SATHEE: Matrices and Determinants 3 Question 2

Matrices and Determinants 3 Question 2

3. Let $A$ and $B$ be two invertible matrices of order $3 \times 3$ . If $\det (A B A^{T}) = 8$ and $\det (A B^{- 1}) = 8$ , then $\det (B A^{- 1} B^{T})$ is equal to

(2019 Main, 11 Jan II)

(a) 1

(b) $\frac{1}{4}$

(d) 16

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Answer:

Correct Answer: 3. (c)

Solution:

Given, $| A B A^{T} | = 8$

$\begin{array}{lrr} \Rightarrow & | A | | B | | A^{T} | = 8 & [∵ | X Y | = | X | | Y |] \\ ∴ & | A |^{2} | B | = 8 & \dots (i) [∵ | A^{T} | = | A |] \end{array}$

Also, we have $| A B^{- 1} | = 8 \Rightarrow | A | | B^{- 1} | = 8$

$\Rightarrow \frac{| A |}{| B |} = 8 …(ii) ∵ | A^{- 1} | \neq {A |}^{- 1} = \frac{1}{| A |}$

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, $| B A^{- 1} B^{T} | = | B | \frac{1}{| A |} | B | = \frac{1}{2} \frac{1}{4} \frac{1}{2} = \frac{1}{16}$

Matrices and Determinants 3 Question 2

3. Let $A$ and $B$ be two invertible matrices of order $3 \times 3$ . If $\det (A B A^{T}) = 8$ and $\det (A B^{- 1}) = 8$ , then $\det (B A^{- 1} B^{T})$ is equal to

Answer:

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Matrices and Determinants 3 Question 2

3. Let A and B be two invertible matrices of order 3×3. If det(ABAT)=8 and det(AB−1)=8, then det(BA−1BT) is equal to

Answer:

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3. Let $A$ and $B$ be two invertible matrices of order $3 \times 3$ . If $\det (A B A^{T}) = 8$ and $\det (A B^{- 1}) = 8$ , then $\det (B A^{- 1} B^{T})$ is equal to