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Question: If transpose of a matrix A is \(A{{A}^{T}}=I\) and \(\det \left( A \right)=1\), then (A). Every el...

If transpose of a matrix A is AAT=IA{{A}^{T}}=I and det⁡(A)=1\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

Explanation

Solution

Hint: In the given expression AAT=IA{{A}^{T}}=I, pre-multiply by A−1{{A}^{-1}} on both the sides. Then you will get the expression as AT=A−1{{A}^{T}}={{A}^{-1}}. We know that A−1=adjAdet⁡(A){{A}^{-1}}=\dfrac{adjA}{\det (A)}, it is given in the question that det⁡(A)=1\det \left( A \right)=1 so plugging the value of det⁡(A)=1\det \left( A \right)=1 in A−1=adjAdet⁡(A){{A}^{-1}}=\dfrac{adjA}{\det (A)} we get A−1=adjA{{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:
AAT=IA{{A}^{T}}=I
Pre – multiplying A−1{{A}^{-1}} on both the sides we get,
A−1AAT=A−1{{A}^{-1}}A{{A}^{T}}={{A}^{-1}}
From the properties of matrices, we know that A−1A=I{{A}^{-1}}A=I and multiplying anything with identity can give you the same result so rewriting the above equation as:
AT=A−1{{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=adjAdet⁡(A){{A}^{-1}}=\dfrac{adjA}{\det \left( A \right)}
Substituting this value of inverse in the eq. (1) we get,
AT=adjAdet⁡(A){{A}^{T}}=\dfrac{adjA}{\det \left( A \right)}
It is given in the question that det⁡(A)=1\det \left( A \right)=1 so plugging the value of det⁡(A)\det \left( A \right) in the above equation we get, AT=adjA{{A}^{T}}=adjA
Taking the transpose on both the sides we get,
(AT)T=(adjA)T ⇒A=Co-factor of A \begin{aligned} & {{\left( {{A}^{T}} \right)}^{T}}={{\left( adjA \right)}^{T}} \\\ & \Rightarrow A=\text{Co-factor of A} \\\ \end{aligned}
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.
(AT)T=A{{\left( {{A}^{T}} \right)}^{T}}=A
Explaining what this equation A=Co-factor of AA=\text{Co-factor of A} means:
Let us assume a matrix A as:
A=(a11 a12 a21 a22 )A=\left( \begin{aligned} & {{a}_{11}}\text{ }{{a}_{12}} \\\ & {{a}_{21}}\text{ }{{a}_{22}} \\\ \end{aligned} \right)
Writing co – factor of matrix A we get,
Co−factor(A)=(C11 C12 C21 C22 )Co-factor\left( A \right)=\left( \begin{aligned} & {{C}_{11}}\text{ }{{C}_{12}} \\\ & {{C}_{21}}\text{ }{{C}_{22}} \\\ \end{aligned} \right)
Equating matrix A and its co – factor we get,
(a11 a12 a21 a22 )\left( \begin{aligned} &{{a}_{11}}\text{ }{{a}_{12}} \\\ & {{a}_{21}}\text{ }{{a}_{22}} \\\ \end{aligned} \right) = (C11 C12 C21 C22 ) \left( \begin{aligned} & {{C}_{11}}\text{ }{{C}_{12}} \\\ & {{C}_{21}}\text{ }{{C}_{22}} \\\ \end{aligned} \right)
Now, it will be more clear to you how each element of the matrix A is equal to its co – factor.