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Multiplication with Vectors

If A=[ -1 ...

Question

If $A = ⎡ ⎢ ⎣ \begin{matrix} - 1 5 - 2 \end{matrix} \begin{matrix} 271 \end{matrix} \begin{matrix} 391 \end{matrix} ⎤ ⎥ ⎦$ and $B = ⎡ ⎢ ⎣ \begin{matrix} - 4 11 \end{matrix} \begin{matrix} 123 \end{matrix} \begin{matrix} - 5 01 \end{matrix} ⎤ ⎥ ⎦$ , then verify that

$(i) {(A + B)}^{T} = A^{T} + B^{T}$
$(i i) {(A - B)}^{T} = A^{T} - B^{T}$

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Solution

We have:
$A^{T} = ⎡ ⎢ ⎣ \begin{matrix} - 1 & 5 & - 2 2 & 7 & 1 3 & 9 & 1 \end{matrix} ⎤ ⎥ ⎦ B^{T} = ⎡ ⎢ ⎣ \begin{matrix} - 4 & 1 & 1 1 & 2 & 3 - 5 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦$
(i) $A + B = ⎡ ⎢ ⎣ \begin{matrix} - 1 & 5 & - 2 2 & 7 & 1 3 & 9 & 1 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} - 4 & 1 & 1 1 & 2 & 3 - 5 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦$
$∴ (A + B) = ⎡ ⎢ ⎣ \begin{matrix} - 5 & 3 & - 2 6 & 9 & 9 - 1 & 4 & 2 \end{matrix} ⎤ ⎥ ⎦$
$A^{T} + B^{T} = ⎡ ⎢ ⎣ \begin{matrix} - 1 & 2 & 3 5 & 7 & 9 - 2 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} - 4 & 1 & - 5 1 & 2 & 0 1 & 3 & 1 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} - 5 & 6 & - 1 3 & 9 & 4 - 2 & 9 & 2 \end{matrix} ⎤ ⎥ ⎦$

Hence, we have verified that ${(A + B)}^{T} = A^{T} + B^{T}$
(ii) $A - B = ⎡ ⎢ ⎣ \begin{matrix} - 1 5 - 2 \end{matrix} \begin{matrix} 271 \end{matrix} \begin{matrix} 391 \end{matrix} ⎤ ⎥ ⎦ - ⎡ ⎢ ⎣ \begin{matrix} - 4 11 \end{matrix} \begin{matrix} 123 \end{matrix} \begin{matrix} - 5 01 \end{matrix} ⎤ ⎥ ⎦$
$∴ {(A - B)}^{T} = ⎡ ⎢ ⎣ \begin{matrix} 318 \end{matrix} \begin{matrix} 459 \end{matrix} \begin{matrix} - 3 - 2 0 \end{matrix} ⎤ ⎥ ⎦$
$A^{T} - B^{T} = ⎡ ⎢ ⎣ \begin{matrix} - 1 5 - 2 \end{matrix} \begin{matrix} 271 \end{matrix} \begin{matrix} 391 \end{matrix} ⎤ ⎥ ⎦ - ⎡ ⎢ ⎣ \begin{matrix} - 4 11 \end{matrix} \begin{matrix} 123 \end{matrix} \begin{matrix} - 5 01 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 318 \end{matrix} \begin{matrix} 459 \end{matrix} \begin{matrix} - 3 - 2 0 \end{matrix} ⎤ ⎥ ⎦$
Hence, we have proved that ${(A - B)}^{T} = A^{T} - B^{T}$ .

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A = [\begin{matrix} 1 & - 1 & 0 \\ 2 & 1 & 3 \\ 1 & 2 & 1 \end{matrix}] and B = [\begin{matrix} 1 & 2 & 3 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{matrix}] .

Find A^T, B^T and verify that
(i) (A + B)^T = A^T + B^T
(ii) (AB)^T = B^T A^T
(iii) (2A)^T = 2A^T.

A = [\begin{matrix} 2 & - 3 \\ - 7 & 5 \end{matrix}] and B = [\begin{matrix} 1 & 0 \\ 2 & - 4 \end{matrix}]

, verify that
(i) (2A)^T = 2A^T
(ii) (A + B)^T = A^T + B^T
(iii) (A − B)^T = A^T − B^T
(iv) (AB)^T = B^T A^T

If $A^{T} = ⎡ ⎢ ⎣ \begin{matrix} 3 & 4 - 1 & 2 0 & 1 \end{matrix} ⎤ ⎥ ⎦ and B = [\begin{matrix} - 1 & 2 & 1 1 & 2 & 3 \end{matrix}]$ ,

then find $A^{T} - B^{T}$

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

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

A^{T} - B^{T} .

I f

2 A + B^{T} = [\begin{matrix} 2 & - 3 4 & 7 \end{matrix}]

A^{T} - B = [\begin{matrix} 4 & - 5 0 & 1 \end{matrix}]

A =

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