# Chapter 04.01: Prerequisites to Simultaneous Linear Equations

## Learning Objectives

After successful completion of this lesson, you should be able to:

1)  Define what a matrix is.

2)  Identify special types of matrices.

## What does a matrix look like?

Matrices are everywhere. If you have used a spreadsheet such as Excel or wrote numbers in a table, you have used a matrix. Matrices make the presentation of numbers clearer and make calculations easier to program. Look at the matrix below about the sale of tires in a Blowoutr’us store – given by quarter and make of tires.

$\begin{matrix} Tirestone\\ Michigan\\ Copper\\ \end{matrix} \stackrel{\mbox{Q1. Q2. Q3. Q4.}}{\begin{bmatrix} 25 & 20 & 3 & 2 \\ 5 & 10 &15 &25 \\ 6 & 16 &7 & 27 \\ \end{bmatrix}}$

If one wants to know how many Copper tires were sold in Quarter $$4$$, we go along the row Copper and column Q4 and find that it is $$27$$.

## So, what is a matrix?

A matrix is a rectangular array of elements. The elements can be symbolic expressions or/and numbers. Matrix $$\lbrack A\rbrack$$ is denoted by

$\displaystyle \lbrack A\rbrack = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\a_{21} & a_{22} &\cdots & a_{2n} \\ \vdots & & & \vdots \\a_{m1} & a_{m2} & \cdots & a_{\text{mn}} \\\end{bmatrix}$

Row $$i$$ of $$\lbrack A\rbrack$$ has $$n$$ elements and is

$\left\lbrack a_{i1}a_{i2}\ldots a_{{in}} \right\rbrack$

and column $$j$$ of $$\lbrack A\rbrack$$ has $$m$$ elements and is

$\begin{bmatrix} a_{1j} \\ a_{2j} \\ \vdots \\ a_{{mj}} \\\end{bmatrix}$

Each matrix has rows and columns, and this defines the size of the matrix. If a matrix $$\lbrack A\rbrack$$ has $$m$$ rows and $$n$$ columns, the size of the matrix is denoted by $$m \times n$$. The matrix $$\lbrack A\rbrack$$ may also be denoted by $$\lbrack A\rbrack_{m \times n}$$ to show that $$\lbrack A\rbrack$$ is a matrix with $$m$$ rows and $$n$$ columns.

Each entry in the matrix is called the entry or element of the matrix and is denoted by $$a_{{ij}}$$ where $$i$$ is the row number, and $$j$$ is the column number of the element.

The matrix for the tire sales example could be denoted by the matrix $$A$$ as

$\lbrack A\rbrack = \begin{bmatrix} 25 & 20 & 3 & 2 \\ 5 & 10 & 15 & 25 \\ 6 & 16 & 7 & 27 \\ \end{bmatrix}$

There are $$3$$ rows and $$4$$ columns, so the size of the matrix is $$3 \times 4$$. In the above $$\lbrack A\rbrack$$ matrix, $$a_{34} = 27$$.

## What are the special types of matrices?

Vector: A vector is a matrix that has only one row or one column. There are two types of vectors – row vectors and column vectors.

## Row Vector

If a matrix $$\lbrack B\rbrack$$ has one row, it is called a row vector $$\lbrack B\rbrack = \lbrack b_{1}\ b_{2}\ldots\ldots b_{n}\rbrack$$ and $$n$$ is the dimension of the row vector.

### Example 1

Give an example of a row vector.

Solution

$\lbrack B \rbrack =\lbrack 25\ \ 20\ \ 3\ \ 2\ \ 0\rbrack$

is an example of a row vector of dimension $$5$$.

## Column vector

If a matrix $$\lbrack C\rbrack$$ has one column, it is called a column vector

$\lbrack C\rbrack = \begin{bmatrix} c_{1} \\ \vdots \\ \vdots \\ c_{m} \\ \end{bmatrix}$

and $$m$$ is the dimension of the vector.

### Example 2

Give an example of a column vector.

Solution

$\lbrack C\rbrack = \begin{bmatrix} 25 \\ 5 \\ 6 \\ \end{bmatrix}$

is an example of a column vector of dimension $$3$$.

## Submatrix

If some row(s) or/and column(s) of a matrix $$\lbrack A\rbrack$$ are deleted (no rows or columns may be deleted), the remaining matrix is called a submatrix of $$\lbrack A\rbrack$$.

