Discussion Forum

 If  $A = \begin{bmatrix}1 & 2 & 2\\2 & 1 &-2 \\ a&2&b\end{bmatrix}$ is a matrix satisfying the equation AA^T =9l, where I  is 3x3 identity matrix , then the ordered pair (a,b)  is equal to

Answer:

Explanation:

Given   ,  $A = \begin{bmatrix}1 & 2 & 2\\2 & 1 &-2 \\ a&2&b\end{bmatrix}$

              $A^{T} = \begin{bmatrix}1 & 2 & a\\2 & 1 &2 \\ 2&-2&b\end{bmatrix}$

$A.A^{T} =  \begin{bmatrix}1 & 2 & 2\\2 & 1 &-2 \\ a&2&b\end{bmatrix}\begin{bmatrix}1 & 2 & a\\2 & 1 &2 \\ 2&-2&b\end{bmatrix}$

$=  \begin{bmatrix}9 & 0 & a+4+2b\\0 & 9 &2a+2-2b \\ a+4+2b&2a+2-2b&a^{2}+4+b^{2}\end{bmatrix}$

 It is given that A.A^T  =9I

  =   $ \begin{bmatrix}9 & 0 & a+4+2b\\0 & 9 &2a+2-2b \\ a+4+2b&2a+2-2b&a^{2}+4+b^{2}\end{bmatrix}$

  = 9  $ \begin{bmatrix}1 & 0 & 0\\0 & 1 &0 \\ 0&0&1\end{bmatrix}$

   $\Rightarrow$   $\begin{bmatrix}9 & 0 & a+4+2b\\0 & 9 &2a+2-2b \\ a+4+2b&2a+2-2b&a^{2}+4+b^{2}\end{bmatrix}= \begin{bmatrix}9 & 0 & 0\\0 & 9 &0 \\ 0&0&9\end{bmatrix}$

On comparing we get, a+4+2b=0

  $\Rightarrow$         a+2b=-4      ......(i)

   2a+2-2b=0

 $\Rightarrow$     a-b=-1

 and     $a^{2}+4+b^{2}=9$           .........(iii)

   On solving Eqs. (i) and (ii) , we get,

                    a=-2, b=-1

This satisfies Eq.(iii)

   Hence, (a,b)= (-2,-1)