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Book Free DemoDistributive property:
The distributive property for the multiplication of the matrix can be classified into two segments.
If \(A\), \(B\), \(C\) are \(m×n\), \(m×n\) and \(n × p\) matrices then:
2. Right distributive property \(\rightarrow A(B+C) = AB+ AC\)
Let's go through each property individually with an example.
Left distributive property - \((A+B)C = AC+ BC\):
Example:
Consider the matrices \(A = \begin{bmatrix}
1 & 2 \\
3 & 4
\end{bmatrix}, B = \begin{bmatrix}
5 & 6\\
7 & 8
\end{bmatrix}C = \begin{bmatrix}
9 & 10\\
11 & 12
\end{bmatrix}\) and then check \((A+B)C = AC+ BC\)
1 & 2 \\
3 & 4
\end{bmatrix}, B = \begin{bmatrix}
5 & 6\\
7 & 8
\end{bmatrix}C = \begin{bmatrix}
9 & 10\\
11 & 12
\end{bmatrix}\) and then check \((A+B)C = AC+ BC\)
Solution:
Let's find the left side of the expression \((A+B)C \).
\((A + B) C = \left (\begin{bmatrix}
1 & 2 \\
3 & 4
\end{bmatrix}+ \begin{bmatrix}
5 & 6\\
7 & 8
\end{bmatrix}\right ) \begin{bmatrix}
9 & 10\\
11 & 12
\end{bmatrix}\)
1 & 2 \\
3 & 4
\end{bmatrix}+ \begin{bmatrix}
5 & 6\\
7 & 8
\end{bmatrix}\right ) \begin{bmatrix}
9 & 10\\
11 & 12
\end{bmatrix}\)
\((A + B) C = \begin{bmatrix}
1 + 5 & 2 + 6\\
3 + 7& 4 + 8
\end{bmatrix} \begin{bmatrix}
9 & 10\\
11 & 12
\end{bmatrix}\)
1 + 5 & 2 + 6\\
3 + 7& 4 + 8
\end{bmatrix} \begin{bmatrix}
9 & 10\\
11 & 12
\end{bmatrix}\)
\((A + B) C = \begin{bmatrix}
6 & 8\\
10 & 12
\end{bmatrix} \begin{bmatrix}
9 & 10\\
11 & 12
\end{bmatrix}\)
6 & 8\\
10 & 12
\end{bmatrix} \begin{bmatrix}
9 & 10\\
11 & 12
\end{bmatrix}\)
Now do the multiply \(A+B\) matrix with \(C\) matrix.
\((A + B) C = (\begin{bmatrix}
(6 × 9) + (8×11) & (6 × 10) + (8×12) \\
(10 × 9) + (12×11) & (10 × 10) + (12×12)
\end{bmatrix}\)
(6 × 9) + (8×11) & (6 × 10) + (8×12) \\
(10 × 9) + (12×11) & (10 × 10) + (12×12)
\end{bmatrix}\)
\((A + B) C = (\begin{bmatrix}
(6 × 9) + (8×11) & (6 × 10) + (8×12) \\
(10 × 9) + (12×11) & (10 × 10) + (12×12)
\end{bmatrix}\begin{bmatrix}
54 + 88 & 60 + 96\\
90 + 132 & 100 + 144
\end{bmatrix}\)
(6 × 9) + (8×11) & (6 × 10) + (8×12) \\
(10 × 9) + (12×11) & (10 × 10) + (12×12)
\end{bmatrix}\begin{bmatrix}
54 + 88 & 60 + 96\\
90 + 132 & 100 + 144
\end{bmatrix}\)
\(= \begin{bmatrix}
142 & 156\\
222 & 244
\end{bmatrix}\)……..(1)
142 & 156\\
222 & 244
\end{bmatrix}\)……..(1)
Similarly, we can find \(AC+ BC\) as follows.
\(AC = \begin{bmatrix}
1 & 2\\
3 & 4
\end{bmatrix}\begin{bmatrix}
9 & 10\\
11 & 12
\end{bmatrix}\)
1 & 2\\
3 & 4
\end{bmatrix}\begin{bmatrix}
9 & 10\\
11 & 12
\end{bmatrix}\)
\(AC = \begin{bmatrix}
(1 × 9) + (2 × 11) & (1 × 10) + (2×12)\\
(3× 9) + (4×11) & (3× 10) + (4×12)
\end{bmatrix} = \begin{bmatrix}
31 & 34\\
71& 78
\end{bmatrix}\)
(1 × 9) + (2 × 11) & (1 × 10) + (2×12)\\
(3× 9) + (4×11) & (3× 10) + (4×12)
\end{bmatrix} = \begin{bmatrix}
31 & 34\\
71& 78
\end{bmatrix}\)
Then multiply \(B\) and \(C\) matrices.
\(BC = \begin{bmatrix}
5 & 6\\
7 & 8
\end{bmatrix}\begin{bmatrix}
9 & 10\\
11 & 12
\end{bmatrix}\)
5 & 6\\
7 & 8
\end{bmatrix}\begin{bmatrix}
9 & 10\\
11 & 12
\end{bmatrix}\)
\(BC = \begin{bmatrix}
(5 × 9) + (6 × 11) & (5 × 10) + (6×12)\\
(7× 9) + (8×11) & (7× 10) + (8×12)
\end{bmatrix} = \begin{bmatrix}
111 & 122\\
151 & 166
\end{bmatrix}\)
(5 × 9) + (6 × 11) & (5 × 10) + (6×12)\\
(7× 9) + (8×11) & (7× 10) + (8×12)
\end{bmatrix} = \begin{bmatrix}
111 & 122\\
151 & 166
\end{bmatrix}\)
Now add the matrices \(AC\) and \(BC\).
