Exercise 2.6 — Question 1

Topic: Solving 3×3 Non-Homogeneous Systems Using Cramer's Rule

This question involves solving a system of three linear equations in three unknowns using Cramer's Rule.

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Key Concepts

Cramer's Rule

For a system $AX=B$ where $A$ is a $3\times 3$ coefficient matrix and $|A|\ne 0$, the unique solution is:

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

where:

• $|A|$ = determinant of the coefficient matrix
• $|A_x|$ = determinant obtained by replacing the 1st column of $A$ with $B$
• $|A_y|$ = determinant obtained by replacing the 2nd column of $A$ with $B$
• $|A_z|$ = determinant obtained by replacing the 3rd column of $A$ with $B$

Evaluating a 3×3 Determinant Using 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:

$x+2y+3z=6$ $2x-y+z=1$ $x+y-z=-1$

Step 1: Write the coefficient matrix and constant vector

$A=\left(\begin{array}{ccc}1& 2& 3\\ 2& -1& 1\\ 1& 1& -1\end{array}\right),B=\left(\begin{array}{c}6\\ 1\\ -1\end{array}\right)$

Step 2: Evaluate $|A|$ (cofactor expansion along Row 1)

$|A|=1\left|\begin{array}{cc}-1& 1\\ 1& -1\end{array}\right|-2\left|\begin{array}{cc}2& 1\\ 1& -1\end{array}\right|+3\left|\begin{array}{cc}2& -1\\ 1& 1\end{array}\right|$

$=1(1-1)-2(-2-1)+3(2+1)=0+6+9=15$

Since $|A|=15\ne 0$, a unique solution exists.

Step 3: Evaluate $|A_x|$ (replace column 1 with $B$)

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

$|A_x|=6(1-1)-2(-1+1)+3(1-1)=0-0+0=15$

Step 4: Evaluate $|A_y|$ (replace column 2 with $B$)

$A_y=\left(\begin{array}{ccc}1& 6& 3\\ 2& 1& 1\\ 1& -1& -1\end{array}\right)$

$|A_y|=1(-1+1)-6(-2-1)+3(-2-1)=0+18-9=9$

Step 5: Evaluate $|A_z|$ (replace column 3 with $B$)

$A_z=\left(\begin{array}{ccc}1& 2& 6\\ 2& -1& 1\\ 1& 1& -1\end{array}\right)$

$|A_z|=1(1-1)-2(-2-1)+6(2+1)=0+6+18=24$

Step 6: Apply Cramer's Rule

$x=\frac{|A_x|}{|A|}=\frac{15}{15}=1,y=\frac{|A_y|}{|A|}=\frac{9}{15}=\frac{3}{5},z=\frac{|A_z|}{|A|}=\frac{24}{15}=\frac{8}{5}$

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Key Conditions