Question

matrix question solved

Answer 1

I had solved both the questions...

Hope it is helpful to you..

Answer 2

Step-by-step explanation:

Given:

$(i)$ $A=\left[\begin{array}{ccc}2&3&5\end{array}\right]$  $B=\left[\begin{array}{ccc}1\\2\\3\end{array}\right]$

$(ii)$ $A=\left[\begin{array}{ccc}2&1&3\\4&1&0\\\end{array}\right]$   $B=\left[\begin{array}{ccc}1&-1\\0&2\\5&0\end{array}\right]$

To find:

$(i)$  $AB$

$(ii)$ $AB$

     $BA$

Solution:

$(i)$ $AB=$ $\left[\begin{array}{ccc}2&3&5\end{array}\right]$  $\left[\begin{array}{ccc}1\\2\\3\end{array}\right]$

   $AB_{11}= 2$ × $1+3$ × $2+5$ × $3=2+6+15$

∴     $AB=23$

(ii) $AB=$ $\left[\begin{array}{ccc}2&1&3\\4&1&0\\\end{array}\right]$  $\left[\begin{array}{ccc}1&-1\\0&2\\5&0\end{array}\right]$

Here,

$AB_{11}=2$ × $1+1$ × $0+3$ × $5=17$

$AB_{12}=2$ × $(-1)+1$ × $2+3$ × $0=0$

$AB_{21}=4$ × $1+1$ × $0$ + $0$ × $5$ $=4$

$AB_{22}=4$ × $(-1)+$ $1$ × $2+0$ × $0=-2$

So the matrix becomes,

$AB=\left[\begin{array}{ccc}17&0\\4&-2\\\end{array}\right]$

Now,

$BA=$ $\left[\begin{array}{ccc}1&-1\\0&2\\5&0\end{array}\right]$ $\left[\begin{array}{ccc}2&1&3\\4&1&0\\\end{array}\right]$

Here,

$BA_{11} =1$ × $2+(-1)$ × $4=-2$                        

$BA_{12} =$ $1$ × $1$  $+(-1)$ × $1=0$              

$BA_{13}=1$ × $3+(-1)$ × $0=3$

$BA_{21} =0$ × $1+2$ × $4=8$

$BA_{22}=0$ × $1$ $+2$ × $1=2$

$BA_{23}=$ $0$ × $3+2$ × $0=0$

$BA_{31}= 5$ × $2+0$ × $4=10$

$BA_{32} =5$ ×$1 +0$ × $1=5$

$BA_{33} =5$ × $3+0$ × $0=15$

So the matrix becomes,

$BA=\left[\begin{array}{ccc}-2&0&3\\8&2&0\\10&5&15\end{array}\right]$

Answer:

Hence the answer is $(i)$  $AB=23$

$(ii)$ $AB=\left[\begin{array}{ccc}17&0\\4&-2\\\end{array}\right]$

$BA=\left[\begin{array}{ccc}-2&0&3\\8&2&0\\10&5&15\end{array}\right]$