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Finding Inverse Using Elementary Transformations

If A = [ [ ...

Question

If $A = [\begin{matrix} 4 & 3 1 & 2 \end{matrix}] B = [\begin{matrix} 2 & 1 1 & 2 \end{matrix}]$ Verify ${(A B)}^{'} = B^{'} A^{'}$

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Solution

$A = [\begin{matrix} 4 & 3 1 & 2 \end{matrix}] B = [\begin{matrix} 2 & 1 1 & 2 \end{matrix}]$

$A B =$ $[\begin{matrix} 11 & 10 4 & 5 \end{matrix}]$

$(A B)^{'}$ = $[\begin{matrix} 11 & 4 10 & 5 \end{matrix}]$

$B^{'} A^{'}$ = $[\begin{matrix} 8 + 3 & 2 + 2 4 + 6 & 1 + 4 \end{matrix}]$ = $[\begin{matrix} 11 & 4 10 & 5 \end{matrix}]$
Thus,
$(A B)^{'} = B^{'} A^{'}$

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Similar questions

A = ⎡ ⎢ ⎣ \begin{matrix} - 2 45 \end{matrix} ⎤ ⎥ ⎦, B = [\begin{matrix} 1 & 3 & - 6 \end{matrix}]

(A B)^{'} = B^{'} A^{'}

A = ⎡ ⎢ ⎣ \begin{matrix} - 2 45 \end{matrix} ⎤ ⎥ ⎦

B = [13 - 6]

(A B)^{'} = B^{'} A^{'}

A = ⎡ ⎢ ⎣ \begin{matrix} 2 - 4 1 \end{matrix} ⎤ ⎥ ⎦, B = [53 - 1]

then verify that

(A B)^{'} = B^{'} A^{'}

A = [\begin{matrix} 2 & - 1 & 2 1 & 3 & - 4 \end{matrix}]

B = ⎡ ⎢ ⎣ \begin{matrix} 1 & - 2 - 3 & 0 5 & 4 \end{matrix} ⎤ ⎥ ⎦

then verify that

(A B)^{'} = B^{'} A^{'}

A = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 2 & 2 3 & - 1 \end{matrix} ∣ ∣ ∣ ∣

B = ∣ ∣ ∣ ∣ \begin{matrix} 2 & 1 1 & 2 - 1 & 0 \end{matrix} ∣ ∣ ∣ ∣

, then find matrix

C

A + 2 B + C = 0

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