\(AC + BC = \begin{bmatrix}
31 & 34\\
71& 78
\end{bmatrix} + \begin{bmatrix}
111 & 122\\
151 & 166
\end{bmatrix} = \begin{bmatrix}
142& 156\\
222&244
\end{bmatrix}\)………\((2)\)
31 & 34\\
71& 78
\end{bmatrix} + \begin{bmatrix}
111 & 122\\
151 & 166
\end{bmatrix} = \begin{bmatrix}
142& 156\\
222&244
\end{bmatrix}\)………\((2)\)
Here \((1) = (2)\)
Therefore, from the \((1)\) and \((2)\) we understand that the given matrices satisfices the left distributive property \((A+B)C = AC+ BC\) of matrix multiplication.
Right distributive property - \(A(B+C) = AB+ AC\)
Example:
Let's take the above \(A\), \(B\) and \(C\) matrices to check \(A(B+C) = AB+ AC\).
\(A (B+ C) = \begin{bmatrix}
1 & 2 \\
3 & 4
\end{bmatrix} \left (\begin{bmatrix}
5 & 6\\
7 & 8
\end{bmatrix}+ \begin{bmatrix}
9 & 10\\
11 & 12
\end{bmatrix}\right )\)
1 & 2 \\
3 & 4
\end{bmatrix} \left (\begin{bmatrix}
5 & 6\\
7 & 8
\end{bmatrix}+ \begin{bmatrix}
9 & 10\\
11 & 12
\end{bmatrix}\right )\)
\(A(B+ C) = \begin{bmatrix}
1 & 2 \\
3& 4
\end{bmatrix} \begin{bmatrix}
5 +9 &6 + 10\\
7 + 11 & 8 + 12
\end{bmatrix}\)
1 & 2 \\
3& 4
\end{bmatrix} \begin{bmatrix}
5 +9 &6 + 10\\
7 + 11 & 8 + 12
\end{bmatrix}\)
\(A(B+ C) = \begin{bmatrix}
1 & 2 \\
3 & 4
\end{bmatrix} \begin{bmatrix}
14 & 16\\
18 & 20
\end{bmatrix}\)
1 & 2 \\
3 & 4
\end{bmatrix} \begin{bmatrix}
14 & 16\\
18 & 20
\end{bmatrix}\)
\(A(B+ C) = \begin{bmatrix}
(1 × 14) + (2 × 18) & (1 × 16) + (2×20)\\
(3× 14) + (4×18) & (3× 16) + (4×20)
\end{bmatrix} = \begin{bmatrix}
14 + 36 & 16 + 40\\
42 + 72 & 48+80
\end{bmatrix}\)
(1 × 14) + (2 × 18) & (1 × 16) + (2×20)\\
(3× 14) + (4×18) & (3× 16) + (4×20)
\end{bmatrix} = \begin{bmatrix}
14 + 36 & 16 + 40\\
42 + 72 & 48+80
\end{bmatrix}\)
\(A(B+ C)= \begin{bmatrix}
50 & 56\\
114 & 128
\end{bmatrix}\)……..(1)
50 & 56\\
114 & 128
\end{bmatrix}\)……..(1)
Similarly, let's find \(AB+ AC\).
\(AB = \begin{bmatrix}
1 & 2\\
3 & 4
\end{bmatrix}\begin{bmatrix}
5 & 6\\
7 & 8
\end{bmatrix}\)
1 & 2\\
3 & 4
\end{bmatrix}\begin{bmatrix}
5 & 6\\
7 & 8
\end{bmatrix}\)
\(AB = \begin{bmatrix}
(1 × 5) + (2 × 7) & (1 × 6) + (2×8)\\
(3× 5) + (4×7) & (3× 6) + (4×8)
\end{bmatrix} = \begin{bmatrix}
19 & 22\\
43 & 50
\end{bmatrix}\)
(1 × 5) + (2 × 7) & (1 × 6) + (2×8)\\
(3× 5) + (4×7) & (3× 6) + (4×8)
\end{bmatrix} = \begin{bmatrix}
19 & 22\\
43 & 50
\end{bmatrix}\)
\(AC = \begin{bmatrix}
1 & 2\\
3 & 4
\end{bmatrix}\begin{bmatrix}
9 & 10\\
11 & 12
\end{bmatrix}\)
1 & 2\\
3 & 4
\end{bmatrix}\begin{bmatrix}
9 & 10\\
11 & 12
\end{bmatrix}\)
\(AC = \begin{bmatrix}
(1 × 9) + (2 × 11) & (1 × 10) + (2×12)\\
(3× 9) + (4×11) & (3× 10) + (4×12)
\end{bmatrix} = \begin{bmatrix}
31 & 34\\
71& 78
\end{bmatrix}\)
(1 × 9) + (2 × 11) & (1 × 10) + (2×12)\\
(3× 9) + (4×11) & (3× 10) + (4×12)
\end{bmatrix} = \begin{bmatrix}
31 & 34\\
71& 78
\end{bmatrix}\)
Now add \(AB\) and \(AC\)
\(AB + AC = \begin{bmatrix}
19 & 22\\
43 & 50
\end{bmatrix}+ \begin{bmatrix}
31 & 34\\
71 & 78
\end{bmatrix}\)
19 & 22\\
43 & 50
\end{bmatrix}+ \begin{bmatrix}
31 & 34\\
71 & 78
\end{bmatrix}\)
\(=\begin{bmatrix}
48 & 56\\
114 & 128
\end{bmatrix}\)…….(2)
48 & 56\\
114 & 128
\end{bmatrix}\)…….(2)
Here \((1) = (2)\)
Therefore, from the \((1)\) and \((2)\) we understand that the given matrices satisfies the right distributive property \((A(B+C) = AB+ AC)\) of matrix multiplication.
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