Question

The characteristic equation of a matrix A is ${{\lambda }^{3}}-5{{\lambda }^{2}}-3\lambda +2I=0$ then $\left| adjA \right|$ =
(a) 4
(b) 25
(c) 9
(d) 30

Answer 1

Hint: There are few things which we have to keep in mind so that we can solve this question. The sum of eigen values $({{\lambda }_{1}}+{{\lambda }_{2}}\,+{{\lambda }_{3}})$ is equal to the sum of diagonal elements and the product of eigen values $({{\lambda }_{1}}\times {{\lambda }_{2}}\times {{\lambda }_{3}})$ will be equal to determinant of A. Also A(adjA) = $\left| A \right|{{I}_{n}}$. Also, there is an identity: $\left| \left| A \right| \right|={{\left| A \right|}^{n}}$.

Complete step-by-step answer:
Now, we can start the solution. ${{\lambda }^{3}}-5{{\lambda }^{2}}-3\lambda +2I=0$ is the characteristic equation of A, so the value of A will satisfy it. So, after replacing A with $\lambda ,$we get:
${{A}^{3}}-5{{A}^{2}}-3A+2I=0...........(i)$
The above cubic equation has three roots which are called eigen values. These eigen values are represented by ${{\lambda }_{1}},{{\lambda }_{2}}$ and ${{\lambda }_{3}}.$ We also know that in the cubic equation of type ${{x}^{3}}+b{{x}^{2}}+cx+d=0$, sum of roots = -b and product of roots = -d. Therefore, from equation (i), we can say that:
${{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}}=5.......(ii)$
${{\lambda }_{1}}\times {{\lambda }_{2}}\times {{\lambda }_{3}}=-2.........(iii)$
Also, we know that A(adjA) = $\left| A \right|{{I}_{n}}$. Taking determinant on both sides, we get $\left| A(adjA) \right|=\left| \left| A \right|{{I}_{n}} \right|$. We can also write this is as:
$\left| A \right|\left| adjA \right|={{\left| A \right|}^{n}}\left| {{I}_{n}} \right|..........(iv)$
$\Rightarrow \left| adjA \right|={{\left| A \right|}^{n-1}}\left| {{I}_{n}} \right|$
The value of $\left| {{I}_{n}} \right|$ is equal to 1 because it is a unit matrix of order n and irrespective of the order, the value of its determinant will always be equal to 1 i.e. $\left| {{I}_{n}} \right|=1$. Therefore, we get, $\left| adjA \right|={{\left| A \right|}^{n-1}}..........(v)$
We know that product of eigen values $({{\lambda }_{1}}\times {{\lambda }_{2}}\times {{\lambda }_{3}})$= $\left| A \right|$. Therefore, from equation (iii) and equation (v), we get $\left| adjA \right|={{(-2)}^{n-1}}$. Now, as we know the order of the above matrix is 3, the value of n becomes 3. So the value of n – 1 = 3 – 1 = 2. Putting this value in the above equation, we get: $\left| adjA \right|={{(-2)}^{2}}\Rightarrow 4$.
So, the value of $\left| adjA \right|$ is equal to 4. Hence, (a) is the correct option.

Note: The student must be aware of the basic concepts to solve this question. The characteristic equation of a matrix, eigen values, adjoint of a matrix, unit matrix etc.. must be known thoroughly. The mistake that the student can make is by assuming the sum of roots = -b and product of roots = -d in the cubic equation obtained.