## Square matrix

If the number of rows $$m$$ of a matrix is equal to the number of columns $$n$$ of a matrix $$\lbrack A\rbrack$$, that is, $$m = n$$, then $$\lbrack A\rbrack$$ is called a square matrix. The entries $$a_{11},a_{22},...,a_{{nn}}$$ are called the diagonal elements of a square matrix. Sometimes the diagonal of the matrix is also called the principal or main of the matrix.

### Example 3

Give an example of a square matrix.

Solution

$\lbrack A\rbrack = \begin{bmatrix} 25 & 20 & 3 \\ 5 & 10 & 15 \\ 6 & 15 & 7 \\ \end{bmatrix}$

is a square matrix as it has the same number of rows and columns, that is, $$3$$. The diagonal elements of $$\lbrack A\rbrack$$ are $a_{11} = 25,\ \ a_{22} = 10,\ \ a_{33} = 7.$

## Upper triangular matrix

A $$n \times n$$ matrix for which $$a_{{ij}} = 0,\ \ i > j$$ for all $$i,j$$ is called an upper triangular matrix. That is, all the elements below the diagonal entries are zero.

### Example 4

Give an example of an upper triangular matrix.

Solution

$\lbrack A\rbrack = \begin{bmatrix} 10 & - 7 & 0 \\ 0 & - 0.001 & 6 \\ 0 & 0 & 15005 \\ \end{bmatrix}$

is an upper triangular matrix.

## Lower triangular matrix

A $$n \times n$$ matrix for which $$a_{{ij}} = 0,\ \ j > i$$ for all $$i,j$$ is called a lower triangular matrix. That is, all the elements above the diagonal entries are zero.

### Example 5

Give an example of a lower triangular matrix.

Solution

$\lbrack A\rbrack = \begin{bmatrix} 1 & 0 & 0 \\ 0.3 & 1 & 0 \\ 0.6 & 2.5 & 1 \\ \end{bmatrix}$

is a lower triangular matrix.

## Diagonal matrix

A square matrix with all non-diagonal elements equal to zero is called a diagonal matrix, that is, only the diagonal entries of the square matrix can be non-zero ($$a_{ij} = 0,\ \ i \neq j$$).

### Example 6

Give examples of a diagonal matrix.

Solution

$\lbrack A\rbrack = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 2.1 & 0 \\ 0 & 0 & 5 \\ \end{bmatrix}$

is a diagonal matrix.

Any or all the diagonal entries of a diagonal matrix can be zero. For example

$\lbrack A\rbrack = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 2.1 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}$

is also a diagonal matrix.

## Identity matrix

A diagonal matrix with all diagonal elements equal to $$1$$ is called an identity matrix, ($$a_{{ij}} = 0,\ \ i \neq j)$$ for all $$i,j$$ and $$a_{{ii}} = 1$$ for all $$i$$).

### Example 7

Give an example of an identity matrix.

Solution

$\lbrack A\rbrack = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$

is an identity matrix.

## Learning Objectives

After successful completion of this lesson, you should be able to:

1)  Add one matrix to another

2)  Subtract one matrix from another

3)  Multiply one matrix by another

## How do you add two matrices?

Two matrices $$\left\lbrack A \right\rbrack$$ and $$\left\lbrack B \right\rbrack$$ can be added if they are of the same size. The addition is then shown as

$\left\lbrack C \right\rbrack = \left\lbrack A \right\rbrack + \left\lbrack B \right\rbrack$

where

$c_{{ij}} = a_{{ij}} + b_{{ij}}$

### Example 1

$\left\lbrack A \right\rbrack = \begin{bmatrix}5 & 2 & 3 \\1 & 2 & 7 \\\end{bmatrix}$ $\left\lbrack B \right\rbrack = \begin{bmatrix}6 & 7 & - 2 \\3 & 5 & 19 \\\end{bmatrix}$

Solution

$\begin{split} \left\lbrack C \right\rbrack &= \left\lbrack A \right\rbrack + \left\lbrack B \right\rbrack\\ &= \begin{bmatrix} 5 & 2 & 3 \\ 1 & 2 & 7 \\ \end{bmatrix} + \begin{bmatrix} 6 & 7 & - 2 \\ 3 & 5 & 19 \\ \end{bmatrix}\\ &= \begin{bmatrix} 5 + 6 & 2 + 7 & 3 - 2 \\ 1 + 3 & 2 + 5 & 7 + 19 \\ \end{bmatrix}\\ &= \begin{bmatrix} 11 & 9 & 1 \\ 4 & 7 & 26 \\ \end{bmatrix} \end{split}$

