Matrices And Determinants Ques 79
Let $A$ and $B$ be two invertible matrices of order $3 \times 3$. If $\operatorname{det}\left(A B A^{T}\right)=8$ and $\operatorname{det}\left(A B^{-1}\right)=8$, then $\operatorname{det}\left(B A^{-1} B^{T}\right)$ is equal to
(2019 Main, 11 Jan II)
(a) $1$
(b) $\frac{1}{4}$
(c) $\frac{1}{16}$
(d) $16$
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Answer:
Correct Answer: 79.(c)
Solution:
- Given, $\left|A B A^{T}\right|=8$
$ \begin{array}{lrr} \Rightarrow & |A||B|\left|A^{T}\right|=8 & {[\because|X Y|=|X||Y|]} \\ \therefore & |A|^{2}|B|=8 & \ldots(\mathrm{i})\left[\because\left|A^{T}\right|=|A|\right] \end{array} $
Also, we have $\quad\left|A B^{-1}\right|=8 \Rightarrow|A|\left|B^{-1}\right|=8$
$ \Rightarrow \quad \frac{|A|}{|B|}=8 \quad \text {…(ii) } [\because\left|A^{-1}\right| \neq\left. A\right|^{-1}=\frac{1}{|A|}] $
On multiplying Eqs. (i) and (ii), we get
$ \begin{aligned} & & |A|^{3}=8 \cdot 8=4^{3} \\ \Rightarrow & & |A|=4 \\ \Rightarrow & & |B|=\frac{|A|}{8}=\frac{4}{8}=\frac{1}{2} \end{aligned} $
Now, $\quad\left|B A^{-1} B^{T}\right|=|B| \frac{1}{|A|}|B|=(\frac{1}{2}) \quad \frac{1}{4} \quad (\frac{1}{2})=\frac{1}{16}$
BETA
