Question
If \[A = \left[ \begin{array}{l}
1\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,2\\
2\,\,\,\,\,\,\,1\,\,\,\,\,\, - 2\\
a\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,b
\end{array} \right]\,\,\] is a matrix satisfying the equation $$A{A^T} = 9I,$$ where $$I$$ 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 :
$$(- 2, - 1)$$
Solution :
\[\left[ \begin{array}{l}
1\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,2\\
2\,\,\,\,\,\,\,1\,\,\,\,\,\, - 2\\
a\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,b
\end{array} \right]\left[ \begin{array}{l}
1\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,a\\
2\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,2\\
2\,\,\,\,\, - 2\,\,\,\,\,\,\,\,\,b
\end{array} \right] = \left[ \begin{array}{l}
9\,\,\,\,\,\,\,0\,\,\,\,\,\,\,0\\
0\,\,\,\,\,\,\,\,9\,\,\,\,\,\,0\\
0\,\,\,\,\,\,\,\,0\,\,\,\,\,\,9
\end{array} \right]\]
\[ \Rightarrow \,\,\left[ \begin{array}{l}
\,\,1 + 4 + 4\,\,\,\,\,\,\,\,\,\,\,\,2 + 2 - 4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a + 4 + 2b\\
\,2 + 2 - 4\,\,\,\,\,\,\,\,\,\,\,\,\,4 + 1 + 4\,\,\,\,\,\,\,\,\,\,\,\,\,2a + 2 - 2b\\
a + 4 + 2b\,\,\,\,\,\,\,\,2a + 2 - 2b\,\,\,\,\,\,\,\,\,\,\,{a^2} + 4 + {b^2}
\end{array} \right] = \left[ \begin{array}{l}
9\,\,\,\,\,\,\,0\,\,\,\,\,\,\,0\\
0\,\,\,\,\,\,\,\,9\,\,\,\,\,\,0\\
0\,\,\,\,\,\,\,\,0\,\,\,\,\,\,9
\end{array} \right]\]
$$⇒ a + 4 + 2b = 0$$
$$\eqalign{
& \Rightarrow \,\,a + 2b = - 4\,\,\,\,\,\,.....\left( {\text{i}} \right) \cr
& 2a + 2 - 2b = 0 \cr
& \Rightarrow \,\,2a - 2b = - 2 \cr
& \Rightarrow \,\,a - b = - 1\,\,\,\,\,\,\,.....\left( {{\text{ii}}} \right) \cr} $$
On solving (i) and (ii) we get
$$\eqalign{
& - 1 + b + 2b = - 4 \cr
& b = - 1\,\,\,{\text{and }}a = - 2 \cr
& \left( {a,b} \right) = \left( { - 2, - 1} \right) \cr} $$