Matrix Operations

Before we learn about the matrix operations, let's first understand what a matrix is. A matrix is a rectangular arrangement of symbols along rows and columns that might be real or complex numbers. Matrix operations mainly include three algebraic operations namely, the addition of matrices, subtraction of matrices, and multiplication of matrices. Matrix analysis is used in the study of optics to account for reflection and refraction. Matrix analysis is also useful in quantum physics, electrical circuits, and resistor conversion of electrical energy.

This Story also Contains

1. Operations on Matrices
2. Addition of matrices:
3. Subtraction of matrices:
4. Scalar multiplication:
5. Matrix multiplication:
6. Solved Examples Based On Matrix Operations

Below is an example of a matrix structure of 3 rows and 4 columns:

$\left[\begin{array}{llll}a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34}\end{array}\right]_{3 \times 4}$

In general form, the above matrix is represented by $A=\left[a_{i j}\right]$

$a_{11}$, $a_{12}$,.. etc. are called the elements of the matrix.

$a_{ij}$ belongs to the ith row and jth column and is called the $(i,j)$ th element of the matrix.

In this article, we will cover the concept of operation on matrices. This category falls under the broader category of matrices, which is a crucial Chapter in class 12 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE and more.

Operations on Matrices

The addition, subtraction, and multiplication of matrices are the three basic algebraic matrix operations.

Condition for performing Addition and Subtraction operations:

• The order of the matrix should be identical for performing addition and subtraction operations.

Condition for performing Multiplication operations

• The first matrix's number of rows and the second matrix's number of columns should be the same.

Addition of matrices:

Two matrices can be added only when they are of the same order

If two matrices of A and B are of the same order, they are said to be conformable for addition.

If A and B are matrices of order m × n, then their sum will also be a matrix of the same order and in addition, corresponding elements of A and B get added.

So if $A=\left[a_{i j}\right]_{m \times n}, B=\left[b_{i j}\right]_{m \times n}$ Then, $A+B=\left[a_{i j}+b_{i j}\right]_{m \times n}$ for all $\mathrm{i}, \mathrm{j}$
Example:
$
\begin{aligned}
\mathrm{A} & =\left[\begin{array}{lll}
10 & 20 & 30 \\
20 & 30 & 40 \\
30 & 40 & 50
\end{array}\right], \quad \mathrm{B}=\left[\begin{array}{lll}
50 & 40 & 30 \\
40 & 30 & 20 \\
30 & 20 & 10
\end{array}\right] \\
\mathrm{A}+\mathrm{B} & =\left[\begin{array}{lll}
10+50 & 20+40 & 30+30 \\
20+40 & 30+30 & 40+20 \\
30+30 & 40+20 & 50+10
\end{array}\right]=\left[\begin{array}{lll}
60 & 60 & 60 \\
60 & 60 & 60 \\
60 & 60 & 60
\end{array}\right]
\end{aligned}
$

Properties of matrix addition:

i) Matrix addition is commutative, A + B = B + A

ii) Matrix addition is associative, A + (B+C) = (A+B) + C

iii) Additive identity exists, which means there exists a matrix O (null matrix) such that A + O = A = O + A (Here O has the same order as A)

iv) Existence of additive inverse means there exists a matrix B such that A + B = O = B + A

v) Cancellation property:

If A + B = A + C then B = C

If A + C = B + C then A = B

Note: All matrices taken in the above property explanation have the same order which is m × n.

Subtraction of matrices:

Two matrices can be subtracted only when they are of the same order. If A and B are matrices of order m × n then their difference will also be a matrix of the same order and in subtraction, corresponding elements of A and B get subtracted. So if

$
\mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right]_{\mathrm{m} \times \mathrm{n}}, \mathrm{B}=\left[\mathrm{b}_{\mathrm{ij}}\right]_{\mathrm{m} \times \mathrm{n}} \text { Then, } \mathrm{A}-\mathrm{B}=\left[\mathrm{a}_{\mathrm{ij}}-\mathrm{b}_{\mathrm{ij}}\right]_{\mathrm{m} \times \mathrm{n} \text { for all } \mathrm{i}, \mathrm{j}}
$

Example:
$
\begin{aligned}
\mathrm{A} & =\left[\begin{array}{lll}
10 & 20 & 30 \\
20 & 30 & 40 \\
30 & 40 & 50
\end{array}\right], \quad \mathrm{B}=\left[\begin{array}{lll}
50 & 40 & 30 \\
40 & 30 & 20 \\
30 & 20 & 10
\end{array}\right] \\
\mathrm{A}-\mathrm{B} & =\left[\begin{array}{lll}
10-50 & 20-40 & 30-30 \\
20-40 & 30-30 & 40-20 \\
30-30 & 40-20 & 50-10
\end{array}\right]=\left[\begin{array}{ccc}
-40 & -20 & 0 \\
-20 & 0 & 20 \\
0 & 20 & 40
\end{array}\right]
\end{aligned}
$

Properties of matrix Subtraction:

i) Matrix subtraction is not commutative, $A-B \neq B-A$
ii) Matrix subtraction is not associative, $A-(B-C) \neq(A-B)-C$

iii) Cancellation property:

If A - B = A - C then B = C

If A - C = B - C then A = B

Note: All matrices taken in the above property explanation have the same order which is m × n.