## How do you subtract two matrices?

Two matrices $$\left\lbrack A \right\rbrack$$ and $$\left\lbrack B \right\rbrack$$ can be subtracted if they are the same size. The subtraction is then shown as

$\left\lbrack D \right\rbrack = \left\lbrack A \right\rbrack - \left\lbrack B \right\rbrack$

where

$d_{{ij}} = a_{{ij}} - b_{{ij}}$

### Example 2

Subtract matrix $$\left\lbrack B \right\rbrack$$ from matrix $$\left\lbrack A \right\rbrack$$.

$\left\lbrack A \right\rbrack = \begin{bmatrix} 5 & 2 & 3 \\ 1 & 2 & 7 \\ \end{bmatrix}$

$\left\lbrack B \right\rbrack = \begin{bmatrix} 6 & 7 & - 2 \\ 3 & 5 & 19 \\ \end{bmatrix}$

Solution

$\begin{split} \left\lbrack D \right\rbrack &= \left\lbrack A \right\rbrack - \left\lbrack B \right\rbrack\\ &= \begin{bmatrix} 5 & 2 & 3 \\ 1 & 2 & 7 \\ \end{bmatrix} - \begin{bmatrix} 6 & 7 & - 2 \\ 3 & 5 & 19 \\ \end{bmatrix}\\ &= \begin{bmatrix} \left( 5 - 6 \right) & \left( 2 - 7 \right) & \left( 3 - \left( - 2 \right) \right) \\ \left( 1 - 3 \right) & \left( 2 - 5 \right) & \left( 7 - 19 \right) \\ \end{bmatrix}\\ &= \begin{bmatrix} - 1 & - 5 & 5 \\ - 2 & - 3 & - 12 \\ \end{bmatrix}\end{split}$

## How do I multiply two matrices?

Two matrices $$\left\lbrack A \right\rbrack$$ and $$\left\lbrack B \right\rbrack$$ can be multiplied only if the number of columns of $$\left\lbrack A \right\rbrack$$ is equal to the number of rows of $$\left\lbrack B \right\rbrack$$ to give

$\left\lbrack C \right\rbrack_{m \times n} = \left\lbrack A \right\rbrack_{m \times p}\left\lbrack B \right\rbrack_{p \times n}$

If $$\left\lbrack A \right\rbrack$$ is a $$m \times p$$ matrix and $$\left\lbrack B \right\rbrack$$ is a $$p \times n$$ matrix, the resulting matrix $$\left\lbrack C \right\rbrack$$ is a $$m \times n$$ matrix.

So how does one calculate the elements of $$\left\lbrack C \right\rbrack$$ matrix?

$\begin{split} c_{{ij}} &= \sum_{k = 1}^{p}{a_{{ik}}b_{{kj}}}\\ &= a_{i1}b_{1j} + a_{i2}b_{2j} + \ldots + a_{{ip}}b_{{pj}} \end{split}$

for each $$i = 1,\ 2,\ \ldots\ \ ,\ m$$ and $$j = 1,\ 2,\ \ldots\ \ ,\ n$$.

To put it in simpler terms, the $$i^{{th}}$$ row and $$j^{{th}}$$ column of the $$\left\lbrack C \right\rbrack$$ matrix in $$\left\lbrack C \right\rbrack = \left\lbrack A \right\rbrack\left\lbrack B \right\rbrack$$ is calculated by multiplying the $$i^{{th}}$$ row of $$\left\lbrack A \right\rbrack$$ by the $$j^{{th}}$$ column of $$\left\lbrack B \right\rbrack$$. That is

$\begin{split} c_{{ij}} &= \left\lceil a_{i1} \ \ a_{i2}\ \ \ldots\ \ \ a_{{ip}} \right\rceil\begin{bmatrix} \begin{matrix} b_{1j} \\ b_{2j} \\ \end{matrix} \\ \begin{matrix} \vdots \\ b_{{pj}} \\ \end{matrix} \\ \end{bmatrix}\\ &= a_{i1}b_{1j} + a_{i2}b_{2j} + \ldots + a_{{ip}}b_{{pj}}\\ &= \sum_{k = 1}^{p}{a_{{ik}}b_{{kj}}}\end{split}$

