Question

Please answer this question . wrong answer will be reported​

Answer 1

Step-by-step explanation:

We have

$A = \left[ \begin {array}{c} 4 &0 & - 1 \end{array} \right] \: \: \: B = \left[ \begin {array}{r} 2 &3 & - 2 \end{array} \right]$

$\implies \: A B= \left[ \begin {array}{c} 4 &0 & - 1 \end{array} \right] \left[ \begin {array}{r} 2 &3 & - 2 \end{array} \right] = \left[ \begin {array}{c} 4 \times 2 + 0 \times 3 + ( - 1) \times ( - 2) \end{array} \right]$

$\implies \: A B = \left[ \begin {array}{c} 8+ 0 + 2\end{array} \right]$

$\implies \: A B = \left[ \begin {array}{c} 10 \end{array} \right]$

Now,

$A^{ \prime} = \left[ \begin {array}{r} 4 &0 & - 1 \end{array} \right] \: \: \: B^{ \prime} = \left[ \begin {array}{c} 2 &3 & - 2 \end{array} \right]$

Now,

$B^{ \prime} A^{ \prime} = \left[ \begin {array}{c} 2 &3 & - 2 \end{array} \right]\left[ \begin {array}{r} 4 &0 & - 1 \end{array} \right]$

$\implies \: B^{ \prime} A^{ \prime} = \left[ \begin {array}{r} 4 \times 2 + 0 \times 3 +( - 1 )\times ( - 2) \end{array} \right]$

$\implies \: B^{ \prime} A^{ \prime} = \left[ \begin {array}{r} 8 + 0 + 2\end{array} \right]$

$\implies \: B^{ \prime} A^{ \prime} = \left[ \begin {array}{r} 10 \end{array} \right]$

So,

$\implies (AB)^{ \prime} = B^{ \prime} A^{ \prime}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:A = \begin{gathered}\left[ \begin {array}{c} 4 \\0 \\ - 1 \end{array} \right] \end{gathered}$

and

$\rm :\longmapsto\:B = \begin{gathered} \left[ \begin {array}{r} 2 &3 & - 2 \end{array} \right]\end{gathered}$

Consider,

$\rm :\longmapsto\:AB$

$\rm \:  =  \: \begin{gathered}\left[ \begin {array}{c} 4 \\0 \\ - 1 \end{array} \right] \end{gathered} \times \begin{gathered} \left[ \begin {array}{r} 2 &3 & - 2 \end{array} \right]\end{gathered}$

$\rm \:  =  \: \begin{gathered}\sf\left[\begin{array}{ccc}8&12&8\\0&0&0\\ - 2& - 3&2\end{array}\right]\end{gathered}$

Hence,

$\rm :\longmapsto\: {(AB)}^{'}$

$\rm \:  =  \: \begin{gathered}\sf \left[\begin{array}{ccc}8&0& - 2\\12&0&-3\\8&0&2\end{array}\right]\end{gathered}$

Now,

$\rm :\longmapsto\:A = \begin{gathered}\left[ \begin {array}{c} 4 \\0 \\ - 1 \end{array} \right] \end{gathered}$

$\bf\implies \:A' = \begin{gathered} \left[ \begin {array}{r} 4 &0 & - 1 \end{array} \right]\end{gathered}$

and

$\rm :\longmapsto\:B = \begin{gathered} \left[ \begin {array}{r} 2 &3 & - 2 \end{array} \right]\end{gathered}$

$\bf\implies \:B' = \begin{gathered}\left[ \begin {array}{c} 2 \\3 \\ - 2 \end{array} \right] \end{gathered}$

Now, Consider

$\rm :\longmapsto\:B'A'$

$\rm \:  =  \: \begin{gathered}\left[ \begin {array}{c} 2 \\3 \\ - 2 \end{array} \right] \end{gathered} \times \begin{gathered} \left[ \begin {array}{r} 4 &0 & - 1 \end{array} \right]\end{gathered}$

$\rm \:  =  \: \begin{gathered}\sf \left[\begin{array}{ccc}8&0& - 2\\12&0&-3\\8&0&2\end{array}\right]\end{gathered}$

So,

$\bf\implies \: {(AB)}' = B'A'$

Hence, Verified.

Additional Information :-

1. Symmetric Matrix : A square matrix A is said to be symmetric iff A' = A

2. Skew Symmetric Matrix : A square matrix A is said to be skew symmetric iff A' = - A

3. If A and B are symmetric matrix then AB + BA is symmetric.

4. If A and B are symmetric matrix then AB - BA is skew symmetric.

5. Null matrix of order n × n is always symmetric as well as skew symmetric.

6. The main diagonal elements of skew symmetric matrix are 0.