UPSKILL MATH PLUS
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Learn moreMatrices also follow the certain properties such as whole numbers and integers. Generally, there are four properties of a matrix in terms of addition and scalar multiplication as follows:
1. Commutative property
2. Associative property
3. Scalar identity for unit matrix
4. Distributive property
Let's dive into each property individually with an example.
Let \(A\), \(B\), \(C\) be \(m×n\) matrices and \(p\) and \(q\) be two non-zero scalars (numbers). Then we have the following properties.
Commutative property of matrix addition:
Changing the order of the matrices does not change the result of the matrices.
\(A + B = B + A\)
Example:
Consider the matrices \( A = \begin{bmatrix}
1 & 2 \\
3 & 4
\end{bmatrix}, B = \begin{bmatrix}
5 & 6\\
7 & 8
\end{bmatrix}\) then verify that \(A + B = B + A\)
1 & 2 \\
3 & 4
\end{bmatrix}, B = \begin{bmatrix}
5 & 6\\
7 & 8
\end{bmatrix}\) then verify that \(A + B = B + A\)
Solution:
First, we find the sum of \(A + B\) matrices.
\( A + B = \begin{bmatrix}
1 & 2 \\
3 & 4
\end{bmatrix}+ \begin{bmatrix}
5 & 6\\
7 & 8
\end{bmatrix} = \begin{bmatrix}
1+5 & 2+6\\
3+7 & 4+8
\end{bmatrix}=\begin{bmatrix}
6& 8\\
10& 12
\end{bmatrix}\)………..(1)
1 & 2 \\
3 & 4
\end{bmatrix}+ \begin{bmatrix}
5 & 6\\
7 & 8
\end{bmatrix} = \begin{bmatrix}
1+5 & 2+6\\
3+7 & 4+8
\end{bmatrix}=\begin{bmatrix}
6& 8\\
10& 12
\end{bmatrix}\)………..(1)
Similarly, let's find \(B + A\).
\( B + A = \begin{bmatrix}
5 & 6 \\
7 & 8
\end{bmatrix}+ \begin{bmatrix}
1 & 2\\
3 & 4
\end{bmatrix} = \begin{bmatrix}
5+1 & 6+2\\
7+3 & 8+4
\end{bmatrix}=\begin{bmatrix}
6& 8\\
10& 12
\end{bmatrix}\)………..(2)
5 & 6 \\
7 & 8
\end{bmatrix}+ \begin{bmatrix}
1 & 2\\
3 & 4
\end{bmatrix} = \begin{bmatrix}
5+1 & 6+2\\
7+3 & 8+4
\end{bmatrix}=\begin{bmatrix}
6& 8\\
10& 12
\end{bmatrix}\)………..(2)
From \((1), (2)\) we can see that \(A + B = B + A\). Thus, the given matrices are satisfy the Commutative property of matrices.
Associative property:
1. Associative property of matrix addition:
It says that rearranging parenthesis in the matrix expression will not change the result of the matrices.
\((A + B)+ C = A + (B+C)\)
Example:
If \( A = \begin{bmatrix}
1 & 2 \\
3 & 4
\end{bmatrix}, B = \begin{bmatrix}
5 & 6\\
7 & 8
\end{bmatrix}, C = \begin{bmatrix}
9 & 10\\
11 & 12
\end{bmatrix}\) then verify that \((A + B)+ C = A + (B+C)\).
1 & 2 \\
3 & 4
\end{bmatrix}, B = \begin{bmatrix}
5 & 6\\
7 & 8
\end{bmatrix}, C = \begin{bmatrix}
9 & 10\\
11 & 12
\end{bmatrix}\) then verify that \((A + B)+ C = A + (B+C)\).
Solution:
First we find the sum of \(A + B\) matrices then add it's result with the \(C) matrix.
