# Equivalent Matrices – Explanation & Examples

Equivalent Matrices are $2$ matrices that have same size and shape. There are certain conditions that must be met for matrices to be equivalent (or equal) to each other. First of all, let’s check the definition of equivalent matrices:

Equivalent matrices are matrices whose dimension (or order) are same and corresponding elements within the matrices are equal.

In this article, we are going to look at what equivalent matrices are, what makes $2$ matrices equal to each other, and some examples that shows the use of equivalent matrices in solving equations.

## What are equivalent matrices?

$2$ matrices are said to be equivalent if they satisfy the conditions shown below:

• Each matrix has the same number of rows
• Each matrix has the same number of columns
• The corresponding elements (or entries) of each matrix are equal to each other

All of these $3$ conditions must be met in order to call two matrices equivalent, or equal.

## How to tell if two matrices are equivalent?

Consider the $2$ matrices shown below:

$A = \begin{bmatrix} 3 & { – 1 } \\ 6 & 5 \end {bmatrix}$

$B = \begin{bmatrix} 3 & { – 1 } \\ 6 & 3 \end {bmatrix}$

First, we have Matrix $A$. This is a matrix that has $2$ rows and $2$ columns. The order (or dimension) of the matrix is $2 \times 2$.

Recall that we can identify a specific element of a matrix by the notation $A_{ i j }$, where $A$ is the name of the matrix, $i$ is the row number and $j$ is the column number. The elements of matrix $A$ are:

$A_{ 1 1 } = 3$

$A_{ 1 2 } = { – 1 }$

$A_{ 2 1 } = 6$

$A_{ 2 2 } = 5$

Secondly, we have Matrix $B$. This is a matrix that has $2$ rows and $2$ columns. The order (or dimension) of the matrix is $2 \times 2$.

The elements of matrix $B$ are:

$B_{ 1 1 } = 3$

$B_{ 1 2 } = { – 1 }$

$B_{ 2 1 } = 6$

$B_{ 2 2 } = 3$

To tell if Matrix $A$ is equivalent to Matrix $B$ or not, we try to satisfy the $3$ conditions we saw before.

• Both matrices have $2$ rows. So, the $1 ^ { st }$ condition is met.
• Both matrices have $2$ columns. So, the $2 ^ { nd }$ condition is met.
• The element in the $2 ^ { nd }$ row and $2 ^ { nd }$ column are different for Matrix $A$ and Matrix $B$. So, the $3 ^ { rd }$ condition is not met.

For two matrices to be equivalent, we need to fulfill All the $3$ conditions of matrix equivalence. Even if $1$ condition isn’t met, the matrices are not equivalent.

### Solving equations using equivalent matrices

We can solve simple equations using the concept of equivalent matrices. Check the $2$ matrices below:

$M = \begin{pmatrix} 5 & { – 2 } \\ x & 3 \end {pmatrix}$

$N = \begin{pmatrix} 5 & y \\ 6 & 3 \end {pmatrix}$

Given $M = N$, can you find the values for $x$ and $y$ ?

Since Matrix M equals (is equivalent to) Matrix N, all the corresponding elements are same.

To solve for $x$, we can note that:

$M_{ 2 1 } = N_{ 2 1 }$

Thus, $x = 6$.

To solve for $y$, we can note that:

$N_{ 1 2 } = M_{ 1 2 }$

Thus, $y = { – 2 }$.

This showed how we can equate corresponding elements to solve for single variables when two matrices are given as equal. We can extend this idea to solve for complex linear equations as well.

## Summary

Let’s summarize what we have learned:

• Equivalent matrices are matrices whose dimension (or order) are same and corresponding elements within the matrices are equal.
• $3$ conditions must be met for two matrices to be equivalent to each other.
• The number of rows of each matrix should be the same
• The number of columns of each matrix should be the same
• Corresponding elements of each matrix should be equal
• When we are given 2 matrices with unknowns, we can equate the corresponding elements in both matrices to solve for the single variable or form an equation and then solve for the unknown.

Below, we show some examples to elucidate the concept of equivalent matrices and solving simple and complex linear equations using equivalent matrices.

#### Example 1

Given the $7$ matrices below, answer the true/false questions that follow:

$A = \begin{pmatrix} 3 & -4 \\ 6 & { – 3 } \end {pmatrix}$

$B = \begin{pmatrix} 5 & { – 6 } & 0 \\ { – 2 } & { 6 } & 9 \end {pmatrix}$

$C = \begin{pmatrix} 11 & { – 5 } & 1 \\ 12 & { – 3 } & 1 \\ { 0 } & { – 9 } & 3 \end {pmatrix}$

$D = \begin{pmatrix} a & b \\ c & d \\ e & f \\ g & h \end {pmatrix}$

$E = \begin{pmatrix} 3 & -4 \\ 6 & { – 3 } \end {pmatrix}$

$F = \begin{pmatrix} 11 & 12 & 0 \\ { – 5 } & { – 3 } & { – 9} \\ 1 & 1 & 3 \end {pmatrix}$

