Question

Let A be an invertible matrix then which of the following is/are true?
(A) $\left| {{A^{ - 1}}} \right| = {\left| A \right|^{ - 1}}$
(B) ${\left( {{A^2}} \right)^{ - 1}} = {\left( {{A^{ - 1}}} \right)^2}$
(C) ${\left( {{A^T}} \right)^{ - 1}} = {\left( {{A^{ - 1}}} \right)^T}$
(D) None of these

Answer 1

As, it is given in the question that A is an invertible matrix. So, check all the options by assuming different invertible matrices that can be equal to the identity matrix (use $A{A^{ - 1}} = I$) to prove the given options.

Formula used:
Here, we will use the formula $\;detA \times detB{\rm{ }} = {\rm{ }}det\left( {AB} \right).$

Complete step-by-step answer:
Here, we will solve for all the options to verify which is correct. As, it is given A be an invertible matrix which means determinant denoted by det should not be equal to zero and when it is able to be multiplied by its inverse.
So, let us check for option (A) $\left| {{A^{ - 1}}} \right| = {\left| A \right|^{ - 1}}$
Now, we will use formula to $\;detA \times detB{\rm{ }} = {\rm{ }}det\left( {AB} \right).$ Here, A = A and B = ${A^{ - 1}}$
By substituting the values in the formula we get, $\;\det A \times \det {A^{ - 1}}{\rm{ }} = {\rm{ }}\det \left( {A{A^{ - 1}}} \right).$
As, we know $A{A^{ - 1}} = I$ where I is the identity matrix. (because A is invertible)
So, we get $\;\det A \times \det {A^{ - 1}}{\rm{ }} = {\rm{ detI}}$
As, we know ${\rm{ detI = 1}}$
Here , $\;\det A \times \det {A^{ - 1}}{\rm{ }} = {\rm{ 1}}$
So, by calculating we get: $\;\det {A^{ - 1}}{\rm{ }} = {\rm{ }}\dfrac{{\rm{1}}}{{\det A}}$ which is equal to $\;\det \left( {{A^{ - 1}}} \right){\rm{ }} = {\rm{ }}{\left( {\det A} \right)^{ - 1}} \Leftrightarrow \left| {{A^{ - 1}}} \right| = {\left| A \right|^{ - 1}}$
Hence, option (A) $\left| {{A^{ - 1}}} \right| = {\left| A \right|^{ - 1}}$ is correct.
So, now let us check for option (B) ${\left( {{A^2}} \right)^{ - 1}} = {\left( {{A^{ - 1}}} \right)^2}$
As, we know A is invertible that means $A{A^{ - 1}} = {A^{ - 1}}A = I$ where I is the identity matrix.
So, ${A^2}$ is also invertible.
Given $A{A^{ - 1}} = I$
On, squaring both sides we get,
${\left( {A{A^{ - 1}}} \right)^2} = {I^2}$
$\left( {{A^2}} \right){\left( {{A^{ - 1}}} \right)^2} = {I^2}$
As we ${I^2} = 1$
Therefore, $\left( {{A^2}} \right){\left( {{A^{ - 1}}} \right)^2} = 1$
So, by calculating we get, ${\left( {{A^{ - 1}}} \right)^2} = \dfrac{1}{{{A^2}}}$ which is equal to ${\left( {{A^{ - 1}}} \right)^2} = {\left( {{A^2}} \right)^{ - 1}}$
Hence, option (B) ${\left( {{A^2}} \right)^{ - 1}} = {\left( {{A^{ - 1}}} \right)^2}$ is also correct.
So, also let us check for option (C) ${\left( {{A^T}} \right)^{ - 1}} = {\left( {{A^{ - 1}}} \right)^T}$
As, we know A is invertible which means $\left( {{A^T}} \right){\left( {{A^{ - 1}}} \right)^T} = {\left( {{A^{ - 1}}A} \right)^T} = {I^T} = I$ --- equation 1, where T is the transpose and I is the identity matrix.
Also, we can say that ${\left( {{A^{ - 1}}} \right)^T}\left( {{A^T}} \right) = {\left( {A{A^{ - 1}}} \right)^T} = {I^T} = I$----equation 2
From equation 1 and 2 we can conclude that $\left( {{A^T}} \right){\left( {{A^{ - 1}}} \right)^T} = {\left( {{A^{ - 1}}} \right)^T}\left( {{A^T}} \right) = I$
Given, $\left( {{A^T}} \right){\left( {{A^{ - 1}}} \right)^T} = I$
As, we know ${\rm{ I = 1}}$
Therefore, $\left( {{A^T}} \right){\left( {{A^{ - 1}}} \right)^T} = 1$
So, by calculating we get, ${\left( {{A^{ - 1}}} \right)^T} = \dfrac{1}{{\left( {{A^T}} \right)}}$ which is equal to ${\left( {{A^{ - 1}}} \right)^T} = {\left( {{A^T}} \right)^{ - 1}}$
Hence, option (C) ${\left( {{A^T}} \right)^{ - 1}} = {\left( {{A^{ - 1}}} \right)^T}$ is also correct.
So, all the three options are correct, that is A, B and C.

Note: In these types of questions you need to verify the left hand side is equal to the right hand side with the help of given information. And select the invertible matrix according to the given part equal to the identity matrix and hence use the formula accordingly to calculate the result.