Question

Let $A$ be an invertible matrix. Which of the following is not true?
A) ${A^{ - 1}} = {\left| A \right|^{ - 1}}$
B) ${({A^2})^{ - 1}} = {({A^{ - 1}})^2}$
C) ${({A^T})^{ - 1}} = {({A^{ - 1}})^T}$
D) None of these

Answer 1

Hint: Given that $A$ is invertible. So ${A^{ - 1}}$ exists and we can use the definition of inverse. First option can be shown wrong by choosing a general $2 \times 2$ matrix. Option B can be proved by the definition of inverse. Option C can be proved using the result of transpose of the product of two matrices.

Useful formula:
$A$ is an invertible matrix if and only if there exists ${A^{ - 1}}$such that $A{A^{ - 1}} = {A^{ - 1}}A = I$, where $I$ is the identity matrix.
For a $2 \times 2$ matrix, A = \left( {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}} \\\ {{a_{21}}}&{{a_{22}}} \end{array}} \right),
$\left| A \right| = {a_{11}}{a_{22}} - {a_{21}}{a_{12}}$ and {A^{ - 1}} = \left( {\begin{array}{*{20}{c}} {{a_{22}}}&{ - {a_{12}}} \\\ { - {a_{21}}}&{{a_{11}}} \end{array}} \right).
${(AB)^T} = {B^T}{A^T}$, where ${A^T}$ represents the transpose of a matrix $A$.

Complete step-by-step answer:
Given that $A$ is an invertible matrix.
So ${A^{ - 1}}$ exists and $A{A^{ - 1}} = {A^{ - 1}}A = I$, where $I$ is the identity matrix.
$\left| A \right|$ denotes the determinant of a matrix $A$.

A) ${A^{ - 1}} = {\left| A \right|^{ - 1}}$
For a $2 \times 2$ matrix, A = \left( {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}} \\\ {{a_{21}}}&{{a_{22}}} \end{array}} \right), $\left| A \right| = {a_{11}}{a_{22}} - {a_{21}}{a_{12}}$
Also {A^{ - 1}} = \left( {\begin{array}{*{20}{c}} {{a_{22}}}&{ - {a_{12}}} \\\ { - {a_{21}}}&{{a_{11}}} \end{array}} \right)
But $\left| A \right| = {a_{11}}{a_{22}} - {a_{21}}{a_{12}}$, which is a number and its inverse is given by ${\left| A \right|^{ - 1}} = \dfrac{1}{{{a_{11}}{a_{22}} - {a_{21}}{a_{12}}}}$.
Therefore, ${A^{ - 1}} \ne {\left| A \right|^{ - 1}}$.
This gives option A is not true.

B) ${({A^2})^{ - 1}} = {({A^{ - 1}})^2}$
Since $A$ is invertible, ${A^2}$ is also invertible.
So there exists ${\left( {{A^2}} \right)^{ - 1}}$ such that $({A^2}){({A^2})^{ - 1}} = {({A^2})^{ - 1}}({A^2}) = I$
$A$ is invertible gives $A{A^{ - 1}} = {A^{ - 1}}A = I$.
Squaring we get, ${(A{A^{ - 1}})^2} = {({A^{ - 1}}A)^2} = {I^2}$
$\Rightarrow {A^2}{({A^{ - 1}})^2} = {({A^{ - 1}})^2}{A^2} = I$
This gives ${({A^{ - 1}})^2}$ is the inverse of ${A^2}$.
So we can write ${({A^2})^{ - 1}} = {({A^{ - 1}})^2}$.
This gives option B is true.

C) ${({A^T})^{ - 1}} = {({A^{ - 1}})^T}$
We know that ${(AB)^T} = {B^T}{A^T}$.
So we have,
${({A^{ - 1}}A)^T} = {A^T}{({A^{ - 1}})^T}$
We know that ${A^{ - 1}}A = I$ and ${I^T} = I$.
This gives,
$I = {A^T}{({A^{ - 1}})^T} - - - (i)$
Also we have,
${(A{A^{ - 1}})^T} = {({A^{ - 1}})^T}{A^T}$
$\Rightarrow I = {({A^{ - 1}})^T}{A^T} - - - (ii)$
From $(i)$ and $(ii)$ we have,
$\Rightarrow {A^T}{({A^{ - 1}})^T} = I = {({A^{ - 1}})^T}{A^T}$
This gives ${({A^{ - 1}})^T}$ is the inverse of ${A^T}$.
This can be written as ${({A^T})^{ - 1}} = {({A^{ - 1}})^T}$.
So option C is true.
The question is to find the wrong statement.
Therefore the answer is option A.

Note: We see that the first option is false. That is ${A^{ - 1}} \ne {\left| A \right|^{ - 1}}$. But we have ${\left| A \right|^{ - 1}} = \left| {{A^{ - 1}}} \right|$.That is determinant of the inverse of a matrix and the inverse of its determinant are the same. Here ${\left| A \right|^{ - 1}}$ actually represents $\dfrac{1}{{\left| A \right|}}$.