Question

Let A and B be two invertible matrices of order \[3\times 3.\] If \[\det \left( AB{{A}^{T}} \right)=8\] and \[\det \left( AB{{A}^{-1}} \right)=8,\] then \[\det \left( B{{A}^{-1}}{{B}^{T}} \right)\] is equal to:
\[\left( a \right)16\]
\[\left( b \right)\dfrac{1}{16}\]
\[\left( c \right)\dfrac{1}{4}\]
\[\left( d \right)1\]

Answer 1

Hint: We need to simplify and find the value of the determinant of A and B to get to the solution. We will use \[\det \left( AB \right)=\det \left( A \right).\det \left( B \right).\] Next, we will use \[\det \left( A \right)=\det \left( {{A}^{T}} \right)\] and lastly we will use the property \[\det \left( {{A}^{-1}} \right)=\dfrac{1}{\det \left( A \right)}\] to simplify. At last, we will get, |A| = 4 and \[\left| B \right|=\dfrac{1}{2}.\] We will use them to get the value of \[\det \left( B{{A}^{-1}}{{B}^{T}} \right).\] We have that A and B are invertible matrix. As we know that \[\det \left( AB \right)=\det \left( A \right).\det \left( B \right),\] so we get,
\[\det \left( AB{{A}^{T}} \right)=\det \left( A \right).\det \left( B \right).\det \left( {{A}^{T}} \right)\]
We also have that, \[\det \left( {{A}^{T}} \right)=\det \left( A \right).\] So, using this above, we get,
\[\det \left( AB{{A}^{T}} \right)=\det A.\det B.\det A\]
\[\Rightarrow \det \left( AB{{A}^{T}} \right)={{\left| A \right|}^{2}}\left| B \right|\]
As \[\det \left( AB{{A}^{T}} \right)=8,\] so we get,
\[\Rightarrow {{\left| A \right|}^{2}}\left| B \right|=8.......\left( i \right)\]

Complete step by step answer:

Now we have,
\[\det \left( A{{B}^{-1}} \right)=8\]
So, \[\det \left( A{{B}^{-1}} \right)=\left| A \right|\left| {{B}^{-1}} \right|\]
As, \[\left| {{B}^{-1}} \right|=\dfrac{1}{\left| B \right|}.\]
So, \[\det \left( A{{B}^{-1}} \right)=\dfrac{\left| A \right|}{\left| B \right|}\]
As, \[\det \left( A{{B}^{-1}} \right)=8,\]so we get,
\[\dfrac{\left| A \right|}{\left| B \right|}=8......\left( ii \right)\]
Using (ii), we get,
\[\Rightarrow \left| A \right|=8\left| B \right|\]
Using this in (i), we get,
\[\Rightarrow {{\left| A \right|}^{2}}\left| B \right|=8\]
\[\Rightarrow {{\left( 8\left| B \right| \right)}^{2}}\left| B \right|=8\]
\[\Rightarrow {{\left| B \right|}^{3}}=\dfrac{8}{{{8}^{2}}}\]
Simplifying further, we get,
\[\Rightarrow \left| B \right|=\dfrac{1}{2}\]
Now putting \[\left| B \right|=\dfrac{1}{2}\] in \[\left| A \right|=8\left| B \right|.\] We get,
\[\left| A \right|=8\times \dfrac{1}{2}=4\]
We have |A| = 4 and \[\left| B \right|=\dfrac{1}{2}.\] So, we get,
\[\det \left( B{{A}^{-1}}{{B}^{T}} \right)=\det B.\det {{A}^{-1}}.\det {{B}^{-1}}\]
\[\Rightarrow \det \left( B{{A}^{-1}}{{B}^{T}} \right)=\left| B \right|.\left| {{A}^{-1}} \right|.\left| B \right|\]
\[\Rightarrow \det \left( B{{A}^{-1}}{{B}^{T}} \right)=\dfrac{{{\left| B \right|}^{2}}}{\left| A \right|}\]
Putting the value, we get,
\[\Rightarrow \det \left( B{{A}^{-1}}{{B}^{T}} \right)={{\left( \dfrac{1}{2} \right)}^{2}}\times \dfrac{1}{4}\]
\[\Rightarrow \det \left( B{{A}^{-1}}{{B}^{T}} \right)=\dfrac{1}{16}\]
Hence, the right option is (b).

Note:
The determinants are commutative in nature. So we can say that,
\[\det \left( AB{{A}^{T}} \right)=\left| A \right|\left| B \right|\left| {{A}^{T}} \right|\]
As \[\left| A \right|=\left| {{A}^{T}} \right|.\] So, \[\left| A \right|\left| B \right|\left| A \right|.\]
As the determinant commutative, so we get,
\[\left| B \right|\left| A \right|=\left| A \right|\left| B \right|\]
So, we get,
\[=\left| A \right|\left| A \right|\left| B \right|\]
\[={{\left| A \right|}^{2}}\left| B \right|\]
This thing is not true for all multiplying matrices A with B. AB is not equal to BA always. But, \[\left| A \right|\left| B \right|=\left| B \right|\left| A \right|\] always.