Question

If $A=\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& -2\\ a& 2& b\end{array}\right]$, where I is a matrix satisfying the equation $A{A}^T=9I,$ is $3\times 3$ identity matrix, then the ordered pair (a, b) is equal to

• A. (2, -1)
• B. (-2, 1)
• C. (2,1)
• D. (-2, -1)

Answer 1

The correct option is D (-2, -1)

Given,
$A=\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& -2\\ a& 2& b\end{array}\right]A^T=\left[\begin{array}{ccc}1& 2& a\\ 2& 1& 2\\ 2& -2& b\end{array}\right]andA{A}^T=\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& -2\\ a& 2& b\end{array}\right]\left[\begin{array}{ccc}1& 2& a\\ 2& 1& 2\\ 2& -2& b\end{array}\right]=\left[\begin{array}{ccc}9& 0& a+4+2b\\ 0& 9& 2a+2-2b\\ a+4+2b& 2a+2-2b& a^2+4+b^2\end{array}\right]$
It is given that, $A{A}^T=9I$
$\Rightarrow \left[\begin{array}{ccc}9& 0& a+4+2b\\ 0& 9& 2a+2-2b\\ a+4+2b& 2a+2-2b& a^2+4+b^2\end{array}\right]=9\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\Rightarrow \left[\begin{array}{ccc}9& 0& a+4+2b\\ 0& 9& 2a+2-2b\\ a+4+2b& 2a+2-2b& a^2+4+b^2\end{array}\right]=\left[\begin{array}{ccc}9& 0& 0\\ 0& 9& 0\\ 0& 0& 9\end{array}\right]$
On comparing, we get
$a+4+2b=0\Rightarrow a+2b=-\mathrm{4......}\left(i\right)$
$2a+2-2b=0\Rightarrow a-b=-\mathrm{1......}\left(ii\right)$
and $a^2+4+b^2=9$
On solving Eqs. (i) and (ii), we get
(a = 2, b = - 1)
This satisfies Eq. (iii)
Hence, (a, b) $\equiv$ (-2, -1)