Given the matrices
Question:

Given the matrices

$A=\left[\begin{array}{rrr}2 & 1 & 1 \\ 3 & -1 & 0 \\ 0 & 2 & 4\end{array}\right], B=\left[\begin{array}{rrr}9 & 7 & -1 \\ 3 & 5 & 4 \\ 2 & 1 & 6\end{array}\right]$ and $C=\left[\begin{array}{rrr}2 & -4 & 3 \\ 1 & -1 & 0 \\ 9 & 4 & 5\end{array}\right]$

Verify that (A + B) + C = A + (B + C).

Solution:

Here,

LHS $=(A+B)+C$

$=\left(\left[\begin{array}{ccc}2 & 1 & 1 \\ 3 & -1 & 0 \\ 0 & 2 & 4\end{array}\right]+\left[\begin{array}{ccc}9 & 7 & -1 \\ 3 & 5 & 4 \\ 2 & 1 & 6\end{array}\right]\right)+\left[\begin{array}{ccc}2 & -4 & 3 \\ 1 & -1 & 0 \\ 9 & 4 & 5\end{array}\right]$

$=\left(\left[\begin{array}{cc}2+9 & 1+7 & 1-1 \\ 3+3 & -1+5 & 0+4 \\ 0+2 & 2+1 & 4+6\end{array}\right]\right)+\left[\begin{array}{ccc}2 & -4 & 3 \\ 1 & -1 & 0 \\ 9 & 4 & 5\end{array}\right]$

$=\left[\begin{array}{ccc}11 & 8 & 0 \\ 6 & 4 & 4 \\ 2 & 3 & 10\end{array}\right]+\left[\begin{array}{ccc}2 & -4 & 3 \\ 1 & -1 & 0 \\ 9 & 4 & 5\end{array}\right]$

$=\left[\begin{array}{cc}11+2 & 8-4 & 0+3 \\ 6+1 & 4-1 & 4+0 \\ 2+9 & 3+4 & 10+5\end{array}\right]$

$=\left[\begin{array}{ccc}13 & 4 & 3 \\ 7 & 3 & 4 \\ 11 & 7 & 15\end{array}\right]$

$\mathrm{RHS}=A+(B+C)$

$=\left[\begin{array}{ccc}2 & 1 & 1 \\ 3 & -1 & 0 \\ 0 & 2 & 4\end{array}\right]+\left(\left[\begin{array}{ccc}9 & 7 & -1 \\ 3 & 5 & 4 \\ 2 & 1 & 6\end{array}\right]+\left[\begin{array}{ccc}2 & -4 & 3 \\ 1 & -1 & 0 \\ 9 & 4 & 5\end{array}\right]\right)$

$=\left[\begin{array}{ccc}2 & 1 & 1 \\ 3 & -1 & 0 \\ 0 & 2 & 4\end{array}\right]+\left(\left[\begin{array}{cc}9+2 & 7-4 & -1+3 \\ 3+1 & 5-1 & 4+0 \\ 2+9 & 1+4 & 6+5\end{array}\right]\right)$

$=\left[\begin{array}{ccc}2 & 1 & 1 \\ 3 & -1 & 0 \\ 0 & 2 & 4\end{array}\right]+\left[\begin{array}{ccc}11 & 3 & 2 \\ 4 & 4 & 4 \\ 11 & 5 & 11\end{array}\right]$

$=\left[\begin{array}{ccc}2+11 & 1+3 & 1+2 \\ 3+4 & -1+4 & 0+4 \\ 0+11 & 2+5 & 4+11\end{array}\right]$

$=\left[\begin{array}{ccc}13 & 4 & 3 \\ 7 & 3 & 4 \\ 11 & 7 & 15\end{array}\right]$

$\therefore \mathrm{LHS}=\mathrm{RHS}$

Hence proved.

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