Discussion Forum



1)

If $\triangle=\begin{bmatrix}1 & 5&6 \\0 & 1&7\\0&0&1 \end{bmatrix}$

and $\triangle ^{'}=\begin{bmatrix}1 & 0&1 \\3 & 0&3\\4&6&100 \end{bmatrix}$


A) $\triangle^{2}-3\triangle'=0$

B) $(\triangle+\triangle')^{2}-3(\triangle+\triangle')+2=0$

C) $(\triangle+\triangle')^{2}-3(\triangle+\triangle')+5=0$

D) $\triangle+3\triangle'+1$

Answer:

Option B

Explanation:

Given, 

$\triangle=\begin{bmatrix}1 & 5&6 \\0 & 1&7\\0&0&1 \end{bmatrix}$

     = $1(1-0)-5(0-0)+6(0-0)$

   =$1-0+0$

$\triangle =1$

 If $\triangle=1$ and $\triangle ^{'}=\begin{bmatrix}1 & 0&1 \\3 & 0&3\\4&6&100 \end{bmatrix}$

             =$1(0-18)-0(300-12)+1(18-0)$

=$-18-0+18$

    $\triangle '=0$

  Now, $(\triangle +\triangle')^{2}-3(\triangle+\triangle')+2$

 =$(1+0)^{2}-3(1+0)+2$

  =-2+2=0