Question

Let A be any 3x3 invertible matrix. Then which one of the following is not always true?
$\begin{align}
  & \text{a) }adj\left( adj\left( A \right) \right)\text{ }=\text{ }\left| A \right|\cdot {{\left( adj(A) \right)}^{-1}} \\
 & \text{b) }adj\left( adj\left( A \right) \right)\text{ }=\text{ }{{\left| A \right|}^{2}}\cdot {{\left( adj(A) \right)}^{-1}} \\
 & \text{c) }adj\left( adj\left( A \right) \right)\text{ }=\text{ }\left| A \right|\cdot A \\
 & \text{d) }adj\left( A \right)\text{ }=\text{ }\left| A \right|\cdot {{A}^{-1}} \\
\end{align}$

Answer 1

Hint: In this, to check that which option does not hold we will use the following two properties of Adjoint of the matrix. Adjoint of a matrix is a matrix formed by the cofactors of the matrix.
\[\begin{align}
  & \text{1) A}\left( \text{adj}\left( \text{A} \right) \right)\text{=}\left( \text{adj}\left( \text{A} \right) \right)\text{A=}\left| \text{A} \right|\centerdot \text{I} \\
 & \text{2) }\!\!|\!\!\text{ adj}\left( \text{A} \right)\text{ }\!\!|\!\!\text{ = }{{\left| \text{A} \right|}^{n-1}} \\
\end{align}\]

Complete step by step answer:
 Given that the matrix A is 3x3 invertible matrix then,
\[\left( \text{adj}\left( \text{A} \right) \right)\text{A=}\left| \text{A} \right|\centerdot \text{I}....\text{(1)}\]
Replacing matrix A by adj(A) in equation (1), we get
\[\left( \text{adj}\left( \text{adj}\left( \text{A} \right) \right) \right)\text{adj}\left( \text{A} \right)\text{=}\left| \text{adj}\left( \text{A} \right) \right|\cdot \text{I}....\text{(2)}\]
Since the matrix is 3x3 \[\Rightarrow \text{ }\!\!|\!\!\text{ adj}\left( \text{A} \right)\text{ }\!\!|\!\!\text{ = }{{\left| \text{A} \right|}^{2}}\]
Equation (2), becomes
\[\left( \text{adj}\left( \text{adj}\left( \text{A} \right) \right) \right)\text{adj}\left( \text{A} \right)\text{=}{{\left| \text{A} \right|}^{2}}\cdot \text{I}....\text{(3)}\]
Since A is invertible matrix implies that adj(A) is a invertible matrix.
Apply \[{{\left( \text{adj}\left( \text{A} \right) \right)}^{\text{-1}}}\]on both sides of equation (3), we get\\
\[\left( \text{adj}\left( \text{adj}\left( \text{A} \right) \right) \right)\text{adj}\left( \text{A} \right){{\left( \text{adj}\left( \text{A} \right) \right)}^{-1}}\text{=}{{\left| \text{A} \right|}^{2}}{{\left( \text{adj}\left( \text{A} \right) \right)}^{-1}}\]
Since \[\text{adj}\left( \text{A} \right){{\left( \text{adj}\left( \text{A} \right) \right)}^{\text{-1}}}=\text{I}\]
\[\left( \text{adj}\left( \text{adj}\left( \text{A} \right) \right) \right)\text{I=}{{\left| \text{A} \right|}^{2}}{{\left( \text{adj}\left( \text{A} \right) \right)}^{-1}}\]
\[\text{adj}\left( \text{adj}\left( \text{A} \right) \right)\text{=}{{\left| \text{A} \right|}^{2}}{{\left( \text{adj}\left( \text{A} \right) \right)}^{-1}}\]
This proves that option b is always true.

