2.6 Q-10

Exercise 2.6 — Question 10

Problem

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

$x+2y+z=8$ $2x-y+3z=13$ $3x+y-2z=1$

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Method 1: Matrix Inversion Method

Write the system as $AX=B$:

$A=\left(\begin{array}{ccc}1& 2& 1\\ 2& -1& 3\\ 3& 1& -2\end{array}\right),X=\left(\begin{array}{c}x\\ y\\ z\end{array}\right),B=\left(\begin{array}{c}8\\ 13\\ 1\end{array}\right)$

Step 1: Find $|A|$

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

$=1(2-3)-2(-4-9)+1(2+3)$

$=1(-1)-2(-13)+1(5)=-1+26+5=30$

Since $|A|=30\ne 0$, the inverse $A^{-1}$ exists and the system has a unique solution.

Step 2: Find the Cofactor Matrix

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

$C_{12}=-\left|\begin{array}{cc}2& 3\\ 3& -2\end{array}\right|=-(-4-9)=13$

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

$C_{21}=-\left|\begin{array}{cc}2& 1\\ 1& -2\end{array}\right|=-(-4-1)=5$

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

$C_{23}=-\left|\begin{array}{cc}1& 2\\ 3& 1\end{array}\right|=-(1-6)=5$

$C_{31}=+\left|\begin{array}{cc}2& 1\\ -1& 3\end{array}\right|=+(6+1)=7$

$C_{32}=-\left|\begin{array}{cc}1& 1\\ 2& 3\end{array}\right|=-(3-2)=-1$

$C_{33}=+\left|\begin{array}{cc}1& 2\\ 2& -1\end{array}\right|=+(-1-4)=-5$

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

$\text{adj}(A)=\left(\begin{array}{ccc}{C}_{11}& C_{21}& C_{31}\\ C_{12}& C_{22}& C_{32}\\ C_{13}& C_{23}& C_{33}\end{array}\right)=\left(\begin{array}{ccc}-1& 5& 7\\ 13& -5& -1\\ 5& 5& -5\end{array}\right)$

$A^{-1}=\frac{1}{|A|}\text{adj}(A)=\frac{1}{30}\left(\begin{array}{ccc}-1& 5& 7\\ 13& -5& -1\\ 5& 5& -5\end{array}\right)$

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

$X=\frac{1}{30}\left(\begin{array}{ccc}-1& 5& 7\\ 13& -5& -1\\ 5& 5& -5\end{array}\right)\left(\begin{array}{c}8\\ 13\\ 1\end{array}\right)$

$x=\frac{1}{30}[(-1)(8)+(5)(13)+(7)(1)]=\frac{-8+65+7}{30}=\frac{60}{30}=2$

$y=\frac{1}{30}[(13)(8)+(-5)(13)+(-1)(1)]=\frac{104-65-1}{30}=\frac{30}{30}=3$

$z=\frac{1}{30}[(5)(8)+(5)(13)+(-5)(1)]=\frac{40+65-5}{30}=\frac{90}{30}=3$

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

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Method 2: Cramer's Rule

For the system $AX=B$ with $|A|=30$:

Compute $|A_x|$ (replace column 1 with $B$):

$A_x=\left(\begin{array}{ccc}8& 2& 1\\ 13& -1& 3\\ 1& 1& -2\end{array}\right)$

$|A_x|=8[(-1)(-2)-(3)(1)]-2[(13)(-2)-(3)(1)]+1[(13)(1)-(-1)(1)]$

$=8(2-3)-2(-26-3)+1(13+1)=8(-1)-2(-29)+14=-8+58+14=60$

Compute $|A_y|$ (replace column 2 with $B$):

$A_y=\left(\begin{array}{ccc}1& 8& 1\\ 2& 13& 3\\ 3& 1& -2\end{array}\right)$

$|A_y|=1[(13)(-2)-(3)(1)]-8[(2)(-2)-(3)(3)]+1[(2)(1)-(13)(3)]$

$=1(-26-3)-8(-4-9)+1(2-39)=-29+104-37=90$

Compute $|A_z|$ (replace column 3 with $B$):

$A_z=\left(\begin{array}{ccc}1& 2& 8\\ 2& -1& 13\\ 3& 1& 1\end{array}\right)$

$|A_z|=1[(-1)(1)-(13)(1)]-2[(2)(1)-(13)(3)]+8[(2)(1)-(-1)(3)]$

$=1(-1-13)-2(2-39)+8(2+3)=-14+74+40=90$

Apply Cramer's Rule:

$x=\frac{|A_x|}{|A|}=\frac{60}{30}=2$

$y=\frac{|A_y|}{|A|}=\frac{90}{30}=3$

$z=\frac{|A_z|}{|A|}=\frac{90}{30}=3$

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

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