Question

Given the matrices
A = $\left[\begin{array}{ccc}2& 1& 1\\ 3& -1& 0\\ 0& 2& 4\end{array}\right]$, B = $\left[\begin{array}{ccc}9& 7& -1\\ 3& 5& 4\\ 2& 1& 6\end{array}\right]$ and C = $\left[\begin{array}{ccc}2& -4& 3\\ 1& -1& 0\\ 9& 4& 5\end{array}\right]$
Verify that (A + B) + C = A + (B + C).

Answer 1

$$\begin{gathered} \mathrm{Here}, \\ \mathrm{LHS}=\left(A+B\right)+C \\ =\left(\left[\begin{array}{ccc}2& 1& 1\\ 3& -1& 0\\ 0& 2& 4\end{array}\right]+\left[\begin{array}{ccc}9& 7& -1\\ 3& 5& 4\\ 2& 1& 6\end{array}\right]\right)+\left[\begin{array}{ccc}2& -4& 3\\ 1& -1& 0\\ 9& 4& 5\end{array}\right] \\ =\left(\left[\begin{array}{ccc}2+9& 1+7& 1-1\\ 3+3& -1+5& 0+4\\ 0+2& 2+1& 4+6\end{array}\right]\right)+\left[\begin{array}{ccc}2& -4& 3\\ 1& -1& 0\\ 9& 4& 5\end{array}\right] \\ =\left[\begin{array}{ccc}11& 8& 0\\ 6& 4& 4\\ 2& 3& 10\end{array}\right]+\left[\begin{array}{ccc}2& -4& 3\\ 1& -1& 0\\ 9& 4& 5\end{array}\right] \\ =\left[\begin{array}{ccc}11+2& 8-4& 0+3\\ 6+1& 4-1& 4+0\\ 2+9& 3+4& 10+5\end{array}\right] \\ =\left[\begin{array}{ccc}13& 4& 3\\ 7& 3& 4\\ 11& 7& 15\end{array}\right] \\ \mathrm{RHS}=A+\left(B+C\right) \\ =\left[\begin{array}{ccc}2& 1& 1\\ 3& -1& 0\\ 0& 2& 4\end{array}\right]+\left(\left[\begin{array}{ccc}9& 7& -1\\ 3& 5& 4\\ 2& 1& 6\end{array}\right]+\left[\begin{array}{ccc}2& -4& 3\\ 1& -1& 0\\ 9& 4& 5\end{array}\right]\right) \\ =\left[\begin{array}{ccc}2& 1& 1\\ 3& -1& 0\\ 0& 2& 4\end{array}\right]+\left(\left[\begin{array}{ccc}9+2& 7-4& -1+3\\ 3+1& 5-1& 4+0\\ 2+9& 1+4& 6+5\end{array}\right]\right) \\ =\left[\begin{array}{ccc}2& 1& 1\\ 3& -1& 0\\ 0& 2& 4\end{array}\right]+\left[\begin{array}{ccc}11& 3& 2\\ 4& 4& 4\\ 11& 5& 11\end{array}\right] \\ =\left[\begin{array}{ccc}2+11& 1+3& 1+2\\ 3+4& -1+4& 0+4\\ 0+11& 2+5& 4+11\end{array}\right] \\ =\left[\begin{array}{ccc}13& 4& 3\\ 7& 3& 4\\ 11& 7& 15\end{array}\right] \\ \therefore \mathrm{LHS}=\mathrm{RHS} \end{gathered}$$

Hence proved.