If A = [ lll 3 & 3 & 2 4 & 2 & 0 ] and B= [ rrr 2 & -1 & 2 1 & 2 & 4 ] verify that 1. (A′)′ = A 2. (A + B)′ = A′ + B′ 3. (kB)′ = kB′, where k is any constant.

1. We have A = [ lll 3 & 3 & 2 4 & 2 & 0 ] A^ = [ cc 3 & 4 3 & 2 2 & 0 ] (A')' = [ ccc 3 & 3 & 2 4 & 2 & 0 ] = A Thus (A′)′ = A 2. We have A = [ lll 3 & 3 & 2 4 & 2 & 0 ], B= [ rrr 2 & -1 & 2 1 & 2 & 4 ] A+B= [ rrr 5 & 3-1 & 4 5 & 4 & 4 ] Therefore (A + B)' = [ ccc 5 & 5 3 -1 & 4 4 & 4 ] Now A' = [ cc 3 & 4 3 & 2 2 & 0 ], B^ = [ cc 2 & 1 -1 & 2 2 & 4 ] So A' + B' = [ ccc 5 & 5 3 -1 & 4 4 & 4 ] Thus (A + B)′ = A′ + B′ 3. We have, kB = k [ rrr 2 & -1 & 2 1 & 2 & 4 ]= [ ccc 2 k & -k & 2 k k & 2 k & 4 k ] (kB)' = [ cc 2 k & k -k & 2 k 2 k & 4 k ]=k [ cc 2 & 1 -1 & 2 2 & 4 ]=k B^ Thus (kB)' = kB'

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

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Question

If A = $\left[\begin{array}{lll} {3} & {\sqrt{3}} & {2} \\ {4} & {2} & {0} \end{array}\right] \text { and } B=\left[\begin{array}{rrr} {2} & {-1} & {2} \\ {1} & {2} & {4} \end{array}\right]$ verify that

(A′)′ = A
(A + B)′ = A′ + B′
(kB)′ = kB′, where k is any constant.

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Answer

We have
A = $\left[\begin{array}{lll} {3} & {\sqrt{3}} & {2} \\ {4} & {2} & {0} \end{array}\right] $
$\Rightarrow \mathrm{A}^{\prime}=\left[\begin{array}{cc} {3} & {4} \\ {\sqrt{3}} & {2} \\ {2} & {0} \end{array}\right] $
$\Rightarrow$ (A')' = $\left[\begin{array}{ccc} {3} & {\sqrt{3}} & {2} \\ {4} & {2} & {0} \end{array}\right]$ = A
Thus (A′)′ = A
We have
A = $\left[\begin{array}{lll} {3} & {\sqrt{3}} & {2} \\ {4} & {2} & {0} \end{array}\right], B=\left[\begin{array}{rrr} {2} & {-1} & {2} \\ {1} & {2} & {4} \end{array}\right] \Rightarrow A+B=\left[\begin{array}{rrr} {5} & {\sqrt{3}-1} & {4} \\ {5} & {4} & {4} \end{array}\right]$
Therefore (A + B)' = $\left[\begin{array}{ccc} {5} & {5} \\ {\sqrt{3}} {-1} & {4} \\ {4} & {4} \end{array}\right]$
Now A' = $\left[\begin{array}{cc} {3} & {4} \\ {\sqrt{3}} & {2} \\ {2} & {0} \end{array}\right], B^{\prime}=\left[\begin{array}{cc} {2} & {1} \\ {-1} & {2} \\ {2} & {4} \end{array}\right]$
So A' + B' = $\left[\begin{array}{ccc} {5} & {5} \\ {\sqrt{3}} {-1} & {4} \\ {4} & {4} \end{array}\right]$
Thus (A + B)′ = A′ + B′
We have,
$kB$ = k$\left[\begin{array}{rrr} {2} & {-1} & {2} \\ {1} & {2} & {4} \end{array}\right]=\left[\begin{array}{ccc} {2 k} & {-k} & {2 k} \\ {k} & {2 k} & {4 k} \end{array}\right]$
$\therefore$ $(kB)'$ = $\left[\begin{array}{cc} {2 k} & {k} \\ {-k} & {2 k} \\ {2 k} & {4 k} \end{array}\right]=k\left[\begin{array}{cc} {2} & {1} \\ {-1} & {2} \\ {2} & {4} \end{array}\right]=k \mathrm{B}^{\prime}$
Thus $(kB)' = kB'$