Answer to Question #110826 in Linear Algebra for Hetisani Sewela

Question

Is the matrix
A =


1 1 1
2 3 0
3 8 3

 .
invertible? (5)
4.2 Solve the following system by Gauss-Jordan elimination
x + y + z = 2
2x + 3y + 2z = 5
3x + 8y + z = 11
(10)
4.3 Now solve the system in 4.2 by applying Cramer’s Rule.

Answer 1

It is necessary to determine whether the matrix

$A=\begin{pmatrix} 1 & 1&1 \\ 2 & 3&0 \\ 3 & 8& 3 \end{pmatrix}$

is invertible.

To calculate the inverse matrix, we write down matrix A with additional identity matrix:

$\left(\begin{matrix} 1 & 1 & 1 \vert& 1 & 0 & 0 \\ 2 & 3 & 0\vert & 0 & 1 & 0 \\ 3 & 8 & 3\vert & 0 & 0 & 1 \end{matrix}\right)$

To find the inverse matrix, we will use elementary transformations over the rows of the matrix, turn the left part of the resulting matrix into a single one.

Subtract the 1st row, multiplied by 2; subtract the 1st row multiplied by 3 from the 3rd row

$\left(\begin{matrix} 1 & 1 & 1& \vert& 1 & 0 & 0 \\ 0 & 1& -2&\vert & -2 & 1 & 0 \\ 0 &5 & 0&\vert & -3 & 0 & 1 \end{matrix}\right)$

subtract the second row multiplied by 1 from the 1st row; subtract the 2nd row multiplied by 5 from the 3rd row

$\left(\begin{matrix} 1 & 0 & 3& \vert& 3 & -1 & 0 \\ 0 & 1& -2&\vert & -2 & 1 & 0 \\ 0 &0 & 10&\vert & 7 & -5 & 1 \end{matrix}\right)$

The 3rd line is divisible by 10

$\left(\begin{matrix} 1 & 0 & 3& \vert& 3 & -1 & 0 \\ 0 & 1& -2&\vert & -2 & 1 & 0 \\ 0 &0 & 1&\vert & 0.7 & -0.5 & 0.1 \end{matrix}\right)$

subtract from the 1st row the 3rd row multiplied by 3; to the 2nd row add the 3rd row multiplied by 2

$\left(\begin{matrix} 1 & 0 & 0& \vert& 0.9 & 0.5 & -0.3 \\ 0 & 1& 0&\vert & -0.6 & 0 & 0.2 \\ 0 &0 & 1&\vert & 0.7 & -0.5 & 0.1 \end{matrix}\right)$

Answer:

$\left(\begin{matrix} 1 & 0 & 0& \vert& 0.9 & 0.5 & -0.3 \\ 0 & 1& 0&\vert & -0.6 & 0 & 0.2 \\ 0 &0 & 1&\vert & 0.7 & -0.5 & 0.1 \end{matrix}\right)$

$A^{-1}=\begin{pmatrix} 0.9 & 0.5&-0.3 \\ -0.6 &0 &0.2 \\ 0.7 & -0.5& 0.1 \end{pmatrix}$

4.2 Solve the following system by eliminating Gauss-Jordan

x + y + z = 2

2x + 3y + 2z = 5

3x + 8y + z = 11 

Expansion of the matrix of systems to the step form is given:

$\begin{pmatrix} 1 & 1&1&\vert2 \\ 2 & 3&2 &\vert5 \\ 3 & 8& 1&\vert11 \end{pmatrix} R_2-2*R_1\mapsto R_2 \begin{pmatrix} 1 & 1&1&\vert2 \\ 0 & 1&0 &\vert1 \\ 3 & 8& 1&\vert11 \end{pmatrix} R_3-3*R_1\mapsto R_3\\ \begin{pmatrix} 1 & 1&1&\vert2 \\ 0 & 1&0 &\vert1 \\ 0 &5& -2&\vert5 \end{pmatrix} R_3-5*R_2\mapsto R_3 \begin{pmatrix} 1 & 1&1&\vert2 \\ 0 & 1&0 &\vert1 \\ 0 & 0& -2&\vert0 \end{pmatrix} \frac{R_3}{-2}\mapsto R_3\\ \begin{pmatrix} 1 & 1&1&\vert2 \\ 0 & 1&0 &\vert1 \\ 0 & 0& 1&\vert0 \end{pmatrix} R_1-1*R_3\mapsto R_1\\ \begin{pmatrix} 1 & 0&0&\vert1 \\ 0 & 1&0 &\vert1 \\ 0 & 0& 1&\vert0 \end{pmatrix}$

$\begin{cases} x=1 \\ y=1 (1)\\ z=0 \end{cases}$

From equation 3 of system (1) we find the variable z

z=0

From equation 2 of system (1) we find the variable y

y=1

From equation 1 of system (1) we find the variable x

x=1

Answer: x=1; y=1; z=0.

The overall solution $X=\begin{pmatrix} 1 \\ 1 \\ 0 \\ \end{pmatrix}$

4.3 Now solve the system in 4.2, applying the Cramer's rule.

$$\begin{cases} x+y+z=2 \\ 2x+3y+2z=5 \\ 3x+8y+z=11 \end{cases}\\ \Delta=\begin{vmatrix} 1 & 1&1 \\ 2 & 3&2\\ 3&8&1 \end{vmatrix}=-2 \\ \Delta_1=\begin{vmatrix} 2 & 1&1 \\ 5 & 3&2\\ 11&8&1 \end{vmatrix}=-2 \\ \Delta_2=\begin{vmatrix} 1 & 2&1 \\ 2 & 5&2\\ 3&11&1 \end{vmatrix}=-2 \\ \Delta_3=\begin{vmatrix} 1 & 1&2 \\ 2 & 3&5\\ 3&8&11 \end{vmatrix}=0 \\ x=\frac{\Delta_1}{\Delta}=\frac{-2}{-2}=1\\ y=\frac{\Delta_2}{\Delta}=\frac{-2}{-2}=1\\ z=\frac{\Delta_3}{\Delta}=\frac{0}{-2}=0\\$$

Answer: x=1; y=1; z=0