Question

find
a) A +2B
b) (A+2B)^T​

Answer 1

GIVEN :–

$\\ \: \: { \huge{.}} \: \: \: \sf {A}^{T} = \: \begin{bmatrix} - 2&3 \\ \\ 1&2 \end{bmatrix}$

$\\ \: \: { \huge{.}} \: \: \: \sf And \: \: \: B = \: \begin{bmatrix} - 1&0 \\ \\ 1&2 \end{bmatrix}$

TO FIND :–

$\\ \: \: \: \sf (a) \: \: \: A + 2B = ?$

$\\ \: \: \: \sf (b) \: \: \:( A + 2B) ^{T} = ?$

SOLUTION :–

$\\ \implies \sf {A}^{T} = \: \begin{bmatrix} - 2&3 \\ \\ 1&2 \end{bmatrix}$

• So that 'A' –

$\\ \implies \sf {A} = \: \begin{bmatrix} - 2&1 \\ \\ 3&2 \end{bmatrix}$

$\\ \: \: \: \sf (a) \: \: \: A + 2B = ?$

$\\ \: \sf\: \implies \: A + 2B = \begin{bmatrix} - 2&1 \\ \\ 3&2 \end{bmatrix} + 2\begin{bmatrix} - 1&0 \\ \\ 1&2 \end{bmatrix}$

$\\ \: \sf\: \implies \: A + 2B = \begin{bmatrix} - 2&1 \\ \\ 3&2 \end{bmatrix} + \begin{bmatrix} - 2&0 \\ \\ 2&4 \end{bmatrix}$

$\\ \: \sf\: \implies \: A + 2B = \begin{bmatrix} - 4&1 \\ \\ 5&6 \end{bmatrix}$

$\\ \: \: \: \sf (b) \: \: \:( A + 2B) ^{T} = ?$

• Using part (a) –

$\\ \: \sf\: \implies \: A + 2B = \begin{bmatrix} - 4&1 \\ \\ 5&6 \end{bmatrix}$

• Take transpose –

$\\ \: \sf\: \implies \: (A + 2B)^{T} = \begin{bmatrix} - 4&5 \\ \\ 1&6 \end{bmatrix}$

Answer 2

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

$\star\: {A}^{T} = \sf{\begin{bmatrix} - 2 &3 \\ 1 & 2\end{bmatrix}}$

$\star \: {B } = \sf{\begin{bmatrix} -1 &0 \\1 & 2\end{bmatrix}}$

${\bf{\blue{\underline{To\:Find:}}}}$

$(1). \: \: {\sf{ A + 2B}}$

$(2). \: \: {\sf{( A+ 2B) ^{T} }}$

${\bf{\blue{\underline{Now:}}}}$

$: \implies {A}^{T} = \sf{\begin{bmatrix} - 2 &3 \\ 1 & 2\end{bmatrix}}$

$: \implies {A} = \sf{\begin{bmatrix} - 2 &1 \\ 3 & 2\end{bmatrix}}$

$: \implies {A} + 2b= \sf{\begin{bmatrix} - 2 &1 \\ 3 & 2\end{bmatrix} } +2 \sf{\begin{bmatrix} - 1 &0 \\ 1 & 2\end{bmatrix} }$

$: \implies {A} + 2b= \sf{\begin{bmatrix} - 2 &1 \\ 3 & 2\end{bmatrix} } + \sf{\begin{bmatrix} - 2&0 \\ 2 & 4\end{bmatrix} }$

$: \implies {A} + 2b= \sf{\begin{bmatrix} -4 &1 \\ 5 & 6\end{bmatrix} }$

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$: \implies {A} + 2b= \sf{\begin{bmatrix} -4 &1 \\ 5 & 6\end{bmatrix} }$

Now transpose of Matrix,

$: \implies ({A} + 2b)^{t} = \sf{\begin{bmatrix} -4&5 \\ 1 & 6\end{bmatrix} }$

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$\star \: \underline {\mathfrak{ \green{Transpose \: of \: matrix :}}}$

• The Matrix obtained from any given matrix A by interchanging it rows and columns is called is transpose and is denoted by A' or A^t.

$\star \: \underline {\mathfrak{ \green{properties:}}}$

$: \implies{\sf{ ( { {A}^{'} )}^{'} = A}}$

$: \implies{\sf{ ( { {A + B})}^{'} = {A}^{'} + {B}^{'} }}$

$: \implies{\sf{ ( { {AB})}^{'} = k{A}^{'} }}$

$: \implies{\sf{ ( { {AB})}^{ - 1} = {B}^{ - 1} {A}^{ - 1} }}$