2.4 Q-1

Exercise 2.4 — Q1: Solving 3×3 Non-Homogeneous Systems

This exercise covers solving a system of three linear equations in three unknowns using the matrix inversion method and Cramer's Rule (SLO M-11-A-19).

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System of Linear Equations

A general $3\times 3$ non-homogeneous system is:

$a_{1}x+b_{1}y+c_{1}z=d_1$ $a_{2}x+b_{2}y+c_{2}z=d_2$ $a_{3}x+b_{3}y+c_{3}z=d_3$

In matrix form: $AX=B$, where

$A=\left(\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right),X=\left(\begin{array}{c}x\\ y\\ z\end{array}\right),B=\left(\begin{array}{c}{d}_1\\ d_2\\ d_3\end{array}\right)$

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Method 1: Matrix Inversion Method

Condition: $|A|\ne 0$ (the system has a unique solution).

Steps:

1. Write the system as $AX=B$.
2. Compute $|A|$ (the determinant of $A$).
3. Find $\text{adj}(A)$ (transpose of the cofactor matrix).
4. Compute $A^{-1}=\frac{1}{|A|}\text{adj}(A)$.
5. The solution is $X=A^{-1}B$.

Key formula: $X=A^{-1}B=\frac{1}{|A|}\text{adj}(A)\cdot B$

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Method 2: Cramer's Rule

Condition: $|A|\ne 0$.

Define:

• $A_x$: replace the $x$-column of $A$ with $B$
• $A_y$: replace the $y$-column of $A$ with $B$
• $A_z$: replace the $z$-column of $A$ with $B$

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

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Determinant Properties Used in Exercise 2.4

Several questions in Exercise 2.4 require showing a determinant equals zero without expanding, using properties:

Proof for skew-symmetric (odd order): If $A^T=-A$ and order $n$ is odd: $|A^T|=|-A|=(-1{)}^n|A|=-|A|$ But $|A^T|=|A|$, so $|A|=-|A|$, giving $|A|=0$.

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