Matrices And Determinants Ques 29
If $A=\left[\begin{array}{ccc}1 & 2 & 2 \\ 2 & 1 & -2 \\ c & 2 & b\end{array}\right]$ is a matrix satisfying the equation $A A^T=9 I$, where, $I$ is $3 \times 3$ identity matrix, then the ordered pair $(a, b)$ is equal to
(2015 Main)
(a) $(2,-1)$
(b) $(-2,1)$
(c) $(2,1)$
(d) $(-2,-1)$
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Answer:
Correct Answer: 29.(d)
Solution:
Formula:
Properties of matrix multiplication:
- Given, $A=\left[\begin{array}{ccc}1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b\end{array}\right], A^T=\left[\begin{array}{ccc}1 & 2 & a \\ 2 & 1 & 2 \\ 2 & -2 & b\end{array}\right]$ and
$ \begin{aligned} A 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+2 b \\ 0 & 9 & 2 a+2-2 b \\ a+4+2 b & 2 a+2-2 b & a^2+4+b^2 \end{array}\right] \end{aligned} $
It is given that, $A A^T=9 I$
$ \begin{gathered} \Rightarrow\left[\begin{array}{ccc} 9 & 0 & a+4+2 b \\ 0 & 9 & 2 a+2-2 b \\ a+4+2 b & 2 a+2-2 b & 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+2 b \\ 0 & 9 & 2 a+2-2 b \\ a+4+2 b & 2 a+2-2 b & a^2+4+b^2 \end{array}\right]=\left[\begin{array}{ccc} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{array}\right] \end{gathered} $
On comparing, we get
$a+4+2 b =0 \quad \Rightarrow a+2 b=-4 \quad$…(i)
$2 a+2-2 b =0 \quad \Rightarrow a-b=-1 \quad$…(ii)
and $a^2+4+b^2=9 \quad$…(iii)
On solving Eqs. (i) and (ii), we get
$ a=-2, b=-1 $
This satisfies Eq. (iii)
Hence, $(a, b)=(-2,-1)$
BETA
