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Distributive 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
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}
(A + B) C = \begin{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}
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}
(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}
= \begin{bmatrix} 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}
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}
Then multiply B and C matrices.
BC = \begin{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}
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)
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 )
A(B+ C) = \begin{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}
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}
A(B+ C)= \begin{bmatrix} 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}
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}
AC = \begin{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}
Now add AB and AC
AB + AC = \begin{bmatrix} 19 & 22\\ 43 & 50 \end{bmatrix}+ \begin{bmatrix} 31 & 34\\ 71 & 78 \end{bmatrix}
=\begin{bmatrix} 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.