Question

If A is a square matrix of order 3 such that $A\left( adj\left( 3A \right) \right)=27I$ , then $\dfrac{\left| adj\left( adj\left( adjA \right) \right) \right|}{\left| adj\left( adjA \right) \right|}$ is
A. 9
B. 81
C. 729
D. 243

Answer 1

In mathematics, a matrix is a rectangular array or tables of numbers, symbols are arranged in rows and columns. In the above question we have a square matrix which means it has the same rows and same column. A square matrix is a matrix with the same numbers or rows and columns. The order of the square matrix is equal to the number of rows (number of columns).

Complete step-by-step solution:
Now, in the above question we have given that $A\left( adj\left( 3a \right) \right)=27I$, here $adj$ means adjoint of square matrix. Let $A=\left[ a_ij \right]$ be a square matrix of order n. The adjoint of a matrix A is the transpose of the cofactor matrix of A.
The formula to find the $adjA=\left| A \right|{{A}^{-1}}$ and we also know that $adj\left( kA \right)={{k}^{n-1}}adjA$
Now we will use these above formula in the given $A\left( adj\left( 3A \right) \right)=27I$, then we get
$\Rightarrow adj\left( 3A \right)={{3}^{3-1}}adjA$
$\Rightarrow adj\left( 3A \right)=9adjA$
Now we will replace $adj\left( 3A \right)$ by $9adjA$ , then we get
$\Rightarrow A\left( 9adjA \right)=27I$
Now divide the both side of the equation by $9$ we get,
$\Rightarrow A adjA=3I$
Now we know we can write $adjA=\left| A \right|{{A}^{-1}}$ then the above equation becomes
$\Rightarrow A\left| A \right|{{A}^{-1}}=3I$
We know that $A{{A}^{-1}}=I$ putting this in above equation we get,
$\begin{aligned} & \Rightarrow \left| A \right|I=3I \\\ & \Rightarrow \left| A \right|=3 \\\ \end{aligned}$
We have to solve $\dfrac{\left| adj\left( adj\left( adjA \right) \right) \right|}{\left| adj\left( adjA \right) \right|}$ by using the value of the $\left| A \right|=3$ , then we get
$\begin{aligned} & \Rightarrow \dfrac{\left| adj\left( adj\left( adjA \right) \right) \right|}{\left| adj\left( adjA \right) \right|}=\dfrac{{{\left| A \right|}^{{{\left( n-1 \right)}^{3}}}}}{{{\left| A \right|}^{{{\left( n-1 \right)}^{2}}}}} \\\ & \Rightarrow \dfrac{\left| adj\left( adj\left( adjA \right) \right) \right|}{\left| adj\left( adjA \right) \right|}={{\left| A \right|}^{n-1}} \\\ & \Rightarrow \dfrac{\left| adj\left( adj\left( adjA \right) \right) \right|}{\left| adj\left( adjA \right) \right|}={{\left| A \right|}^{3-1}}={{\left| A \right|}^{2}} \\\ \end{aligned}$
Now putting the value of $\left| A \right|=3$ then we get
$\Rightarrow \dfrac{\left| adj\left( adj\left( adjA \right) \right) \right|}{\left| adj\left( adjA \right) \right|}={{3}^{2}}=9$
Hence option A is correct which is $9$.

Note: Matrices are used in computer graphics, such that they are used to manipulate 3D models and project them onto a 2-dimensional screen. Matrix is also used in economics to describe systems of economic relationships.