Question

If \[{\rm{X}}\] is a square matrix of order \[{\rm{3}} \times {\rm{3}}\]and \[\lambda \] is a scalar, then find the value of \[adj\left( {\lambda X} \right)\].
A. \[\lambda adj\left( X \right)\]
B. \[{\lambda ^3}adj\left( X \right)\]
C\[{\lambda ^2}adj\left( X \right)\].
D. \[{\lambda ^4}adj\left( X \right)\]

Answer 1

Hint: In the given question, we need to find the value of \[adj\left( {\lambda X} \right)\]. For this, we have to use the property of adjoint of the matrix such as \[adj\left( {\lambda X} \right) = {\lambda ^{n - 1}}\left( {adj\left( X \right)} \right)\] . After, putting the value of the order of a matrix, we will get the value of \[adj\left( {\lambda X} \right)\].

Formula used: The property of adjoint of matrix \[{\rm{X}}\] is given by
\[\left. {adj\left( {\lambda X} \right)} \right| = {\lambda ^{n - 1}}\left. {\left( {adj\left( X \right)} \right)} \right|\]. Here, \[n\] is the order of the matrix \[{\rm{X}}\].

Complete step by step solution:
We know that \[X\] is a matrix of order \[{\rm{3}} \times {\rm{3}}\].
Also, \[\lambda \] is a scalar factor.
Now, we will find the value of \[adj\left( {\lambda X} \right)\].
By using the property of adjoint of matrix \[{\rm{X}}\] , we get
\[adj\left( {\lambda X} \right) = {\lambda ^{n - 1}}\left( {adj\left( X \right)} \right)\]
Here, \[n = 3\]
Thus, we get \[adj\left( {\lambda X} \right) = {\lambda ^{3 - 1}}\left( {adj\left( X \right)} \right)\]
By simplifying, we get
\[adj\left( {\lambda X} \right) = {\lambda ^2}\left( {adj\left( X \right)} \right)\]
Hence, the value of \[adj\left( {\lambda X} \right)\] is \[{\lambda ^2}\left( {adj\left( X \right)} \right)\].
Therefore, the correct option is (C).

Additional information: The adjoint of the matrix is defined as the transpose of a cofactor matrix of the square matrix. The adjoint of the matrix \[A\] is denoted by \[adj\left( A \right)\]. It is essential to determine the adjoint of a matrix to calculate the inverse of a matrix. This can be applied only to square matrices.

Note: Students may make mistakes while applying the property of adjoint of matrix. We can solve this example, in another way which is shown below.
Let \[A\]be any \[{\rm{3}} \times {\rm{3}}\] matrix.
We know that, \[{A^{ - 1}} = \dfrac{{adj\left( A \right)}}{{\left| A \right|}}\]
So, \[adj\left( A \right) = {A^{ - 1}}\left| A \right|\]
Now, we can say that \[adj\left( {\lambda A} \right) = {\left( {\lambda A} \right)^{ - 1}}\left| {\lambda A} \right|\]
Here, \[\lambda A\] is a scaling factor.
So, we get \[adj\left( {\lambda A} \right) = {\lambda ^{ - 1}}{\left( A \right)^{ - 1}}\left| {\lambda A} \right|\]
This gives, \[adj\left( {\lambda A} \right) = \dfrac{1}{\lambda }{\left( A \right)^{ - 1}}\left| {\lambda A} \right|\]
But we know that \[\left| {kA} \right| = {k^n}\left| A \right|\], \[n\]is the order of the matrix \[A\].
Thus, we get \[adj\left( {\lambda A} \right) = \dfrac{1}{\lambda }{\left( A \right)^{ - 1}}\left( {{\lambda ^n}} \right)\left| A \right|\]
Here, \[n = 3\]
Hence, we get \[adj\left( {\lambda A} \right) = \dfrac{1}{\lambda }{\left( A \right)^{ - 1}}\left( {{\lambda ^3}} \right)\left| A \right|\]
By simplifying, we get \[adj\left( {\lambda A} \right) = {\left( A \right)^{ - 1}}\left( {{\lambda ^2}} \right)\left| A \right|\]
Therefore, \[adj\left( {\lambda A} \right) = \left( {{\lambda ^2}} \right){\left( A \right)^{ - 1}}\left| A \right|\]
But \[adj\left( A \right) = {A^{ - 1}}\left| A \right|\]
So, we get \[adj\left( {\lambda A} \right) = \left( {{\lambda ^2}} \right)adj\left( A \right)\]
Therefore, the result of both methods is the same.