3: Find $A^{-1}$

The adjoint (transpose of cofactor matrix):

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

$A^{-1}=\frac{1}{|A|}\cdot \text{adj}(A)=\frac{1}{-6}\left(\begin{array}{ccc}3& -5& 1\\ -3& 1& 1\\ -3& 3& -3\end{array}\right)$

Step 4: Solve $X=A^{-1}B$

$X=\frac{1}{-6}\left(\begin{array}{ccc}3& -5& 1\\ -3& 1& 1\\ -3& 3& -3\end{array}\right)\left(\begin{array}{c}8\\ 9\\ 3\end{array}\right)$

$=\frac{1}{-6}\left(\begin{array}{c}24-45+3\\ -24+9+3\\ -24+27-9\end{array}\right)=\frac{1}{-6}\left(\begin{array}{c}-18\\ -12\\ -6\end{array}\right)=\left(\begin{array}{c}3\\ 2\\ 1\end{array}\right)$

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

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

$|A|=-6$ (computed above)

$|A_x|$: Replace column 1 with $B$

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

$|A_x|=8(2+1)-2(18-3)+1(-9-3)=8(3)-2(15)+1(-12)=24-30-12=-18$

$|A_y|$: Replace column 2 with $B$

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

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

$|A_z|$: Replace column 3 with $B$

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

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

Solution:

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

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

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Key Concepts