\((A + B) + C= \left (\begin{bmatrix}
1 & 2\\
3 & 4
\end{bmatrix}+ \begin{bmatrix}
5 & 6\\
7 & 8
\end{bmatrix}\right ) + \begin{bmatrix}
9 & 10\\
11 & 12
\end{bmatrix}\)
1 & 2\\
3 & 4
\end{bmatrix}+ \begin{bmatrix}
5 & 6\\
7 & 8
\end{bmatrix}\right ) + \begin{bmatrix}
9 & 10\\
11 & 12
\end{bmatrix}\)
\(=\begin{bmatrix}
1+5 & 2+6\\
3+7 & 4+8
\end{bmatrix}+ \begin{bmatrix}
9 & 10\\
11 & 12
\end{bmatrix}\)
1+5 & 2+6\\
3+7 & 4+8
\end{bmatrix}+ \begin{bmatrix}
9 & 10\\
11 & 12
\end{bmatrix}\)
\(=\begin{bmatrix}
6 & 8\\
10 & 12
\end{bmatrix}+ \begin{bmatrix}
9 & 10\\
11 & 12
\end{bmatrix}\)
6 & 8\\
10 & 12
\end{bmatrix}+ \begin{bmatrix}
9 & 10\\
11 & 12
\end{bmatrix}\)
\(=\begin{bmatrix}
6+9 & 8+10\\
10+11 & 12+12
\end{bmatrix}= \begin{bmatrix}
15 & 18\\
21 & 24
\end{bmatrix}\)……….(1)
6+9 & 8+10\\
10+11 & 12+12
\end{bmatrix}= \begin{bmatrix}
15 & 18\\
21 & 24
\end{bmatrix}\)……….(1)
Similarly, let's find \(A + (B + C)\).
\(A + ( B + C) =\begin{bmatrix}
1 & 2\\
3 & 4
\end{bmatrix}+ \begin{bmatrix}
5+9 & 6+10\\
7+11 & 8+12
\end{bmatrix}\)
1 & 2\\
3 & 4
\end{bmatrix}+ \begin{bmatrix}
5+9 & 6+10\\
7+11 & 8+12
\end{bmatrix}\)
\( =\begin{bmatrix}
1 & 2\\
3 & 4
\end{bmatrix}+ \begin{bmatrix}
14 & 16\\
18 & 20
\end{bmatrix}\)
1 & 2\\
3 & 4
\end{bmatrix}+ \begin{bmatrix}
14 & 16\\
18 & 20
\end{bmatrix}\)
\(= \begin{bmatrix}
1+14 & 2+16\\
18+3 & 4+20
\end{bmatrix}=\begin{bmatrix}
15 & 18\\
21 & 24
\end{bmatrix}\)……….(2)
1+14 & 2+16\\
18+3 & 4+20
\end{bmatrix}=\begin{bmatrix}
15 & 18\\
21 & 24
\end{bmatrix}\)……….(2)
2. Associative property of scalar multiplication - \((pq)A = p(Aq)\)
Example:
Verify the associative property if \( A = \begin{bmatrix}
2 & 4\\
6 & 8
\end{bmatrix}\) and \( p = 4\) and \(q = 6\).
2 & 4\\
6 & 8
\end{bmatrix}\) and \( p = 4\) and \(q = 6\).
Solution:
We know the associative property of scalar multiplication is \((pq)A = p(Aq)\). Here, \(p = 4\) and \(q = 6\).
So, \((4 × 6) A = (4 × 6) \begin{bmatrix}
2 & 4\\
6 & 8
\end{bmatrix}\)
2 & 4\\
6 & 8
\end{bmatrix}\)
\(= 24\begin{bmatrix}
2 & 4\\
6 & 8
\end{bmatrix} = \begin{bmatrix}
48 & 96\\
144 & 192
\end{bmatrix}\)……….(1)
2 & 4\\
6 & 8
\end{bmatrix} = \begin{bmatrix}
48 & 96\\
144 & 192
\end{bmatrix}\)……….(1)
Similarly, let's find \(4( A × 6) = 4 (6 ×\begin{bmatrix}
2 & 4\\
6 & 8
\end{bmatrix})\)
2 & 4\\
6 & 8
\end{bmatrix})\)
\(= 4 × \begin{bmatrix}
12 & 24\\
36 & 48
\end{bmatrix} = \begin{bmatrix}
48 & 96\\
144 & 192
\end{bmatrix}\)……..(2)
12 & 24\\
36 & 48
\end{bmatrix} = \begin{bmatrix}
48 & 96\\
144 & 192
\end{bmatrix}\)……..(2)
From the equations \((1)\) and \((2)\), we can see that \((pq)A = p(Aq)\).
Hence, the given matrix follows the associative property of scalar multiplication.
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