If A^ = [ cc 3 & 4 -1 & 2 0 & 1 ] and B= [ rrr -1 & 2 & 1 1 & 2 & 3 ] then verify (A – B)′ = A′ – B′

Given, A^ = [ cc 3 & 4 -1 & 2 0 & 1 ] and B= [ rrr -1 & 2 & 1 1 & 2 & 3 ] Now, A = [ rrr 3 & -1 & 0 4 & 2 & 1 ] A-B= [ ccc 3 & -1 & 0 4 & 2 & 1 ]- [ ccc -1 & 2 & 1 1 & 2 & 3 ] A-B= [ ccc 3-(-1) & -1-2 & 0-1 4-1 & 2-2 & 1-3 ] A-B= [ ccc 4 & -3 & -1 3 & 0 & -2 ] Therefore, (A-B)^ = [ cc 4 & 3 -3 & 0 -1 & -2 ] ~~~~~~A^ -B^ = [ cc 3 & 4 -1 & 2 0 & 1 ]- [ cc -1 & 1 2 & 2 1 & 3 ] A^ -B^ = [ cc 4 & 3 -3 & 0 -1 & -2 ] From equation (1) & (2) we see that (A - B)’ = A’ - B’. Hence verified.

If A^ = [ cc 3 & 4 -1 & 2 0 & 1 ] and B= [ rrr -1 & 2 & 1 1 & 2 &…

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Question

If $A^{\prime}=\left[\begin{array}{cc} {3} & {4} \\ {-1} & {2} \\ {0} & {1} \end{array}\right] \text { and } B=\left[\begin{array}{rrr} {-1} & {2} & {1} \\ {1} & {2} & {3} \end{array}\right]$ then verify (A – B)′ = A′ – B′

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Answer

Given, $A^{\prime}=\left[\begin{array}{cc} {3} & {4} \\ {-1} & {2} \\ {0} & {1} \end{array}\right] \text { and } B=\left[\begin{array}{rrr} {-1} & {2} & {1} \\ {1} & {2} & {3} \end{array}\right]$
Now, A = $\left[\begin{array}{rrr} {3} & {-1} & {0} \\ {4} & {2} & {1} \end{array}\right]$
$\Rightarrow A-B=\left[\begin{array}{ccc} {3} & {-1} & {0} \\ {4} & {2} & {1} \end{array}\right]-\left[\begin{array}{ccc} {-1} & {2} & {1} \\ {1} & {2} & {3} \end{array}\right]$
$\Rightarrow A-B=\left[\begin{array}{ccc} {3-(-1)} & {-1-2} & {0-1} \\ {4-1} & {2-2} & {1-3} \end{array}\right]$
$\Rightarrow A-B=\left[\begin{array}{ccc} {4} & {-3} & {-1} \\ {3} & {0} & {-2} \end{array}\right]$
Therefore, $(A-B)^\prime$ = $ \left[\begin{array}{cc} {4} & {3} \\ {-3} & {0} \\ {-1} & {-2} \end{array}\right] $
$~~~~~~A^{\prime}-B^{\prime}=\left[\begin{array}{cc} {3} & {4} \\ {-1} & {2} \\ {0} & {1} \end{array}\right]-\left[\begin{array}{cc} {-1} & {1} \\ {2} & {2} \\ {1} & {3} \end{array}\right]$
$\Rightarrow \mathrm{A}^{\prime}-\mathrm{B}^{\prime}=\left[\begin{array}{cc} {4} & {3} \\ {-3} & {0} \\ {-1} & {-2} \end{array}\right] $
From equation (1) & (2) we see that
(A - B)’ = A’ - B’. Hence verified.

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