Question

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Answer 1

Here A = $\left[\begin{array}{c c c} 3&2&4 \ \\ 2&1&1 \\ 1&3&5 \end{array} \right]$

L.H.S : ( 5A )' : -

( 5A ) = 5 $\left[\begin{array}{c c c} 3&2&4 \ \\ 2&1&1 \\ 1&3&5 \end{array} \right]$ = $\left[\begin{array}{c c c} 15&10&20 \ \\ 10&5&5 \\ 5&15&25\end{array} \right]$

Now ( 5A )' = $\left[\begin{array}{c c c} 15&10&5 \ \\ 10&5&15 \\ 20&5&25\end{array} \right]$

R.H.S : 5A' : A' = $\left[\begin{array}{c c c} 3&2&1 \ \\ 2&1&3 \\ 4&1&5 \end{array} \right]$

Now 5A' = 5 $\left[\begin{array}{c c c} 3&2&1 \ \\ 2&1&3 \\ 4&1&5 \end{array} \right]$ = $\left[\begin{array}{c c c} 15&10&5\ \\ 10&5&15 \\ 20&5&25\end{array} \right]$

L.H.S = R.H.S

Hence Proved!

Answer 2

Question given:

If A = $\begin{gathered}\left[\begin{array}{c c c} 3&2&4 \ \\ 2&1&1 \\ 1&3&5 \end{array} \right]\end{gathered}$

Find the value of 5A

Solution :

We are given with the statement that ,

A = $\begin{gathered}\left[\begin{array}{c c c} 3&2&4 \ \\ 2&1&1 \\ 1&3&5 \end{array} \right]\end{gathered}$

5A would be 5 into A

5A = 5 × $\begin{gathered}\left[\begin{array}{c c c} 3&2&4 \ \\ 2&1&1 \\ 1&3&5 \end{array} \right]\end{gathered}$

5A = $\begin{gathered}\left[\begin{array}{c c c} 15&10&20 \ \\ 10&5&5 \\ 5&15&25 \end{array} \right]\end{gathered}$

Thus verified