If A- 1 2- 4 1- 5 6 - B- 1 2- 6 4- 7 3 - then verify that -i 2A-B-2A-B-ii A-B- A- B-

Name: EXEMP - Grade 12 - Mathematics - Matrices - Q28_02
Uploaded: 2022-07-04T10:36:42+05:30
Description: (i) Given: A=[[ 1 amp;2; 4 amp;1; 5 amp;6 ]], B=[[ 1 amp;2; 6 amp;4; 7 amp;3 ]] ⇒ (2A)= nbsp;[[ 2 amp;4; 8 amp;2; 10 ...

(i) Given: A=[[ 1 amp;2; 4 amp;1; 5 amp;6 ]], B=[[ 1 amp;2; 6 amp;4; 7 amp;3 ]] ⇒ (2A)= nbsp;[[ 2 amp;4; 8 amp;2; 10 ...

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Standard XII

Mathematics

Binomial Coefficients

If A=[[ 1 2; ...

Question

If $A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 4 & 1 5 & 6 \end{matrix} ⎤ ⎥ ⎦$ , $B = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 6 & 4 7 & 3 \end{matrix} ⎤ ⎥ ⎦$ , then verify that :

(i) $(2 A + B)^{'} = 2 A^{'} + B^{'}$

(ii) $(A - B)^{'} = A^{'} - B^{'}$

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Solution

(i)
Given: $A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 4 & 1 5 & 6 \end{matrix} ⎤ ⎥ ⎦$ , $B = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 6 & 4 7 & 3 \end{matrix} ⎤ ⎥ ⎦$

$\Rightarrow (2 A) = ⎡ ⎢ ⎣ \begin{matrix} 2 & 4 8 & 2 10 & 12 \end{matrix} ⎤ ⎥ ⎦$

$\Rightarrow 2 A + B = ⎡ ⎢ ⎣ \begin{matrix} 2 & 4 8 & 2 10 & 12 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 6 & 4 7 & 3 \end{matrix} ⎤ ⎥ ⎦$

$\Rightarrow 2 A + B = ⎡ ⎢ ⎣ \begin{matrix} 3 & 6 14 & 6 17 & 15 \end{matrix} ⎤ ⎥ ⎦$

Taking transpose on both sides
$\Rightarrow (2 A + B)^{^{'}} = [\begin{matrix} 3 & 14 & 17 6 & 6 & 15 \end{matrix}]$ …(1)
$A^{'} = [\begin{matrix} 1 & 4 & 5 2 & 1 & 6 \end{matrix}]$

$\Rightarrow 2 A^{'} = [\begin{matrix} 2 & 8 & 10 4 & 2 & 12 \end{matrix}]$ ….(2)

And $B^{'} = [\begin{matrix} 1 & 6 & 7 2 & 4 & 3 \end{matrix}]$ .… (3)

Adding equation (2) and (3)

$\Rightarrow 2 A^{'} + B^{'} = [\begin{matrix} 3 & 14 & 17 6 & 6 & 15 \end{matrix}]$ ...(4)

From equation (1) and (4), we get

$(2 A + B)^{'} = 2 A^{'} + B^{'}$

Hence verified.

(ii)

We have, $A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 4 & 1 5 & 6 \end{matrix} ⎤ ⎥ ⎦$ , $B = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 6 & 4 7 & 3 \end{matrix} ⎤ ⎥ ⎦$

$A - B = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 4 & 1 5 & 6 \end{matrix} ⎤ ⎥ ⎦ - ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 6 & 4 7 & 3 \end{matrix} ⎤ ⎥ ⎦$

$\Rightarrow A - B = ⎡ ⎢ ⎣ \begin{matrix} 0 & 0 - 2 & - 3 - 2 & - 3 \end{matrix} ⎤ ⎥ ⎦$

Taking transpose on both sides

$(A - B)^{'} = [\begin{matrix} 0 & - 2 & - 2 0 & - 3 & 3 \end{matrix}]$ ...(1)

$A^{'} = [\begin{matrix} 1 & 4 & 5 2 & 1 & 6 \end{matrix}]$ and $B^{'} = [\begin{matrix} 1 & 6 & 7 2 & 4 & 3 \end{matrix}]$

$\Rightarrow A^{'} - B^{'} = [\begin{matrix} 0 & - 2 & - 2 0 & - 3 & 3 \end{matrix}]$ ...(2)

From equation (1) and (2), we get

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

Hence, verified.

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Binomial Coefficients

Standard XII Mathematics

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