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

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.