Question
If A = $\left[\begin{array}{ccc} {-1} & {2} & {3} \\ {5} & {7} & {9} \\ {-2} & {1} & {1} \end{array}\right]$ and B = $\left[\begin{array}{rrr} {-4} & {1} & {-5} \\ {1} & {2} & {0} \\ {1} & {3} & {1} \end{array}\right]$, then verify (A + B)′ = A′ + B′,

Answer

A = $\left[\begin{array}{ccc} {-1} & {2} & {3} \\ {5} & {7} & {9} \\ {-2} & {1} & {1} \end{array}\right] \text { and } B=\left[\begin{array}{ccc} {-4} & {1} & {-5} \\ {1} & {2} & {0} \\ {1} & {3} & {1} \end{array}\right]$
(A+B)’ = A’+B’
Explanation: We will first calculate L.H.S i.e. (A+B)’ and then consecutively we will calculate R.H.S and verify that both are equal.
So, A + B = $\left[\begin{array}{ccc} {-1} & {2} & {3} \\ {5} & {7} & {9} \\ {-2} & {1} & {1} \end{array}\right]+\left[\begin{array}{ccc} {-4} & {1} & {-5} \\ {1} & {2} & {0} \\ {1} & {3} & {1} \end{array}\right]$
$\Rightarrow A+B=\left[\begin{array}{ccc} {-1+(-4)} & {2+1} & {3+(-5)} \\ {5+1} & {7+2} & {9+0} \\ {-2+1} & {1+3} & {1+1} \end{array}\right]$
$\Rightarrow A+B=\left[\begin{array}{ccc} {-5} & {3} & {-2} \\ {6} & {9} & {9} \\ {-1} & {4} & {2} \end{array}\right]$
Therefore, $(A+B)^\prime=\left[\begin{array}{ccc} {-5} & {6} & {-1} \\ {3} & {9} & {4} \\ {-2} & {9} & {2} \end{array}\right] $ ...(1)
Noe, $A^{\prime}=\left[\begin{array}{ccc} {-1} & {5} & {-2} \\ {2} & {7} & {1} \\ {3} & {9} & {1} \end{array}\right] \text { and } B^{\prime}=\left[\begin{array}{ccc} {-4} & {1} & {1} \\ {1} & {2} & {3} \\ {-5} & {0} & {1} \end{array}\right]$
So, A' + B' = $\left[\begin{array}{ccc} {-1} & {5} & {-2} \\ {2} & {7} & {1} \\ {3} & {9} & {1} \end{array}\right]+\left[\begin{array}{ccc} {-4} & {1} & {1} \\ {1} & {2} & {3} \\ {-5} & {0} & {1} \end{array}\right]$
$\Rightarrow \mathrm{A}^{\prime}+\mathrm{B}^{\prime}=\left[\begin{array}{ccc} {-1+(-4)} & {5+1} & {-2+1} \\ {2+1} & {7+2} & {1+3} \\ {3+(-5)} & {9+0} & {1+1} \end{array}\right]$
$\Rightarrow A^{\prime}+B^{\prime}=\left[\begin{array}{ccc} {-5} & {6} & {-1} \\ {3} & {9} & {4} \\ {-2} & {9} & {2} \end{array}\right] $  ...(2)
From equation (1) &( 2), we see that
(A+B)’ = A’+B’. Hence verified.

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