Monday, February 12, 2024

19.2 - Equality of Matrices

In the previous section, we saw the different types of matrices. In this section, we will see equality of matrices.

Equality of matrices

This can be written in 3 steps:
1. Two matrices $A = \left[a_{ij} \right]$ and $B = \left[b_{ij} \right]$ are said to be equal, if two conditions are satisfied:
(i) A and B are of the same order.
(ii) Corresponding elements of A and B are equal. That is., aij = bij for all i and j.
• For example:
    ♦ $\begin{bmatrix}4 & 3\\8 & 1\end{bmatrix}$ and $\begin{bmatrix}4 & 3\\8 & 1\end{bmatrix}$ are equal matrices.

    ♦ $\begin{bmatrix}4 & 3\\8 & 1\end{bmatrix}$ and $\begin{bmatrix}3 & 4\\8 & 1\end{bmatrix}$ are not equal matrices.

2. If two matrices A and B are equal, then we can write it symbolically as: A = B.

3. If $\begin{bmatrix}a & b\\c & d\\u & v\end{bmatrix}$ = $\begin{bmatrix}3 & 2.5\\9 & \sqrt3\\5 & -1.2\end{bmatrix}$, then:

a = 3, b = 2.5, c = 9, d = √3, u = 5 and v = -1.2

Solved example 19.4
If $\left[\begin{array}{r}           
x+3    &{z+4}    &{2y-7}    \\
-6    &{a-1}    &{0}    \\
b-3    &{-21}    &{0}    \\
\end{array}\right]
=
\left[\begin{array}{r}           
0    &{6}    &{2y-2}    \\
-6    &{-3}    &{2c+2}    \\
2b+4    &{-21}    &{0}    \\
\end{array}\right]$, find the values of a, b, c, x, y and z.
Solution:
• Given that, the two matrices are equal. So their corresponding elements will be equal.
• Thus we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{~~~• }    &{x+3}    & {~=~}    &{0}    \\
{~\color{magenta}    2    }    &{\implies}    &{x}    & {~=~}    &{-3}    \\
{~\color{magenta}    3    }    &{~~~• }    &{z+4}    & {~=~}    &{6}    \\
{~\color{magenta}    4    }    &{\implies}    &{z}    & {~=~}    &{2}    \\
{~\color{magenta}    5    }    &{~~~• }    &{2y-7}    & {~=~}    &{3y-2}    \\
{~\color{magenta}    6    }    &{\implies}    &{y}    & {~=~}    &{-5}    \\
{~\color{magenta}    7    }    &{~~~• }    &{a-1}    & {~=~}    &{-3}    \\
{~\color{magenta}    8    }    &{\implies}    &{a}    & {~=~}    &{-2}    \\
{~\color{magenta}    9    }    &{~~~• }    &{0}    & {~=~}    &{2c+2}    \\
{~\color{magenta}    {10}    }    &{\implies}    &{c}    & {~=~}    &{-1}    \\
{~\color{magenta}    {11}    }    &{~~~• }    &{b-3}    & {~=~}    &{2b+4}    \\
{~\color{magenta}    {12}    }    &{\implies}    &{b}    & {~=~}    &{-7}    \\
\end{array}                           
$

Solved example 19.5
Find the values of a, b, c and d from the following equation:
$\left[\begin{array}{r}           
2a+b    &{a-2b}        \\
5c-d    &{4c+3d}        \\
\end{array}\right]
=
\left[\begin{array}{r}           
4    &{-3}        \\
11    &{24}      \\
\end{array}\right]$
Solution:
• Given that, the two matrices are equal. So their corresponding elements will be equal.
• Thus we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{~~~• }    &{2a+b}    & {~=~}    &{4}    \\
{~\color{magenta}    2    }    &{~~~• }    &{a-2b}    & {~=~}    &{-3}    \\
{~\color{magenta}    3    }    &{(2) \times 2}    &{2a-4b}    & {~=~}    &{-6}    \\
{~\color{magenta}    4    }    &{(1) – (3)}    &{5b}    & {~=~}    &{10}    \\
{~\color{magenta}    5    }    &{\implies}    &{b}    & {~=~}    &{2}    \\
{~\color{magenta}    6    }    &{\implies}    &{a}    & {~=~}    &{1}    \\
{~\color{magenta}    7    }    &{~~~• }    &{5c-d}    & {~=~}    &{11}    \\
{~\color{magenta}    8    }    &{~~~• }    &{4c+3d}    & {~=~}    &{24}    \\
{~\color{magenta}    9    }    &{(7) \times 3}    &{15c – 3d}    & {~=~}    &{33}    \\
{~\color{magenta}    {10}    }    &{(8) + (9)}    &{19c}    & {~=~}    &{57}    \\
{~\color{magenta}    {11}    }    &{\implies}    &{c}    & {~=~}    &{3}    \\
{~\color{magenta}    {12}    }    &{\implies}    &{d}    & {~=~}    &{4}    \\
\end{array}                           
$

Solved example 19.6
Write a scalar matrix of order 4. Assume constant k to be equal to 3.2
Solution:
• Any scalar matrix will be a square matrix. In our present case, the order is to be 4. So there will be 4 rows and 4 columns. Also, in a scalar matrix, all non-diagonal elements will be zero. Thus the required matrix is:

$\left[\begin{array}{c}                
3.2    &{0}    &{0}    &{0}    \\
0    &{3.2}    &{0}    &{0}    \\
0    &{0}    &{3.2}    &{0}    \\
0    &{0}    &{0}    &{3.2}    \\
\end{array}\right]$


The link below gives a few more solved examples:

Exercise 19.1


In the next section, we will see operations on matrices. 

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