$G = \begin{pmatrix} a & b \\ c & d \\ e & c \\ g & h \end {pmatrix}$

1. True/False   Matrix $A$ = Matrix $G$
2. True/False   Matrix $B$ = Matrix $F$
3. True/False   Matrix $D$ = Matrix $G$
4. True/False   Matrix $A$ = Matrix $E$
5. True/False   Matrix $C$ = Matrix $F$

Solution

For two matrices to be equal (or, equivalent), these $3$ conditions have to be fullfilled:

• The number of rows of each matrix should be the same
• The number of columns of each matrix should be the same
• Corresponding elements of each matrix should be equal
1. Just by looking at it, we can say that the matrices aren’t equal. The shapes aren’t same, let alone the elements.
False.
2. These two matrices aren’t equal. Row number differ as well as the elements.
False.
3. At first glance, they look exactly same. But, if you look closely, the element in the $3^{rd}$ row, $3^{rd}$ column is different.
$D_{ 3 3 } = f$
$G_{ 3 3 } = c$
So, the matrices aren’t equal.
False.
4. These two matrices, $A$ and $E$, are in fact equivalent. The $3$ conditions are satisfied. Both has 2 rows and 2 columns. All the corresponding elements in each matrix are equal.
True.
5. There are $3$ rows and $3$ columns of each matrix. The elements are same. BUT the rows and columns are exchanged! All the 9 elements are same but corresponding elements aren’t! Thus, we can’t say Matrix $C$ and Matrix $F$ are equivalent.
False.

#### Example 2

Given Matrix $T$ = Matrix $S$, calculate the values of $x$ and $y$.
$T = \begin{bmatrix} 11 & { 2 + x } & 1 \\ 12 & { – 3 } & 1 \\ { 0 } & { – 9 } & 3 \end {bmatrix}$

$S = \begin{bmatrix} 11 & { – 5 } & 1 \\ 12 & { – 3 } & 1 \\ { 0 } & { – 9 } & { 2y – 3 } \end {bmatrix}$

Solution

Both Matrix $T$ and Matrix $S$ are $3 \times 3$ matrices.

We know in equivalent matrices, corresponding entries are equal.

In Matrix $T$,  $T_{12} = 2+x$.
In Matrix $S$, $S_{12} = -5$.

We can equate both entries and find out the value of $x$. Shown below:

$2 + x = -5$
$x = -5 -2$
$x = -7$

Now,

In Matrix $S$,  $T_{ 33 } = 2y – 3$.
In Matrix $T$, $S_{ 33 } = 3$.

We can equate both entries and find out the value of $y$. Shown below:

$2y – 3 = 3$
$2y = 6$
$y = 3$

### Practice Questions

1. For Matrices $A$ through $E$, identify which pair of matrices are equivalent to each other.
$A = \begin{bmatrix} -2 & 6 \\ 12 & 5 \\ -3 & 7 \end {bmatrix}$
$B = \begin{bmatrix} 0 & 5 \\6 & -8 \end {bmatrix}$
$C = \begin{bmatrix} -2 & 4 \\ 13 & 5 \\ -3 & 5 \end {bmatrix}$
$D = \begin{bmatrix} -2 & 6 \\ 12 & 5 \\ -3 & 7 \end {bmatrix}$
$E = \begin{bmatrix} 0 & 6 \\5 & -8 \end {bmatrix}$
2. Given Matrix $J$ = Matrix $K$, calculate the values of $a$ and $b$.
$J = \begin{bmatrix} 11 & 3 & {4a-b} \\ 1 & { – 1 } & 1 \\ { 0 } & { – 3 } & -7 \end {bmatrix}$

$K = \begin{bmatrix} 11 & 3 & 7 \\ 1 & { – 1 } & 1 \\ { 0 } & { -3 } & { -a+2b } \end {bmatrix}$

1. Looking at the 5 matrices carefully, we can make the following observations:
• Both matrices $A$ and $D$ are equivalent to each other. They are both $3 \times 2$ matrices with each corresponding entry equal to each other.
• Matrix $B$ and $E$ are both $2 \times 2$ matrices. Some of their corresponding entries are equal but not all. Eventhough they look equal, they are not!
• Matrix $C$ looks somewhat equal to Matrix $A$ but upon close observation, we can conclude that they are not equal.
2. Both Matrix $J$ and Matrix $K$ are $3 \times 3$ matrices.

We know in equivalent matrices, corresponding entries are equal.

In Matrix $J$,  $J_{13} = 4a-b$.
In Matrix $K$, $K_{13} = 7$.

Equating, we can write:

$4a – b = 7$

Also,

In Matrix $J$,  $T_{33} = -7$.
In Matrix $K$, $S_{13} = -a+2b$.

Equating, we can write:

$-a+2b = -7$

These 2 equations can be solved simultaneous to get the values of $a$ and $b$.
After solving, we get:

$a = 1$ and $b = -3$.