Question

If $A=\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& -2\\ a& 2& b\end{array}\right]$ is a matrix satisfying $A{A}^T=9I_3$ then the values of $a$ and $b$ are

• A. $1,2$
• B. $-1,2$
• C. -1,-2
• D. $2,1$
• E. $-2,-1$

Answer 1

The correct option is E

$-2,-1$

Step 1. Find the value of $a$ and $b$

Given,

$A=\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& -2\\ \mathrm{a}& 2& \mathrm{b}\end{array}\right]$

$A{A}^{\mathrm{T}}=9I_3$

$\Rightarrow$ $\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& -2\\ \mathrm{a}& 2& \mathrm{b}\end{array}\right]\left[\begin{array}{ccc}1& 2& \mathrm{a}\\ 2& 1& 2\\ 2& -2& \mathrm{b}\end{array}\right]=9\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$\Rightarrow$$\left[\begin{array}{ccc}9& 0& \mathrm{a}+4+2\mathrm{b}\\ 0& 9& 2\mathrm{a}+2-2\mathrm{b}\\ \mathrm{a}+4+2\mathrm{b}& 2\mathrm{a}+2-2\mathrm{b}& {\mathrm{a}}^2+4+{\mathrm{b}}^2\end{array}\right]=\left[\begin{array}{ccc}9& 0& 0\\ 0& 9& 0\\ 0& 0& 9\end{array}\right]$

Step 2. On Comparing corresponding elements, we get

$\mathrm{a}+4+2\mathrm{b}=0$

$\Rightarrow$ $\mathrm{a}+2\mathrm{b}=-4$ ….(1)

$2\mathrm{a}+2-2\mathrm{b}=0$

$\Rightarrow$ $2\mathrm{a}-2\mathrm{b}=-2$ …..(2)

Step 3. On solving equation (1) and (2), we get

$\mathbf{a}\mathbf{=}\mathbf{-}\mathbf{2}$ and $\mathbf{b}\mathbf{=}\mathbf{-}\mathbf{1}$

Hence, option(E) is correct.