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