### Example 3

Given

$\left\lbrack A \right\rbrack = \begin{bmatrix} 5 & 2 & 3 \\ 1 & 2 & 7 \\ \end{bmatrix}$

$\left\lbrack B \right\rbrack = \begin{bmatrix} 3 & - 2 \\ 5 & - 8 \\ 9 & - 10 \\ \end{bmatrix}$

Find

$\left\lbrack C \right\rbrack = \left\lbrack A \right\rbrack\left\lbrack B \right\rbrack$

Solution

$$c_{12}$$ can be found by multiplying the first row of $$\left\lbrack A \right\rbrack$$ by the second column of $$\left\lbrack B \right\rbrack$$,

$\begin{split} c_{12} &= \begin{bmatrix} 5 & 2 & 3 \\ \end{bmatrix}\begin{bmatrix} - 2 \\ - 8 \\ - 10 \\ \end{bmatrix}\\ &= \left( 5 \right)\left( - 2 \right) + \left( 2 \right)\left( - 8 \right) + \left( 3 \right)\left( - 10 \right)\\ &= - 56\end{split}$

Similarly, one can find the other elements of $$\left\lbrack C \right\rbrack$$ to give

$\left\lbrack C \right\rbrack = \begin{bmatrix} 52 & - 56 \\ 76 & - 88 \\ \end{bmatrix}$

## Learning Objectives

After successful completion of this lesson, you should be able to:

1)  Develop simultaneous linear equations model from a physical problem

2)  Set up simultaneous linear equations in matrix form

## Matrix algebra is used for solving systems of equations. Can you illustrate this concept?

Matrix algebra is used to solve a system of simultaneous linear equations. In fact, for many
mathematical procedures such as the solution to a set of nonlinear equations, interpolation, integration, and differential equations, the solutions reduce to a set of simultaneous linear equations. Let us illustrate with an example for interpolation.

### Example 1

The upward velocity of a rocket is given at three different times on the following table.

Table 1. Velocity vs. time data for a rocket

Time, t Velocity, v
$$\text{s}$$ $$(\text{m/s})$$
$$5$$ $$106.8$$
$$8$$ $$177.2$$
$$12$$ $$279.2$$

The velocity data is approximated by a polynomial as

$v\left( t \right) = at^{2} + {bt} + c,\ 5 \leq t \leq 12\;\;\;\;\;\;\;\;\;\;\;\; (E1.1)$

Set up the equations in matrix form to find the coefficients $$a,b,c$$ of the velocity profile.

Solution

The polynomial is going through three data points $$\left( t_{1},v_{1} \right),\left( t_{2},v_{2} \right), \text{ and} \left( t_{3},v_{3} \right)$$ where from Table 1

$t_{1} = 5,\ v_{1} = 106.8$

$t_{2} = 8,\ v_{2} = 177.2$

$t_{3} = 12,\ v_{3} = 279.2$

Requiring that $$v\left( t \right) = at^{2} + {bt} + c$$ passes through the three data points gives

$\begin{split} v\left( t_{1} \right) &= v_{1} = at_{1}^{2} + bt_{1} + c\\ v\left( t_{2} \right) &= v_{2} = at_{2}^{2} + bt_{2} + c\\ v\left( t_{3} \right) &= v_{3} = at_{3}^{2} + bt_{3} + c \end{split}$

Substituting the data $$\left( t_{1},v_{1} \right),\left( t_{2},v_{2} \right),\ and\ \left( t_{3},v_{3} \right)$$ gives

$\begin{split} a\left( 5^{2} \right) + b\left( 5 \right) + c &= 106.8\\ a\left( 8^{2} \right) + b\left( 8 \right) + c &= 177.2\\ a\left( 12^{2} \right) + b\left( 12 \right) + c &= 279.2\end{split}$

or

$\begin{split} 25a + 5b + c &= 106.8\\ 64a + 8b + c &= 177.2\\ 144a + 12b + c &= 279.2 \end{split}$

This set of equations can be rewritten in the matrix form as

$\begin{bmatrix} 25a + & 5b + & c \\ 64a + & 8b + & c \\ 144a + & 12b + & c \\ \end{bmatrix} = \begin{bmatrix} 106.8 \\ 177.2 \\ 279.2 \\ \end{bmatrix}$

