I If A- - 3 -3 2- 4 2 0 - - and B - - 2 -1 2- 1 2 4 - verify that A- - - A-ii If A- - 3 -3 2- 4 2 0 - - and B - - 2 -1 2- 1 2 4 - verify that A- B- - A-B-iii If A- - 3 -3 2- 4 2 0 - - and B - - 2 -1 2- 1 2 4 - verify that kB-kB- where k is any constant-

Name: NCERT - Grade 12 - Mathematics - Matrices - Q52_01
Uploaded: 2022-07-04T10:36:42+05:30
Description: (i) Given: nbsp;A= nbsp;[ 3 amp;√(3) amp;2; nbsp; 4 amp;2 amp;0; nbsp; ] nbsp; To verify: (A')'=A A ' = nbsp; nbsp;[ 3 amp; ...

(i) Given: nbsp;A= nbsp;[ 3 amp;√(3) amp;2; nbsp; 4 amp;2 amp;0; nbsp; ] nbsp; To verify: (A')'=A A ' = nbsp; nbsp;[ 3 amp; ...

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Standard XII

Mathematics

Nilpotent Matrix

i If A= [ 3 ...

Question

$(i)$ If $A = [\begin{matrix} 3 & \sqrt{3} & 2 4 & 2 & 0 \end{matrix}]$ and $B = [\begin{matrix} 2 & - 1 & 2 1 & 2 & 4 \end{matrix}]$ , verify that $(A^{'})^{'} = A .$
$(i i)$ If $A = [\begin{matrix} 3 & \sqrt{3} & 2 4 & 2 & 0 \end{matrix}]$ and $B = [\begin{matrix} 2 & - 1 & 2 1 & 2 & 4 \end{matrix}]$ , verify that $(A + B)^{'} = A^{'} + B^{'} .$
$(i i i)$ If $A = [\begin{matrix} 3 & \sqrt{3} & 2 4 & 2 & 0 \end{matrix}]$ and $B = [\begin{matrix} 2 & - 1 & 2 1 & 2 & 4 \end{matrix}]$ , verify that $(k B)^{'} = (k B^{'})$ where $k$ is any constant.

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Solution

$(i)$ Given: $A = [\begin{matrix} 3 & \sqrt{3} & 2 4 & 2 & 0 \end{matrix}]$
To verify: $(A^{'})^{'} = A$
$A^{'} = {[\begin{matrix} 3 & \sqrt{3} & 2 4 & 2 & 0 \end{matrix}]}^{'} = ⎡ ⎢ ⎣ \begin{matrix} 3 & 4 \sqrt{3} & 2 2 & 0 \end{matrix} ⎤ ⎥ ⎦$
$(A^{'})^{'} = {⎡ ⎢ ⎣ \begin{matrix} 3 & 4 \sqrt{3} & 2 2 & 0 \end{matrix} ⎤ ⎥ ⎦}^{'} = [\begin{matrix} 3 & \sqrt{3} & 2 4 & 2 & 0 \end{matrix}] = A$
Thus, $(A^{'}) = A$

$(i i)$ Given: $A = [\begin{matrix} 3 & \sqrt{3} & 2 4 & 2 & 0 \end{matrix}]$ and $B = [\begin{matrix} 2 & - 1 & 2 1 & 2 & 4 \end{matrix}]$
To verify: $(A + B)^{'} = A^{'} + B^{'}$
Solving $L . H . S .$
$(A + B) = [\begin{matrix} 3 & \sqrt{3} & 2 4 & 2 & 0 \end{matrix}] + [\begin{matrix} 2 & - 1 & 2 1 & 2 & 4 \end{matrix}]$
$\Rightarrow (A + B) = [\begin{matrix} 3 + 2 & \sqrt{3} + (- 1) & 2 + 2 4 + 1 & 2 + 2 & 0 + 4 \end{matrix}]$
$\Rightarrow (A + B) = [\begin{matrix} 5 & \sqrt{3} - 1 & 4 5 & 4 & 4 \end{matrix}]$
Thus, $(A + B)^{'} = ⎡ ⎢ ⎣ \begin{matrix} 5 & 5 \sqrt{3} - 1 & 4 4 & 4 \end{matrix} ⎤ ⎥ ⎦$

$A = [\begin{matrix} 3 & \sqrt{3} & 2 4 & 2 & 0 \end{matrix}]$ and $B = [\begin{matrix} 2 & - 1 & 2 1 & 2 & 4 \end{matrix}]$
Solving $R . H . S .$
$A^{'} + B^{'} = ⎡ ⎢ ⎣ \begin{matrix} 3 & 4 \sqrt{3} & 2 2 & 0 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} 2 & 1 - 1 & 2 2 & 4 \end{matrix} ⎤ ⎥ ⎦$
$\Rightarrow A^{'} + B^{'} = ⎡ ⎢ ⎣ \begin{matrix} 3 + 2 & 4 + 1 \sqrt{3} + (- 1) & 2 + 2 2 + 2 & 0 + 4 \end{matrix} ⎤ ⎥ ⎦$
$\Rightarrow A^{'} + B^{'} = ⎡ ⎢ ⎣ \begin{matrix} 5 & 5 \sqrt{3} + (- 1) & 4 4 & 4 \end{matrix} ⎤ ⎥ ⎦$
So, $L . H . S . = R . H . S .$
Thus, $(A + B)^{'} = A^{'} + B^{'}$
$(i i i)$ $A = [\begin{matrix} 3 & \sqrt{3} & 2 4 & 2 & 0 \end{matrix}]$ and $B = [\begin{matrix} 2 & - 1 & 2 1 & 2 & 4 \end{matrix}]$
To verify: $(k B)^{'} = (k B^{'}), k$ is any constant.
$B = [\begin{matrix} 2 & - 1 & 2 1 & 2 & 4 \end{matrix}]$
$\Rightarrow k B = [\begin{matrix} 2 k & - k & 2 k k & 2 k & 4 k \end{matrix}]$
$\Rightarrow (k B)^{'} = ⎡ ⎢ ⎣ \begin{matrix} 2 k & 2 k - k & 2 k 2 k & 4 k \end{matrix} ⎤ ⎥ ⎦ = k ⎡ ⎢ ⎣ \begin{matrix} 2 & 1 - 1 & 2 2 & 4 \end{matrix} ⎤ ⎥ ⎦ = k B^{'}$
Hence, verified.

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Nilpotent Matrix

Standard XII Mathematics

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(i) If A=[3√32420] and B=[2−12124], verify that (A′)′=A. (ii) If A=[3√32420] and B=[2−12124], verify that (A+B)′=A′+B′. (iii) If A=[3√32420] and B=[2−12124], verify that (kB)′=(kB′) where k is any constant.