Question

Solve the following system of linear equations by using the inverse matrix
method: (10)
x + y + z = 4
-2x - y + 3z = 1
y + 5z = 9

Answer 1

Given system can be written in matrix form as $AX = b$

where $A = \begin{bmatrix} 1 & 1 & 1\\ -2 & -1 & 3 \\ 0 & 1 & 5 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, b = \begin{bmatrix} 4 \\ 1 \\ 9 \end{bmatrix}$.

Now, By expending with the first row $|A| = 1(-5-3)-1(-10)+1(-2) = -8+10-2 = 0$

So, $\rho(A)<3.$

Now, $\begin{vmatrix} 1 & 1 \\ -2 & -1 \end{vmatrix} = -1+2 = 1 \neq 0$ .

So, $\rho(A) = 2$ and number of variable to be let $= 3-\rho(A) = 1.$

We let $z = k.$

So, from given first two equations:

$x+y=4-k \\ -2x-y = 1-3k$

Which can be written as $\begin{bmatrix} 1 & 1\\ -2 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 4-k\\ 1-3k \end{bmatrix}$

i.e. $A_1 X_1 = b_1$ and $|A_1| = 1 \neq 0$ .

So, Inverse method can apply in $A_1 X_1 = b_1$,

$\implies X_1 = A_1^{-1} b_1$

$\implies \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -1 & -1\\ 2 & 1 \end{bmatrix} \begin{bmatrix} 4-k\\ 1-3k \end{bmatrix}$ $= \begin{bmatrix} 4k-5\\ -5k+9 \end{bmatrix}$.

So, Solution of given system of equation is

$x =4k-5, y = -5k+9, z= k$ where $k$ is any real number.