The above equation can be written as a linear combination as follows

$a\begin{bmatrix} 25 \\ 64 \\ 144 \\ \end{bmatrix} + b\begin{bmatrix} 5 \\ 8 \\ 12 \\ \end{bmatrix} + c\begin{bmatrix} 1 \\ 1 \\ 1 \\ \end{bmatrix} = \begin{bmatrix} 106.8 \\ 177.2 \\ 279.2 \\ \end{bmatrix}$

and further using matrix multiplication gives

$\begin{bmatrix} 25 & 5 & 1 \\ 64 & 8 & 1 \\ 144 & 12 & 1 \\ \end{bmatrix}\begin{bmatrix} a \\ b \\ c \\ \end{bmatrix} = \begin{bmatrix} 106.8 \\ 177.2 \\ 279.2 \\ \end{bmatrix}$

The above is an illustration of why matrix algebra is needed. The complete solution to the set of equations is given later in this chapter.

A general set of $$m$$ linear equations and $$n$$ unknowns,

$\begin{split} &a_{11}x_{1} + a_{12}x_{2} + {......} + a_{1n}x_{n} = c_{1}\\ &a_{21}x_{1} + a_{22}x_{2} + {......} + a_{2n}x_{n} = c_{2}\\ &{.......................................}\\ & {.......................................}\\ &a_{m1}x_{1} + a_{m2}x_{2} + ...... + a_{\text{mn}}x_{n} = c_{m} \end{split}$

can be rewritten in the matrix form as

$\begin{bmatrix} a_{11} & a_{12} & . & . & a_{1n} \\ a_{21} & a_{22} & . & . & a_{2n} \\ \vdots & & & & \vdots \\ \vdots & & & & \vdots \\ a_{m1} & a_{m2} & . & . & a_{\text{mn}} \\ \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \\ \vdots \\ \vdots \\ x_{n} \\ \end{bmatrix} = \begin{bmatrix} c_{1} \\ c_{2} \\ \vdots \\ \vdots \\ c_{m} \\ \end{bmatrix}$

Denoting the matrices by $$\left\lbrack A \right\rbrack$$, $$\left\lbrack X \right\rbrack$$, and $$\left\lbrack C \right\rbrack$$, the system of equation is $$\left\lbrack A \right\rbrack\ \left\lbrack X \right\rbrack = \left\lbrack C \right\rbrack$$, where $$\left\lbrack A \right\rbrack$$ is called the coefficient matrix, $$\left\lbrack C \right\rbrack$$ is called the right-hand side vector and $$\left\lbrack X \right\rbrack$$ is called the solution vector.

Sometimes $$\left\lbrack A \right\rbrack\ \left\lbrack X \right\rbrack = \left\lbrack C \right\rbrack$$ system of equations is written in the augmented form, that is,

$[A\ \vdots\ C]= \begin{bmatrix} a_{11} &a_{12} &\cdots &a_{1n}\ \ \vdots&c_1 \\ a_{21} &a_{22} &\cdots &a_{2n}\ \ \vdots&c_2 \\ \vdots&\vdots&\ddots&\ \ \ \ \ \ \ \vdots& \vdots\\ a_{m1} &a_{m2} &\cdots &a_{mn}\ \vdots&c_n \end{bmatrix}$

As an example, for the set of equations $\begin{bmatrix} 25 & 5 & 1 \\ 64 & 8 & 1 \\ 144 & 12 & 1 \\ \end{bmatrix}\begin{bmatrix} a \\ b \\ c \\ \end{bmatrix} = \begin{bmatrix} 106.8 \\ 177.2 \\ 279.2 \\ \end{bmatrix}$ the augmented matrix form is $\begin{bmatrix} 25 & 5 & 1 & | & 106.8 \\ 64 & 8 & 1 & | & 177.2 \\ 144 & 12 & 1 & | & 279.2 \\ \end{bmatrix}$

## Learning Objectives

After successful completion of this lesson, you should be able to:

1)  define the inverse of a matrix

2)  know important statements about the inverse of a matrix

3)  solve a set of equations where the inverse of the coefficient matrix is given

## Can you divide two matrices?

If $$\lbrack A\rbrack\ \lbrack B\rbrack = \lbrack C\rbrack$$ is defined, it might seem intuitive that $$\displaystyle \lbrack A\rbrack = \frac{\left\lbrack C \right\rbrack}{\left\lbrack B \right\rbrack}$$, but matrix division is not defined like that. However, an inverse of a matrix can be defined for certain types of square matrices. The inverse of a square matrix $$\lbrack A\rbrack$$, if existing, is denoted by $$\lbrack A\rbrack^{- 1}$$ such that

$\lbrack A\rbrack\ \lbrack A\rbrack^{- 1} = \lbrack I\rbrack = \lbrack A\rbrack^{- 1}\lbrack A\rbrack$

where $$\lbrack I\rbrack$$ is the identity matrix.

