Question

If ${A_{3 \times 3}}$ and $\det A = 6$, then $\det \left( {2{\text{adj}}A} \right)$ is equal to:
A) 48
B) 8
C) 288
D) 12

Answer 1

Hint: In order to solve the problem we will use basic properties of matrix giving relation between adjoint and determinant of the matrix.

Complete step-by-step answer:

Given that: $\det A = 6$ for the matrix ${A_{3 \times 3}}$

So n = 3 for the matrix and $\left| A \right| = 6$

We have to find $\det \left( {2{\text{adj}}A} \right)$

As we know that $\left| {{\text{adj}}A} \right| = {\left| A \right|^{n - 1}}$ where A is the matrix of $n \times n$ and “a” is a scalar.

Substituting the given values from the problem we get:

$

  \because \left| {{\text{adj}}A} \right| = {\left| A \right|^{n - 1}} \\

   \Rightarrow \left| {{\text{adj}}A} \right| = {\left( 6 \right)^{3 - 1}} \\

   \Rightarrow \left| {{\text{adj}}A} \right| = {\left( 6 \right)^2} \\

   \Rightarrow \left| {{\text{adj}}A} \right| = 36 \\

 $

Also we know that $\left| {aA} \right| = {a^n}\left| A \right|$

So let us find the value of problem statement:

$

  \because \left| {aA} \right| = {a^n}\left| A \right| \\

   \Rightarrow \left| {2A} \right| = {2^n}\left| A \right| \\

   \Rightarrow \left| {2adjA} \right| = {2^n}\left| {adjA} \right| \\

 $

Now let us put the values from above solved part

$

  \because \left| {2adjA} \right| = {2^n}\left| {adjA} \right| \\

   \Rightarrow \left| {2adjA} \right| = {2^3}\left( {36} \right) \\

   \Rightarrow \left| {2adjA} \right| = 8 \times \left( {36} \right) \\

   \Rightarrow \left| {2adjA} \right| = 288 \\

 $

Hence, the value of $\det \left( {2{\text{adj}}A} \right)$ is 288.

So, option C is the correct option.

Note: In order to solve such types of problems students must remember the basic formulae related to matrices and determinants. The determinant is a scalar value that can be computed from the elements of a square matrix and encodes certain properties of the linear transformation described by the matrix. The determinant of a matrix A is denoted det A.