Scalar multiplication:

Let $\mathrm{k}$ be any scalar number, and $A=\left[a_{i j}\right]_{m \times n}$ be a matrix. Then the matrix is obtained by multiplying every element $\mathrm{A}$ by a scalar $\mathrm{k}$ and denoted as kA.
$
\begin{aligned}
& k A=\left[k a_{i j}\right]_{m \times n} \\
& \qquad \mathrm{~A}=\left[\begin{array}{ll}
2 & 6 \\
3 & 7 \\
5 & 8
\end{array}\right] \text { then, } 3 \mathrm{~A}=\left[\begin{array}{ll}
3 \times 2 & 3 \times 6 \\
3 \times 3 & 3 \times 7 \\
3 \times 5 & 3 \times 8
\end{array}\right]=\left[\begin{array}{cc}
6 & 18 \\
9 & 21 \\
15 & 24
\end{array}\right]
\end{aligned}
$

Properties of scalar multiplication:

If $A$ and $B$ are two matrices and $k, l$ are scalar then
i) $k(A+B)=k A+k B$
ii) $k l(A)=k(I A)=l(k A)$
iii) $(k+I) A=k A+I A$
iv) $(-k) A=-(k A)=k(-A)$
v) $1 \mathrm{~A}=\mathrm{A},(-1) \mathrm{A}=-\mathrm{A}$

Note: $A$ and $B$ have the same order $m \times n$.

Matrix multiplication:

Product AB can be found if the number of columns in matrix A and the number of Product $A B$ can be found if the number of columns in matrix $A$ and the number of rows in matrix B are equal. Otherwise, multiplication AB is not possible.

i) $A B$ is defined only if $\operatorname{col}(A)=\operatorname{row}(B)$
ii) $B A$ is defined only if $\operatorname{col}(B)=\operatorname{row}(A)$

If $\begin{aligned} & \mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right]_{\mathrm{m} \times \mathrm{n}} \\ & \mathrm{B}=\left[\mathrm{b}_{\mathrm{ij}}\right]_{\mathrm{n} \times \mathrm{p}} \\ & \mathrm{C}=\mathrm{AB}=\left[\mathrm{c}_{\mathrm{ij}}\right]_{\mathrm{m} \times \mathrm{p}} \\ & \text { Where } c_{\mathrm{ij}}=\sum_{\mathrm{j}=1}^{\mathrm{n}} \mathrm{a}_{\mathrm{ij}} \mathrm{b}_{\mathrm{jk}}, 1 \leq \mathrm{i} \leq \mathrm{m}, 1 \leq \mathrm{k} \leq \mathrm{p} \\ & =a_{i 1} b_{1 k}+a_{i 2} b_{2 k}+a_{i 3} b_{3 k}+\ldots+a_{i n} b_{n k} \\ & \end{aligned}$

Properties of matrix multiplication:

i) Multiplication may or may not be commutative, so AB may or may not be equal to BA.
ii) Matrix multiplication is associative, meaning $A(B C)=(A B) C$
iii) Matrix multiplication is distributive over addition, mean $A(B+C)=A B+A C$ and $(B+C) A=B A+C A$
iv) If matrix multiplication of two matrices gives a null matrix then it doesn't mean that any of those two matrices was a null matrix.

Example:
$
A=\left[\begin{array}{ll}
0 & 2 \\
0 & 0
\end{array}\right] \text { and } B=\left[\begin{array}{ll}
1 & 0 \\
0 & 0
\end{array}\right] \text {, then } A B=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]
$
v) Matrix multiplication $A \times A$ is represented by $A^2$. Thus, $A \cdot A \cdot A \cdot A \ldots \ldots . . n$ times $=A^n$.
vi) if $A$ is $m \times n$ matrix then, $I_m A=A=A I_n$.