In other words, let $$A$$ be a square matrix. If $$\lbrack B\rbrack$$ is another square matrix of the same size such that $$\lbrack B\rbrack\ \lbrack A\rbrack = \lbrack I\rbrack$$, then $$\lbrack B\rbrack$$ is the inverse of $$\lbrack A\rbrack$$. $$\lbrack A\rbrack$$ is then called to be invertible or nonsingular. If $$\lbrack A\rbrack^{- 1}$$ does not exist, $$\lbrack A\rbrack$$ is called noninvertible or singular.

If $$\lbrack A\rbrack$$ and $$\lbrack B\rbrack$$ are two $$n \times n$$ matrices such that $$\lbrack B\rbrack\ \lbrack A\rbrack = \lbrack I\rbrack$$, then these statements are also true

a)  $$[B]$$ is the inverse of $$[A]$$
b)  $$[A]$$ is the inverse of $$[B]$$
c)  $$[A]$$ and $$[B]$$ are both invertible
d)  $$[A] [B]= [I]$$.
e)  $$[A]$$ and $$[B]$$ are both nonsingular
f)  all columns of $$[A]$$ and $$[B]$$ are linearly independent
g)  all rows of $$[A]$$ and $$[B]$$ are linearly independent.

### Example 1

Determine if

$\lbrack B\rbrack = \begin{bmatrix} 3 & 2 \\ 5 & 3 \\ \end{bmatrix}$

is the inverse of

$\lbrack A\rbrack = \begin{bmatrix} - 3 & 2 \\ 5 & - 3 \\ \end{bmatrix}$

Solution

$\begin{split} \lbrack B\rbrack\lbrack A\rbrack &= \begin{bmatrix} 3 & 2 \\ 5 & 3 \\ \end{bmatrix}\begin{bmatrix} - 3 & 2 \\ 5 & - 3 \\ \end{bmatrix}\\ &= \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \\ &= \lbrack I\rbrack\end{split}$

Since

$\left\lbrack B \right\rbrack\left\lbrack A \right\rbrack = \left\lbrack I \right\rbrack,$

$$\lbrack B\rbrack$$ is the inverse of $$\lbrack A\rbrack$$, and $$\lbrack A\rbrack$$ is the inverse of $$\lbrack B\rbrack$$.

But, we can also show that

$\begin{split} \lbrack A\rbrack\lbrack B\rbrack &= \begin{bmatrix} - 3 & 2 \\ 5 & - 3 \\ \end{bmatrix}\begin{bmatrix} 3 & 2 \\ 5 & 3 \\ \end{bmatrix}\\ &= \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}\\ &= \lbrack I\rbrack \end{split}$

to prove that $$\lbrack A\rbrack$$ is the inverse of $$\lbrack B\rbrack$$.

## Can I use the concept of the inverse of a matrix to find the solution of a set of equations [A][X] = [C]?

Yes, if the number of equations is the same as the number of unknowns, the coefficient matrix $$\lbrack A\rbrack$$ is a square matrix.

Given

$\lbrack A\rbrack\ \lbrack X\rbrack = \lbrack C\rbrack$

Then, if $$\lbrack A\rbrack^{- 1}$$ exists, multiplying both sides by $$\lbrack A\rbrack^{- 1}$$.

$\lbrack A\rbrack^{- 1}\lbrack A\rbrack\lbrack X\rbrack = \lbrack A\rbrack^{- 1}\lbrack C\rbrack$

$\lbrack I\rbrack\ \lbrack X\rbrack = \lbrack A\rbrack^{- 1}\lbrack C\rbrack$

$\lbrack X\rbrack = \lbrack A\rbrack^{- 1}\lbrack C\rbrack$

This implies that if we are able to find $$\lbrack A\rbrack^{- 1}$$, the solution vector of $$\lbrack A\rbrack\ \lbrack X\rbrack = \lbrack C\rbrack$$ is simply a multiplication of $$\lbrack A\rbrack^{- 1}$$ and the right-hand side vector, $$\lbrack C\rbrack$$.

