Question

If transpose of a matrix A is $A{{A}^{T}}=I$ and $\det \left( A \right)=1$, then
(A). Every element of A is equal to its co-factor.
(B). Every element of A and its co-factor are additive inverse of each other.
(C). Every element of A and its co-factor are multiplicative inverse of each other.
(D). None of these

Answer 1

Hint: In the given expression $A{{A}^{T}}=I$, pre-multiply by ${{A}^{-1}}$ on both the sides. Then you will get the expression as ${{A}^{T}}={{A}^{-1}}$. We know that ${{A}^{-1}}=\dfrac{adjA}{\det (A)}$, it is given in the question that $\det \left( A \right)=1$ so plugging the value of $\det \left( A \right)=1$ in ${{A}^{-1}}=\dfrac{adjA}{\det (A)}$ we get ${{A}^{-1}}=adjA$. This shows that every element of A is equal to its co-factor.

Complete step-by-step solution -
It is given in the question that:
$A{{A}^{T}}=I$
Pre – multiplying ${{A}^{-1}}$ on both the sides we get,
${{A}^{-1}}A{{A}^{T}}={{A}^{-1}}$
From the properties of matrices, we know that ${{A}^{-1}}A=I$ and multiplying anything with identity can give you the same result so rewriting the above equation as:
${{A}^{T}}={{A}^{-1}}$………..Eq. (1)
 We know that if we want to find the inverse of a matrix then the formula for that is:
${{A}^{-1}}=\dfrac{adjA}{\det \left( A \right)}$
Substituting this value of inverse in the eq. (1) we get,
${{A}^{T}}=\dfrac{adjA}{\det \left( A \right)}$
It is given in the question that $\det \left( A \right)=1$ so plugging the value of $\det \left( A \right)$ in the above equation we get, ${{A}^{T}}=adjA$
Taking the transpose on both the sides we get,
$\begin{align}
  & {{\left( {{A}^{T}} \right)}^{T}}={{\left( adjA \right)}^{T}} \\
 & \Rightarrow A=\text{Co-factor of A} \\
\end{align}$
We know that adj A is the transpose of the co-factor matrix of A so when we do transpose of A then we will be left with co-factor of matrix A only.
The above equation shows that every element of A is equal to the corresponding cofactor of A.
Hence, the correct option is (a).

Note: Some properties of matrices that we have used above:
Double transpose of any matrix will give you the matrix itself.
${{\left( {{A}^{T}} \right)}^{T}}=A$
 Explaining what this equation $A=\text{Co-factor of A}$ means:
Let us assume a matrix A as:
$A=\left( \begin{align}
  & {{a}_{11}}\text{ }{{a}_{12}} \\
 & {{a}_{21}}\text{ }{{a}_{22}} \\
\end{align} \right)$
Writing co – factor of matrix A we get,
$Co-factor\left( A \right)=\left( \begin{align}
  & {{C}_{11}}\text{ }{{C}_{12}} \\
 & {{C}_{21}}\text{ }{{C}_{22}} \\
\end{align} \right)$
Equating matrix A and its co – factor we get,
$\left( \begin{align}
  &{{a}_{11}}\text{ }{{a}_{12}} \\
 & {{a}_{21}}\text{ }{{a}_{22}} \\
\end{align} \right)$ = $ \left( \begin{align}
  & {{C}_{11}}\text{ }{{C}_{12}} \\
 & {{C}_{21}}\text{ }{{C}_{22}} \\
\end{align} \right)$
Now, it will be more clear to you how each element of the matrix A is equal to its co – factor.