Recommended Video Based on Matrix Operations:

Solved Examples Based On Matrix Operations

Example 1: Find A+B if
$
A=\left[\begin{array}{rrr}
-3 & 2 & 4 \\
8 & 3 & 4
\end{array}\right], B=\left[\begin{array}{lll}
4 & 1 & 5 \\
1 & 0 & 2
\end{array}\right]
$

Solution:
As the orders of the matrices A and B are the same $(2 \times 3)$, we can add them
$
\begin{aligned}
& A+B=\left[\begin{array}{rrr}
-3+4 & 2+1 & 4+5 \\
8+1 & 3+0 & 4+2
\end{array}\right] \\
& A+B=\left[\begin{array}{lll}
1 & 3 & 9 \\
9 & 3 & 6
\end{array}\right]
\end{aligned}
$

Hence, the value of
$
A+B \text { is }\left[\begin{array}{lll}
1 & 3 & 9 \\
9 & 3 & 6
\end{array}\right]
$

Example 2: Find A - B if
$
A=\left[\begin{array}{lll}
8 & 6 & 5 \\
5 & 6 & 1
\end{array}\right], B=\left[\begin{array}{lll}
5 & 3 & 4 \\
2 & 4 & 0
\end{array}\right]
$

Solution:
As the order of both matrices are same $(2 \times 3)$, we can subtract them
$
\begin{aligned}
& A=\left[\begin{array}{lll}
8 & 6 & 5 \\
5 & 6 & 1
\end{array}\right], B=\left[\begin{array}{lll}
5 & 3 & 4 \\
2 & 4 & 0
\end{array}\right] \\
& A-B=\left[\begin{array}{lll}
8-5 & 6-3 & 5-4 \\
5-2 & 6-4 & 1-0
\end{array}\right] \\
& A-B=\left[\begin{array}{lll}
3 & 3 & 1 \\
3 & 2 & 1
\end{array}\right]
\end{aligned}
$

Hence, the value of
$
A-B \text { is }\left[\begin{array}{lll}
3 & 3 & 1 \\
3 & 2 & 1
\end{array}\right]
$

Example 3: If $\mathrm{X}$ and $\mathrm{Y}$ are two matrices such that
$
X+2 Y=\left[\begin{array}{ll}
5 & 2 \\
8 & 9
\end{array}\right]_{\text {and }}
$
$X-Y=\left[\begin{array}{cc}2 & -1 \\ 2 & 0\end{array}\right]$, then find the matrix $\mathbf{Y}$.
Solution:
Subtract both the given matrices
$
\begin{aligned}
& (X+2 Y)-(X-Y)=\left[\begin{array}{ll}
5 & 2 \\
8 & 9
\end{array}\right]-\left[\begin{array}{cc}
2 & -1 \\
2 & 0
\end{array}\right] \\
& \Rightarrow 3 Y=\left[\begin{array}{ll}
3 & 3 \\
6 & 9
\end{array}\right] \\
& \Rightarrow Y=\left[\begin{array}{ll}
1 & 1 \\
2 & 3
\end{array}\right]
\end{aligned}
$

Hence, the matrix $\mathrm{Y}$ is $\left[\begin{array}{ll}1 & 1 \\ 2 & 3\end{array}\right]$
Example 4: If a matrix $B$ of $3 \times 2$ order is multiplied by a scalar $\lambda$ then how many times will $\lambda$ be multiplied in the matrix?

Solution:
Scalar multiplication of matrix -
$
\lambda A=\left[\lambda a_{i j}\right]
$
- wherein
$\lambda$ is multiplied by every element of the matrix $A$
Since there are 6 elements it will be multiplied 6 times.
Hence, $\lambda$ be multiplied 6 times in the matrix.

Example 5: The matrix $A^2+4 A-5 I$, where $I$ is identity matrix and $A=\left[\begin{array}{cc}1 & 2 \\ 4 & -3\end{array}\right]$ equals :

Solution:
$
\begin{aligned}
& A^2+4 A-5 I=A \times A+4 A-5 I \\
& =\left[\begin{array}{cc}
1 & 2 \\
4 & -3
\end{array}\right] \times\left[\begin{array}{cc}
1 & 2 \\
4 & -3
\end{array}\right]+4\left[\begin{array}{cc}
1 & 2 \\
4 & -3
\end{array}\right]-5\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \\
& =\left[\begin{array}{cc}
9 & -4 \\
-8 & 17
\end{array}\right]+\left[\begin{array}{cc}
4 & 8 \\
16 & -12
\end{array}\right]-\left[\begin{array}{ll}
5 & 0 \\
0 & 5
\end{array}\right] \\
& =\left[\begin{array}{cc}
9+4-5 & -4+8-0 \\
-8+16-0 & 17-12-5
\end{array}\right]=\left[\begin{array}{ll}
8 & 4 \\
8 & 0
\end{array}\right] \\
& \text { Hence, the required answer is }\left[\begin{array}{ll}
8 & 4 \\
8 & 0
\end{array}\right]
\end{aligned}
$