Question

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Answer 1

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

To Find :  A²  and AB

Solution:

A² = A. A

$A^2=\left[\begin{array}{ccc}2&3&4\\1&2&3\\1&-1&1\end{array}\right]\left[\begin{array}{ccc}2&3&4\\1&2&3\\1&-1&1\end{array}\right]$  

2*2+3*1 +4*1      2*3 + 3*2 + (4*-1)   2*4 + 3*3 +4*1

1*2+2*1 +3*1      1*3 + 2*2 +3*(-1)     1*4 + 2*3 +3*1

1*2+(-1)*1 +1*1      1*3 + (-1)*2 + 1*(-1)   1*4 + (-1)*3 +1*1

$A^2=\left[\begin{array}{ccc}11&8&21\\7&4&13\\2&0&2\end{array}\right]$  

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

  2*1+3*(-1)+4*0    2*3+3*1+4*0    2*2+3*1+4*3

  1*1+2*(-1)+3*0    1*3+2*1+3*0    1*2+2*1+3*3

  1*1+(-1)*(-1)+1*0      1*3+(-1)*1+1*0 1*2+(-1)*1+1*3

$AB=\left[\begin{array}{ccc}-1&9&19\\-1&5&13\\2&2&4\end{array}\right]$  

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