Lesson Objective

Dear learner,

by the end of this lesson you will be able to:

• Define singular matrices
• Determine inverse of any given matrix
• Apply principles of matrices to solve problems
• Solve real life problems by using the concept of matrices.

Key Terms

• Singular matrix
• Non singular matrix
• Invertible matrix

Brainstorming Activity

1.Let $A = \begin{pmatrix}
1 & 4 \\
4 & 6
\end{pmatrix}$ be a matrix and find a matrix B suchthat $AB = BA = I_{2}$

Solution:

lets take $I_{2} = \begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}$ , and assume that $B = \begin{pmatrix}
a & b \\
c & d
\end{pmatrix}$. then:

$AB = \begin{pmatrix}
1 & 4 \\
4 & 6
\end{pmatrix} \begin{pmatrix}
a & b \\
c & d
\end{pmatrix}$ = $\begin{pmatrix}
a+4c& b+4d \\
4a+6c & 4b+6d
\end{pmatrix}$ = $\begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix}$

$\implies$ $a+4c = 1$ and $4a+6c = 0$ $\implies$ $ c = 2/5$ and $a =-3/5$

$\implies$ $b+4d = 0$ and $4b +6d = 1$ $\implies$ $d = -1/10$ and $ b = 2/5$

therefore matrix $B = \begin{pmatrix}
-3/5 & 2/5 \\
2/5& -1/10
\end{pmatrix}$

4.1. Inverse of a Square Matrix

Definition: A square matrix A with order n is said to be invertible or non- singular if and only if there is a square matrix B such that$AB =I_{n} =BA$,  where B is an identity matrix having the same order as A.

For a square matrix A

• The inverse of A is denoted by $A^{-1}$. And $A^{-1} \ne 0$
• A matrix that doesn’t have an inverse is singular

Example 1:

1.Find the inverse of a matrix given below

$B = \begin{pmatrix}1&2&3\\4&5&1\\2&1&4\end{pmatrix}$

Solution:

Use Guass-Jordan elimination method to determine $A^{-1}$

Step 1: take BI like augmented matrix:

$BI$ = $\begin{pmatrix}1&2&3\\4&5&1\\2&1&4\end{pmatrix}$ $\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}$

step 2: reduce BI until B is changed in to an identity matrix

$R_{2}\rightarrow(-4)R_{1} + R_{2}$

$R_{3}\rightarrow(-2)R_{1} + R_{3}$ $\begin{pmatrix}1&2&3&1&0&0\\0&-3&-11&-4&1&0\\0&-3&-2&-2&0&1\end{pmatrix}$

$R_{2}\rightarrow(-1/3)R_{2}$ $\begin{pmatrix}1&2&3&1&0&0\\0&1 & 11/3 & 4/3 & -1/3&0\\0&-3&-2&-2&0&1\end{pmatrix}$

$R_{1}\rightarrow(-2)R_{2} + R_{1}$

$R_{3}\rightarrow 3R_{2} + R_{3}$ $\begin{pmatrix}1 & 0 &-13/3 &-5/3 & 2/3 &0\\0 & 1 & 11/3 & 4/3 & -1/3 & 0\\0 & 0 & 9 & 2 &-1 & 1\end{pmatrix}$

$R_{3}\rightarrow 1/9 R_{3}$ $\begin{pmatrix}1 & 0 &-13/3 & -5/3 & 2/3 & 0\\0 & 1 & 11/3 & 4/3 & -1/3 & 0\\0 & 0 & 1 & 2/9 & -1/9 & 1/9 \end{pmatrix}$

$R_{1}\rightarrow(13/3)R_{3} + R_{1}$

$R_{2}\rightarrow 11/3 R_{3} + R_{2}$ $\begin{pmatrix}1 & 0 & 0 & -19/27 & 2/9 & 13/27 \\0 & 1 & 0 & 14/27 & 2/27 & -11/27 \\0 & 0 & 1 & 2/9 & -1/9 & 1/9\end{pmatrix}$

now matrix B is completely changed in to Identity matrix therefore you get the inverse matrix:

$B^{-1} = \begin{pmatrix}19/27 & 2/9 & 13/27 \\ 14/27 & 2/27 & -11/27 \\ 2/9 & -1/9 & 1/9\end{pmatrix}$

4.2. Solving linear system of equations

Dear learner, you solve system of linear equations in the previous lesson: besides here you see how to solve the linear system using the concept of mtrix invertiblity:

using the concept $AX = B$ where A = the coefficient matrix and B = the constant vector or column matrix whileX = the set of variables.

as $AX = B$ $\implies$ $x = A^{-1}B$

Example 2:

Solve $\begin{cases}3x +y =5\\x +2y = -5\end{cases}$

Solution:

First describe the system in the form AX = B implies X = $A^{-1}B$ where $A = \begin{pmatrix}3&1\\1&2\end{pmatrix}$, $X = \begin{pmatrix}x\\y\end{pmatrix}$ and $B = \begin{pmatrix}5\\3\end{pmatrix}$.

first find $A^{-1}$ applyind elementary eliminations like $A|I = \begin{pmatrix}3&1&1&0\\1&2&0&1\end{pmatrix}$, then