## How do I find the inverse of a matrix?

If $$\lbrack A\rbrack$$ is a $$n \times n$$ matrix, then $$\lbrack A\rbrack^{- 1}$$ is a $$n \times n$$ matrix, and according to the definition of inverse of a matrix

$\lbrack A\rbrack\ \lbrack A\rbrack^{- 1} = \lbrack I\rbrack$

Denoting

$\lbrack A\rbrack = \begin{bmatrix} a_{11} & a_{12} & \cdot & \cdot & a_{1n} \\ a_{21} & a_{22} & \cdot & \cdot & a_{2n} \\ \cdot & \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot & \cdot \\ a_{n1} & a_{n2} & \cdot & \cdot & a_{{nn}} \\ \end{bmatrix}$

$\lbrack A\rbrack^{- 1} = \begin{bmatrix} a_{11}^{\prime} & a_{12}^{\prime} & \cdot & \cdot & a_{1n}^{\prime} \\ a_{21}^{\prime} & a_{22}^{\prime} & \cdot & \cdot & a_{2n}^{\prime} \\ \cdot & \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot & \cdot \\ a_{n1}^{\prime} & a_{n2}^{\prime} & \cdot & \cdot & a_{\text{nn}}^{\prime} \\ \end{bmatrix}$

$\lbrack I\rbrack = \begin{bmatrix} 1 & 0 & \cdot & \cdot & \cdot & 0 \\ 0 & 1 & & & & 0 \\ 0 & & \cdot & & & \cdot \\ \cdot & & & 1 & & \cdot \\ \cdot & & & & \cdot & \cdot \\ 0 & \cdot & \cdot & \cdot & \cdot & 1 \\ \end{bmatrix}$

Using the definition of matrix multiplication, the first column of the $$\lbrack A\rbrack^{- 1}$$ matrix can then be found by solving

$\begin{bmatrix} a_{11} & a_{12} & \cdot & \cdot & a_{1n} \\ a_{21} & a_{22} & \cdot & \cdot & a_{2n} \\ \cdot & \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot & \cdot \\ a_{n1} & a_{n2} & \cdot & \cdot & a_{\text{nn}} \\ \end{bmatrix}\begin{bmatrix} a_{11}^{\prime} \\ a_{21}^{\prime} \\ \cdot \\ \cdot \\ a_{n1}^{\prime} \\ \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ \cdot \\ \cdot \\ 0 \\ \end{bmatrix}$

Similarly, one can find the other columns of the $$\lbrack A\rbrack^{- 1}$$ matrix by changing the right-hand side accordingly.

### Example 2

The upward velocity of the rocket is given by

Table 1. Velocity vs. time data for a rocket

Time, t (s) Velocity, v (m/s)
$$5$$ $$106.8$$
$$8$$ $$177.2$$
$$12$$ $$279.2$$

In an earlier example, we wanted to approximate the velocity profile by

$v\left( t \right) = at^{2} + {bt} + c,\ 5 \leq t \leq 12$

We found that the coefficients $$a,\ b,\text{ and }\ c$$ in $$v\left( t \right)$$ are given by solving

$\begin{bmatrix} 25 & 5 & 1 \\ 64 & 8 & 1 \\ 144 & 12 & 1 \\ \end{bmatrix}\begin{bmatrix} a \\ b \\ c \\ \end{bmatrix} = \begin{bmatrix} 106.8 \\ 177.2 \\ 279.2 \\ \end{bmatrix}$

First, find the inverse of

$\left\lbrack A \right\rbrack = \begin{bmatrix} 25 & 5 & 1 \\ 64 & 8 & 1 \\ 144 & 12 & 1 \\ \end{bmatrix}$

and then use the definition of inverse to find the coefficients $$a,\ b,\ \text{and}\ c$$, and the velocity profile

