2.5 Q-3

Exercise 2.5 — Question 3

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

• The Matrix Inversion Method
• Cramer's Rule
• Row Operations to find the inverse

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Key Concepts

Matrix Form of a Linear System

A system of 3 equations in 3 unknowns can be written as:

$AX=B$

where: $A=\left(\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right),X=\left(\begin{array}{c}x\\ y\\ z\end{array}\right),B=\left(\begin{array}{c}{b}_1\\ b_2\\ b_3\end{array}\right)$

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Method 1: Matrix Inversion Method

Condition: $|A|\ne 0$ (matrix $A$ must be non-singular)

Solution: $X=A^{-1}B$

Steps:

1. Write the system in the form $AX=B$.
2. Find $|A|$ using cofactor expansion.
3. Find $A^{-1}$ using the formula: $A^{-1}=\frac{1}{|A|}\cdot \text{adj}(A)$ where $\text{adj}(A)$ is the transpose of the cofactor matrix.
4. Compute $X=A^{-1}B$.

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Method 2: Finding Inverse via Row Operations (Gauss-Jordan)

Steps:

1. Form the augmented matrix $[A\mid I_3]$.
2. Apply elementary row operations to reduce $A$ to the identity matrix $I_3$.
3. The right side transforms into $A^{-1}$: $[A\mid I]\stackrel{\text{row ops}}{\to }[I\mid A^{-1}]$
4. If a row of zeros appears on the left, $A$ is singular and $A^{-1}$ does not exist.

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Method 3: Cramer's Rule

Condition: $|A|\ne 0$

For the system $AX=B$:

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

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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Evaluating a 3×3 Determinant (Cofactor Expansion)

For matrix $A$, expanding along Row 1:

$|A|=a_{11}{C}_{11}+a_{12}{C}_{12}+a_{13}{C}_{13}$

where the cofactor $C_{ij}=(-1{)}^{i+j}{M}_{ij}$ and $M_{ij}$ is the minor (determinant of the 2×2 submatrix obtained by deleting row $i$ and column $j$).

Sign pattern for cofactors: $\left(\begin{array}{ccc}+& -& +\\ -& +& -\\ +& -& +\end{array}\right)$

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Worked Example

Solve the system using the matrix inversion method: $x+y+z=6$ $2x-y+z=3$ $x+2y-z=2$

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

Step 2: Find $|A|$ (expanding along Row 1): $|A|=1\cdot [(-1)(-1)-(1)(2)]-1\cdot [(2)(-1)-(1)(1)]+1\cdot [(2)(2)-(-1)(1)]$ $=1(1-2)-1(-2-1)+1(4+1)=-1+3+5=7$

Since $|A|=7\ne 0$, the inverse exists.

Step 3: Find cofactors and $\text{adj}(A)$, then $A^{-1}=\frac{1}{7}\text{adj}(A)$.

Step 4: Compute $X=A^{-1}B$ to get $x$, $y$, $z$.