Question

Calculate AC, BC and (A+B)C if A = $\left[ \begin{align}
  & \text{0 6 7} \\
 & \text{-6 0 8} \\
 & \text{7 -8 0} \\
\end{align} \right]$ , B = $\left[ \begin{align}
  & \text{0 1 1} \\
 & \text{1 0 2} \\
 & \text{1 2 0} \\
\end{align} \right]$ , C = $\left[ \begin{align}
  & {2} \\
 & {-2} \\
 & {3} \\
\end{align} \right]$. Also, verify that (A+B) C = AC + BC.

Answer 1

Hint: To solve this problem, we use the basic properties of matrix operations (in this case addition and multiplication). We will then use this to prove (A+B) C = AC + BC.

Complete step-by-step answer:
We solve this problem by understanding how we can perform the addition operation first. For a 3$\times$3 matrix, we add two matrices by adding the corresponding entries of both matrices together. One should keep in mind that both the matrices we add should have the same number of rows and columns. This is explained through the example below -
$\left[ \begin{align}
  & \text{0 6 7} \\
 & \text{-6 0 8} \\
 & \text{7 -8 0} \\
\end{align} \right]$ + $\left[ \begin{align}
  & \text{0 6 7} \\
 & \text{-6 0 8} \\
 & \text{7 -8 0} \\
\end{align} \right]$ = $\left[ \begin{align}
  & \text{0 12 14} \\
 & \text{-12 0 16} \\
 & \text{14 -16 0} \\
\end{align} \right]$ -- (1)
Now, we also understand how to perform matrix multiplication operations. In this case, since we will always multiply 3 $\times $ 3 and 3 $\times $ 1 together, we use the following formula, we have,
$\left[ \begin{align}
  & \text{a b c} \\
 & \text{d e f} \\
 & \text{g h i} \\
\end{align} \right]$$\left[ \begin{align}
  & x \\
 & y \\
 & z \\
\end{align} \right]$ = \[\left[ \begin{align}
  & \left( ax+by+cz \right) \\
 & \left( dx+ey+fz \right) \\
 & \left( gx+hy+iz \right) \\
\end{align} \right]\] -- (2)
We will use this to multiply any two matrices and compare the numbers in these matrices with a, b, c, d, e, f, g, h and i.
Now, we calculate AC using the formula given by (2), we have,
=$\left[ \begin{align}
  & \text{0 6 7} \\
 & \text{-6 0 8} \\
 & \text{7 -8 0} \\
\end{align} \right]$$\left[ \begin{align}
  & {2} \\
 & {-2} \\
 & {3} \\
\end{align} \right]$= \[\left[ \begin{align}
  & 9 \\
 & 12 \\
 & 30 \\
\end{align} \right]\] -- (A)
Also, we similarly calculate BC, we get,
=$\left[ \begin{align}
  & \text{0 1 1} \\
 & \text{1 0 2} \\
 & \text{1 2 0} \\
\end{align} \right]$$\left[ \begin{align}
  & {2} \\
 & {-2} \\
 & {3} \\
\end{align} \right]$= \[\left[ \begin{align}
  & {1} \\
 & {8} \\
 & {-2} \\
\end{align} \right]\] -- (B)
We now calculate (A+B) C, we first start with calculating (A+B), we do this by calculating by using the formula given by (1), we have,
$\left[ \begin{align}
  & \text{0 6 7} \\
 & \text{-6 0 8} \\
 & \text{7 -8 0} \\
\end{align} \right]$ + $\left[ \begin{align}
  & \text{0 1 1} \\
 & \text{1 0 2} \\
 & \text{1 2 0} \\
\end{align} \right]$ = $\left[ \begin{align}
  & \text{0 7 8} \\
 & \text{-5 0 10} \\
 & \text{ 8 -6 0} \\
\end{align} \right]$
Now, we multiply this matrix by matrix C. We have,
= $\left[ \begin{align}
  & \text{0 7 8} \\
 & \text{-5 0 10} \\
 & \text{ 8 -6 0} \\
\end{align} \right]$$\left[ \begin{align}
  & {2} \\
 & {-2} \\
 & {3} \\
\end{align} \right]$ = $\left[ \begin{align}
  & 10 \\
 & 20 \\
 & 28 \\
\end{align} \right]$ -- (C)
Now, to prove that (A+B) C = AC + BC, on the LHS, we have (by (C)) $\left[ \begin{align}
  & 10 \\
 & 20 \\
 & 28 \\
\end{align} \right]$ .
Now, on the RHS, we have AC + BC, thus we add the matrices obtained from (1) and (2), thus, we have,
= AC + BC
= \[\left[ \begin{align}
  & 9 \\
 & 12 \\
 & 30 \\
\end{align} \right]\] + \[\left[ \begin{align}
  & {1} \\
 & {8} \\
 & {-2} \\
\end{align} \right]\]
= \[\left[ \begin{align}
  & 10 \\
 & 20 \\
 & 28 \\
\end{align} \right]\] -- (D)
Now, from (C) and (D), we can conclude that LHS = RHS.

Note: While performing multiplication of two matrices, we should remember that the number of columns of the first matrix should be the same as the number of rows of the second matrix. Similarly, while addition, we should keep in mind that both the matrices should have the same number of rows and columns. If these conditions cannot be met, we cannot perform the required operations.