Question

If A and B are two non-singular matrices of order 3 such that \[A{A^T} = 2{\rm I}\] and \[{A^{ - 1}} = {A^T} - A \cdot adj\left( {2{B^{ - 1}}} \right)\] , then \[\det \left( B \right)\] is equal to
1.\[8\]
2.\[8\sqrt 2 \]
3.\[16\sqrt 2 \]
4.\[32\]

Answer 1

Hint: In the above given question, we are given a matrix A of order 3×3 such that \[A{A^T} = 2{\rm I}\] and \[{A^{ - 1}} = {A^T} - A \cdot adj\left( {2{B^{ - 1}}} \right)\] . We have to determine the determinant of another matrix B that appeared in the second equation. In order to approach the solution, we need to use some of the properties of matrices and determinants.

Complete answer:
Given that, a 3×3 order matrix A.
According to the given question, we have the first equation written as,
\[ \Rightarrow A{A^T} = 2{\rm I}\]
Taking determinants of both the LHS and RHS, we can write is as,
\[ \Rightarrow \left| {A{A^T}} \right| = \left| {2{\rm I}} \right|\]
Since \[I\] is a 3×3 identity matrix, hence
\[ \Rightarrow \left| A \right| \cdot \left| {{A^T}} \right| = {2^3}\left| {\rm I} \right|\]
Now, since the determinant of a transpose matrix is equal to the determinant of the original matrix,
i.e., \[\left| A \right| = \left| {{A^T}} \right|\]
Then we have,
\[ \Rightarrow \left| A \right| \cdot \left| A \right| = 8\left| {\rm I} \right|\]
Since, \[\left| I \right| = 1\] hence
\[ \Rightarrow {\left| A \right|^2} = 8\] ...(1)
Now, the second equation is also given in the question, written as
\[ \Rightarrow {A^{ - 1}} = {A^T} - A \cdot adj\left( {2{B^{ - 1}}} \right)\]
Multiplying both the LHS and the RHS by \[A\] , we can write
\[ \Rightarrow A{A^{ - 1}} = A{A^T} - {A^2} \cdot adj\left( {2{B^{ - 1}}} \right)\]
Now since \[A{A^{ - 1}} = I\] and given that \[A{A^T} = 2{\rm I}\] hence we have,
\[ \Rightarrow I = 2{\rm I} - {A^2} \cdot adj\left( {2{B^{ - 1}}} \right)\]
That can be written as,
\[ \Rightarrow {A^2} \cdot adj\left( {2{B^{ - 1}}} \right) = I\]
Now taking determinant of both the LHS and RHS, we can write the above equation as,
\[ \Rightarrow \left| {{A^2} \cdot adj\left( {2{B^{ - 1}}} \right)} \right| = \left| I \right|\]
That gives us the equation,
\[ \Rightarrow \left| {{A^2}} \right| \cdot \left| {adj\left( {2{B^{ - 1}}} \right)} \right| = 1\]
Now since we can write \[\left| {adj\left( {2{B^{ - 1}}} \right)} \right| = {\left| {2{B^{ - 1}}} \right|^{3 - 1}}\] hence we have,
\[ \Rightarrow \left| {{A^2}} \right| \cdot {\left| {2{B^{ - 1}}} \right|^2} = 1\]
That can be written as,
\[ \Rightarrow \left| {{A^2}} \right| \cdot {\left( {{2^3}} \right)^2}{\left| {{B^{ - 1}}} \right|^2} = 1\]
Now since from (1) we have \[{\left| A \right|^2} = 8\] hence,
\[ \Rightarrow 8 \cdot {2^6}{\left| {{B^{ - 1}}} \right|^2} = 1\]
Also we can write \[{\left| {{B^{ - 1}}} \right|^2} = {\left| B \right|^{ - 2}}\] hence we have,
\[ \Rightarrow 8 \cdot 64{\left| B \right|^{ - 2}} = 1\]
That gives us,
\[ \Rightarrow \dfrac{{8 \cdot 64}}{{{{\left| B \right|}^2}}} = 1\]
By cross multiplying the above equation, we get
\[ \Rightarrow {\left| B \right|^2} = 8 \cdot 64\]
Taking square root of both sides, we get
\[ \Rightarrow \left| B \right| = 2\sqrt 2 \cdot 8\]
Therefore,
\[ \Rightarrow \left| B \right| = 16\sqrt 2 \]
Hence, the determinant of matrix B is \[16\sqrt 2 \] .

Therefore, the correct option is (3).

Note:
A matrix is said to be singular if its determinant is equal to zero. For example, if we have matrix A whose all elements in the first column are zero then it is called a singular matrix.
 Similarly, a non-singular matrix is a matrix which has non-zero value of its determinant. Non-singular matrices are invertible, that is, their inverse matrix \[{A^{ - 1}}\] exists.