Solution

If

$\left\lbrack A \right\rbrack^{- 1} = \begin{bmatrix} a_{11}^{\prime} & a_{12}^{\prime} & a_{13}^{\prime} \\ a_{21}^{\prime} & a_{22}^{\prime} & a_{23}^{\prime} \\ a_{31}^{\prime} & a_{32}^{\prime} & a_{33}^{\prime} \\ \end{bmatrix}$

is the inverse of $$\lbrack A\rbrack$$, then

$\begin{bmatrix} 25 & 5 & 1 \\ 64 & 8 & 1 \\ 144 & 12 & 1 \\ \end{bmatrix}\begin{bmatrix} a_{11}^{\prime} & a_{12}^{\prime} & a_{13}^{\prime} \\ a_{21}^{\prime} & a_{22}^{\prime} & a_{23}^{\prime} \\ a_{31}^{\prime} & a_{32}^{\prime} & a_{33}^{\prime} \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$

gives three sets of equations

$\begin{bmatrix} 25 & 5 & 1 \\ 64 & 8 & 1 \\ 144 & 12 & 1 \\ \end{bmatrix}\begin{bmatrix} a_{11}^{\prime} \\ a_{21}^{\prime} \\ a_{31}^{\prime} \\ \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \\ \end{bmatrix}$

$\begin{bmatrix} 25 & 5 & 1 \\ 64 & 8 & 1 \\ 144 & 12 & 1 \\ \end{bmatrix}\begin{bmatrix} a_{12}^{\prime} \\ a_{22}^{\prime} \\ a_{32}^{\prime} \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 0 \\ \end{bmatrix}$

$\begin{bmatrix} 25 & 5 & 1 \\ 64 & 8 & 1 \\ 144 & 12 & 1 \\ \end{bmatrix}\begin{bmatrix} a_{13}^{\prime} \\ a_{23}^{\prime} \\ a_{33}^{\prime} \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \\ \end{bmatrix}$

Solving the above three sets of equations separately gives

$\begin{bmatrix} a_{11}^{\prime} \\ a_{21}^{\prime} \\ a_{31}^{\prime} \\ \end{bmatrix} = \begin{bmatrix} 0.04762 \\ - 0.9524 \\ 4.571 \\ \end{bmatrix}$

$\begin{bmatrix} a_{12}^{\prime} \\ a_{22}^{\prime} \\ a_{32}^{\prime} \\ \end{bmatrix} = \begin{bmatrix} - 0.08333 \\ 1.417 \\ - 5.000 \\ \end{bmatrix}$

$\begin{bmatrix} a_{13}^{\prime} \\ a_{23}^{\prime} \\ a_{33}^{\prime} \\ \end{bmatrix} = \begin{bmatrix} 0.03571 \\ - 0.4643 \\ 1.429 \\ \end{bmatrix}$

Hence

$\lbrack A\rbrack^{- 1} = \begin{bmatrix} 0.04762 & - 0.08333 & 0.03571 \\ - 0.9524 & 1.417 & - 0.4643 \\ 4.571 & - 5.000 & 1.429 \\ \end{bmatrix}$

Now

$\left\lbrack A \right\rbrack\left\lbrack X \right\rbrack = \left\lbrack C \right\rbrack$

where

$\left\lbrack X \right\rbrack = \begin{bmatrix} a \\ b \\ c \\ \end{bmatrix}$

$\left\lbrack C \right\rbrack = \begin{bmatrix} 106.8 \\ 177.2 \\ 279.2 \\ \end{bmatrix}$

Using the definition of $$\left\lbrack A \right\rbrack^{- 1},$$

$\left\lbrack A \right\rbrack^{- 1}\left\lbrack A \right\rbrack\left\lbrack X \right\rbrack = \left\lbrack A \right\rbrack^{- 1}\left\lbrack C \right\rbrack$

$\left\lbrack X \right\rbrack = \left\lbrack A \right\rbrack^{- 1}\left\lbrack C \right\rbrack$

$\begin{bmatrix} a \\ b \\ c \\ \end{bmatrix} =\begin{bmatrix} 0.04762 & - 0.08333 & 0.03571 \\ - 0.9524 & 1.417 & - 0.4643 \\ 4.571 & - 5.000 & 1.429 \\ \end{bmatrix}\begin{bmatrix} 106.8 \\ 177.2 \\ 279.2 \\ \end{bmatrix}$

Conducting matrix multiplication of the right hand side gives

$\begin{bmatrix} a \\ b \\ c \\ \end{bmatrix} = \begin{bmatrix} 0.2905 \\ 19.69 \\ 1.086 \\ \end{bmatrix}$

So

$v\left( t \right) = 0.2905t^{2} + 19.69t + 1.086,\ 5 \leq t \leq 12$