Exercise 2.5 — Q1: Matrix Inversion Method

Solving a 3×3 Non-Homogeneous System Using Matrix Inversion

A system of three linear equations in three unknowns can be written in matrix form as:

$AX=B$

where:

• $A$ is the $3\times 3$ coefficient matrix
• $X=\left(\begin{array}{c}x\\ y\\ z\end{array}\right)$ is the column vector of unknowns
• $B$ is the column vector of constants

Condition for Unique Solution

The system has a unique solution if and only if $|A|\ne 0$ (i.e., $A$ is non-singular).

If $|A|\ne 0$, then $A^{-1}$ exists and the solution is:

$X=A^{-1}B$

Steps to Solve Using Matrix Inversion

Step 1: Write the system in the form $AX=B$.

Step 2: Find $|A|$ (the determinant of $A$) using cofactor expansion.

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

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

Step 3: If $|A|\ne 0$, find the cofactor matrix $C$, then the adjoint:

$\text{adj}(A)=C^T$

Step 4: Compute the inverse:

$A^{-1}=\frac{1}{|A|}\cdot \text{adj}(A)$

Step 5: Multiply to get the solution:

$X=A^{-1}B$

Finding the Inverse Using Row Operations

An alternative method uses the augmented matrix $[A\mid I]$. Apply elementary row operations to reduce $A$ to the identity matrix $I$. The right side transforms into $A^{-1}$:

$[A\mid I]\stackrel{\text{row ops}}{\to }[I\mid A^{-1}]$

Row operations allowed:

• $R_i\leftrightarrow R_j$ (swap two rows)
• $R_i\to k{R}_i$ (multiply a row by a non-zero scalar $k$)
• $R_i\to R_i+k{R}_j$ (add a multiple of one row to another)

Worked Example

Solve the system: $x+y+z=6$ $2x-y+z=3$ $x+2y-z=2$

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

Step 1: Find $|A|$ (expanding along Row 1):

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

$=1(1-2)-1(-2-1)+1(4+1)=-1+3+5=7$

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

Step 2: Find cofactors:

$C_{11}=+\left|\begin{array}{cc}-1& 1\\ 2& -1\end{array}\right|=1-2=-1$ $C_{12}=-\left|\begin{array}{cc}2& 1\\ 1& -1\end{array}\right|=-(-2-1)=3$ $C_{13}=+\left|\begin{array}{cc}2& -1\\ 1& 2\end{array}\right|=4+1=5$ $C_{21}=-\left|\begin{array}{cc}1& 1\\ 2& -1\end{array}\right|=-(-1-2)=3$ $C_{22}=+\left|\begin{array}{cc}1& 1\\ 1& -1\end{array}\right|=-1-1=-2$ $C_{23}=-\left|\begin{array}{cc}1& 1\\ 1& 2\end{array}\right|=-(2-1)=-1$ $C_{31}=+\left|\begin{array}{cc}1& 1\\ -1& 1\end{array}\right|=1+1=2$ $C_{32}=-\left|\begin{array}{cc}1& 1\\ 2& 1\end{array}\right|=-(1-2)=1$ $C_{33}=+\left|\begin{array}{cc}1& 1\\ 2& -1\end{array}\right|=-1-2=-3$

Step 3: Adjoint (transpose of cofactor matrix):

$\text{adj}(A)=\left(\begin{array}{ccc}-1& 3& 2\\ 3& -2& 1\\ 5& -1& -3\end{array}\right)$

Step 4: Inverse:

$A^{-1}=\frac{1}{7}\left(\begin{array}{ccc}-1& 3& 2\\ 3& -2& 1\\ 5& -1& -3\end{array}\right)$

Step 5: Solution:

$X=A^{-1}B=\frac{1}{7}\left(\begin{array}{ccc}-1& 3& 2\\ 3& -2& 1\\ 5& -1& -3\end{array}\right)\left(\begin{array}{c}6\\ 3\\ 2\end{array}\right)$

$=\frac{1}{7}\left(\begin{array}{c}-6+9+4\\ 18-6+2\\ 30-3-6\end{array}\right)=\frac{1}{7}\left(\begin{array}{c}7\\ 14\\ 21\end{array}\right)=\left(\begin{array}{c}1\\ 2\\ 3\end{array}\right)$

$\therefore x=1,y=2,z=3$