$A^{-1} = \begin{pmatrix}2/5& -1/5 \\-1/5 & 3/5 \end{pmatrix}$.

second find X by $X = A^{-1}B$ $\implies$ $X = \begin{pmatrix}x\\y\end{pmatrix}$ = $\begin{pmatrix}2/5 & -1/5 \\-1/5 & 3/5 \end{pmatrix}$ $\times$ $\begin{pmatrix}5\\3\end{pmatrix}$ = $\begin{pmatrix}\frac{13}{5} \\ -\frac{14}{5}\end{pmatrix}$

4.3. Properties of invertible Matrix

Let A and B be invertible matrices of the same order, then:

• $A^{-1}$ invertible and $(A^{-1})^{-1} = A$
• $A^{T}$ is invertible and $(A^{T})^{-1} = (A^{-1})^{T}$
• AB is invertible and $(AB)^{-1} = B^{-1}A^{-1}$.

NB: An invertible matrix is called non-singular matrix and a matrix which is not invertible is singular.

Example 3:

Let $A = \begin{pmatrix}2&3\\1&2\end{pmatrix}$ and $\begin{pmatrix}0&2\\3&1\end{pmatrix}$ then find:

a). $A^{-1}$

b). $(A^{-1})^{-1}$

Solution:

a). $A^{-1}$ $\implies$ $\begin{pmatrix}2&3&1&0\\1&2&0&1\end{pmatrix}$ is transformed to obtain the identity matrix for A.

$R_{2}\leftrightarrow R_{1}$ $\begin{pmatrix}1&2&0&1\\2&3&1&0\end{pmatrix}$ and

$R_{2}\rightarrow-2R_{1} +R_{2}$ $\begin{pmatrix}1&2&0&1\\0&-1&1&-2\end{pmatrix}$ and

$R_{1}\rightarrow2R_{2} + R_{1}$ $\begin{pmatrix}1&0&2&-3\\0&-1&1&-2\end{pmatrix}$

$\implies$ $A^{-1}$ = $\begin{pmatrix}2 &-3\\ -1& 2\end{pmatrix}$

b). $(A^{-1})^{-1} = A = \begin{pmatrix}2&3\\1&2\end{pmatrix}$

4.4. Application of Matrices

Example 4:

1. In a triangle the smallest angle measures  more than one – third of the largest angle. The middle angle measures . Find the measure of each angle of the triangle?
2. The perimeter of a given triangle is 32 cm. the shortest side is 6cm shorter than the largest side of the given triangle, the longest side is 10cm less than the sum of the other two sides of a triangle. Find the measure of lengths of sides of a triangle?

Solution:

1. Let x be the measure of smallest angle of a triangle, y be the measure of the middle angle and z be the measure of largest angle. From the above information x+y+z =180

x = 1/2z+20

y = 15 +x

The system of linear equations is of the form$\begin{cases}x+y+z =180\\2x -z= 20\\y-x =15\end{cases}$

$\implies$ Augmented matrix$A|B =\begin{pmatrix}1&1&1&180\\2&0&-1&20\\-1&1&0&15\end{pmatrix}$

Row reduction leads to:

$R_{2}\rightarrow-2R_{1}+R_{2}$

$R_{3}\rightarrow R_{1 }+ R_{3}$ $\begin{pmatrix}1&1&1&180\\0&-2&-3&-340\\0&2&1&195\end{pmatrix}$

$\implies$ $R_{3}\rightarrow R_{2} + R_{3}$ $\begin{pmatrix}1&1&1&180\\0&-2&-3&-340\\0&0&-2&-145\end{pmatrix}$

back substitution leads $z = 72.5$ and $-2y -3z = -340$ $\implies$ $y =61.5$ where as x +y +z =180 gives you $x = 46$

therefore, the smmalest angle is $46^{0}$, the middle angle measures $61.5^{0}$ while the largest angle measures $72.5^{0}$.

2.Let P= perimeter of the triangle, x = length of shortest side, y = length of the middle sized side, z = length of lomgest side. Then z = x+y -10, x = z-6 and perimeter p = 32 cm>

$\implies$ $\begin{cases}x+y+z = 180\\x+y-z = 10\\ -x+ z = 6\end{cases}$

the augmented matrix $\begin{pmatrix}1&1&1&32\\1&1&-1&10\\-1&1&0&6\end{pmatrix}$

thus

$R_{2}\rightarrow -R_{1}+R_{2}$

$R_{3}\rightarrow R_{1} +R_{3}$ $\begin{pmatrix}1&1&1&32\\0&0&-2&-22\\0&2&1&38\end{pmatrix}$

$R_{2}\leftrightarrow R_{3}$ $\begin{pmatrix}1&1&1&32\\0&2&1&38\\0&0&-2&-22\end{pmatrix}$

then back substitution leads -2z = -22 implies z = 11, and 2y +z = 38 means y = 13.5 while x +y +z = 180 implies x = 7.5

therefore, the shortest side measures 7.5 cm

the medium sized side measures 11cm and the longest side measures 13.5 cm.

https://www.geogebra.org/m/fup7xwta