From property (1) of adjoint of matrix we get
\[\text{A}\left( \text{adj}\left( \text{A} \right) \right)\text{=}\left| \text{A} \right|\cdot \text{I}....\text{(3)}\]
Replacing matrix A by adj(A) in equation (3), we get
\[\text{adj}\left( \text{A} \right)\left( \text{adj}\left( \text{adj}\left( \text{A} \right) \right) \right)\text{=}\left| \text{adj}\left( \text{A} \right) \right|\cdot \text{I}....\text{(4)}\]
Multiplying equation (4) matrix A from left side, we get
\[\text{A}\cdot \text{adj}\left( \text{A} \right)\left( \text{adj}\left( \text{adj}\left( \text{A} \right) \right) \right)\text{=A}\left\{ \left| \text{adj}\left( \text{A} \right) \right|\cdot \text{I} \right\}\]
\[\left( \text{A}\cdot \text{adj}\left( \text{A} \right) \right)\left( \text{adj}\left( \text{adj}\left( \text{A} \right) \right) \right)\text{=}\left| \text{adj}\left( \text{A} \right) \right|\left\{ \text{A}\cdot \text{I} \right\}\]
From property (1), we get
\[\left( \left| \text{A} \right|\cdot \text{I} \right)\left( \text{adj}\left( \text{adj}\left( \text{A} \right) \right) \right)\text{=}\left| \text{adj}\left( \text{A} \right) \right|\left\{ \text{A}\cdot \text{I} \right\}\]\\
\[\left| \text{A} \right|\left( \text{adj}\left( \text{adj}\left( \text{A} \right) \right) \right)\text{=}\left| \text{adj}\left( \text{A} \right) \right|\cdot \text{A}\]
For 3x3 matrix property (2) is \[\text{ }\!\!|\!\!\text{ adj}\left( \text{A} \right)\text{ }\!\!|\!\!\text{ = }{{\left| \text{A} \right|}^{2}}\]
\[\left| \text{A} \right|\left( \text{adj}\left( \text{adj}\left( \text{A} \right) \right) \right)\text{=}{{\left| \text{A} \right|}^{2}}\cdot \text{A}\]
By cancelling |A| form both sides, we get
\[\text{adj}\left( \text{adj}\left( \text{A} \right) \right)\text{=}\left| \text{A} \right|\cdot \text{A}\]\\
The option c is always true.

Since A is invertible implies \[{{\text{A}}^{\text{-1}}}\] exist.
Applying \[{{\text{A}}^{\text{-1}}}\] on both sides of equation (1), we get
\[\left( \text{adj}\left( \text{A} \right) \right)\text{A}\cdot {{\text{A}}^{-1}}\text{=}\left| \text{A} \right|\cdot {{\text{A}}^{-1}}\]
Since \[\text{A}\cdot {{\text{A}}^{-1}}\text{=I}\]
\[\Rightarrow \left( \text{adj}\left( \text{A} \right) \right)\text{I=}\left| \text{A} \right|\cdot {{\text{A}}^{-1}}\]
\[\Rightarrow \text{adj}\left( \text{A} \right)\text{=}\left| \text{A} \right|\cdot {{\text{A}}^{-1}}\]\\
This proves that option d is always true.
If the option b is always true implies that option a is always not true

So, the correct answer is “Option A”.

Note: proof of property (2) is given below for nxn matrix.
Let A be any nxn matrix.
By property (1), we have
\[\text{A}\left( \text{adj}\left( \text{A} \right) \right)\text{=}\left| \text{A} \right|\cdot \text{I}\]
Taking determinants on both sides, we get
\[\text{ }\!\!|\!\!\text{ A}\left( \text{adj}\left( \text{A} \right) \right)\text{ }\!\!|\!\!\text{ = }\!\!|\!\!\text{ }\left| \text{A} \right|\cdot \text{I }\!\!|\!\!\text{ }\]
Since det(AB) = det(A)det(B)
\[\text{ }\!\!|\!\!\text{ A }\!\!|\!\!\text{ }\!\!|\!\!\text{ adj}\left( \text{A} \right)\text{ }\!\!|\!\!\text{ = }\!\!|\!\!\text{ }\left| \text{A} \right|\cdot \text{I }\!\!|\!\!\text{ }\]
For nxn matrix, det(kA) = kndet(A), where k is constant number
\[\text{ }\!\!|\!\!\text{ A }\!\!|\!\!\text{ }\!\!|\!\!\text{ adj}\left( \text{A} \right)\text{ }\!\!|\!\!\text{ =}{{\left| \text{A} \right|}^{n}}\text{ }\!\!|\!\!\text{ I }\!\!|\!\!\text{ }\]
Since the determinant of the identity matrix is 1.
\[\text{ }\!\!|\!\!\text{ A }\!\!|\!\!\text{ }\!\!|\!\!\text{ adj}\left( \text{A} \right)\text{ }\!\!|\!\!\text{ =}{{\left| \text{A} \right|}^{n}}\cdot 1\]
\[\text{ }\!\!|\!\!\text{ A }\!\!|\!\!\text{ }\!\!|\!\!\text{ adj}\left( \text{A} \right)\text{ }\!\!|\!\!\text{ =}{{\left| \text{A} \right|}^{n}}\]
By cancelling |A| on both sides, we get
\[\text{ }\!\!|\!\!\text{ adj}\left( \text{A} \right)\text{ }\!\!|\!\!\text{ =}{{\left| \text{A} \right|}^{n-1}}\]