GATE CSE 2016 | Set 2 | Question: 06

Question

Suppose that the eigenvalues of matrix $A$ are $1, 2, 4$. The determinant of $\left(A^{-1}\right)^{T}$ is _________.

Comments

What I did

1. (λ-1)(λ-2)(λ-4) = 0, solved this and got the characteristic equation as $\lambda ^3 - 7\lambda^2 + 14\lambda-8 = 0$
2. Replaced $\lambda$ with A → $A^3-7A^2+14A-8=0$
    $A^3-7A^2+14A=8$
3. To get the inverse Multiplied both side by $A^{-1}$
4.  $A^{-1}(A^3-7A^2+14A)=A^{-1}8A$
   $A^{2}-7A+14I = 8I$
   therefore, $A^{-1} = \frac{A^2-7A+14I}{8} $
5. After solving this equation I got $A^{-1} = 1*\frac{1}{2} * \frac{1}{4} = \frac{1}{8}$ 😥

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@faisal_sayyed how did you arrive at 5th point? 
or you just took 1/2 * 1/1 * 1/4 = 1/8? (if yes then what was the point of above caluclations ?) 
 

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you are right @R2-D2

The 5th step is completely incorrect. From the Cayley-Hamilton theorem, we obtain the true matrix inverse expression:

$$A^{-1} = \frac{A^2 - 7A + 14I}{8}$$

Writing $A^{-1} = 1 \times \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$ is wrong.

$A^{-1}$ is a matrix, whereas $1$, $\frac{1}{2}$, and $\frac{1}{4}$ are its individual eigenvalues. You cannot equate a full matrix to a single scalar product.

Answer 1

Determinant of Matrix $A =$ product of eigen values $= 1 \times 2 \times 4 =8$

Determinant of Inverse Matrix of $A, \text{det}(A^{-1}) = \frac{1}{\text{det}(A)} =\frac{1}{8}$

Determinant remains same after the Transpose operation.

So, determinant of $(A^{-1})^T = \text{det}(A^{-1}) = \frac{1}{8} = 0.125$

Comments

Inverse of a matrix comprises of transpose of cofactor matrix divided by det of that matrix, so do we not need to consider the value of transpose of cofactor matrix turned into determinant? pls explain.

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https://www.youtube.com/watch?v=ighcncy6qQ8&t=48s

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We can use any of these 2 properties:

1) det(A$^{-1}$) = 1 / (det(A))

Proof:

det(AA$^{-1}$) = det(I)

i;e

det(A) det(A$^{-1}$) = 1

det(A$^{-1}$) = 1 / (det(A))

2) if $\lambda$ is the eigen value of A, then (1/$\lambda$) is the eigen value of A$^{-1}$

Proof: (source: stackoverflow)

Av=λv

Pre multiply by A$^{-1}$, we have

⟹A$^{-1}$Av=λA$^{-1}$v

⟹Iv=λA$^{-1}$v

⟹(1/λ)v=A$^{-1}$v

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@ShamikBanerjee

you can simply do it like this:

eigen values of Matrix A are :1,2,4

Now,eigen values of A inverse are 1/1,1/2,1/4 

then the det of A inverse will be 1/1*1/2*1/4=0.125 

and then ( A inverse )^T = (0.125)^T =0.125

Answer 2

0.125

Eigen value of A inverse is 1,1/2,1/4. Product of those eigen values gives determinant value.

Transposing a matrix doesn't change is eigen value.

Answer 3

$|A|= \text{Product of eigenvalues of A} =1*2*4=8$

$|(A^{-1})^{T}| = |(A^{T})^{-1}| = \frac{1}{|A^{T}|} = \frac{1}{|A|} = \frac{1}{8} = 0.125$

Property Used:

$(A^{-1})^{T} = (A^{T})^{-1}$

$|A^T|=|A|$

$|A^{-1}| = \frac{1}{|A|}$

Answer 4

see this solution based on simple eigen value property 

Answer 5

Answer: 0.125 

$$\text{If eigenvalues of matrix A are } \lambda_1, \lambda_2 \text{ and } \lambda_3 \text{ then, eigenvalues of } A^{-1} \text{ are } 1/\lambda_1 \text{ , } 1/\lambda_2 \text{ and } 1/\lambda_3.$$

$$\text{If eigenvalues of matrix A are } \lambda_1, \lambda_2 \text{ and } \lambda_3 \text{ then, eigenvalues of } A^{T} \text{ are } \lambda_1 \text{ , } \lambda_2 \text{ and } \lambda_3.$$

$$\text{Determinant of a matrix is product of its eigenvalues.} $$

$$ \text{Thus, eigenvalues of } A^{-1} \text{ are } 1 \text{, } 1/2 \text{ and } 1/4.$$

$$\text{Hence, eigenvalues of matrix } (A^{-1})^T \text{are } 1, 1/2 \text{ and }1/4.$$

$$\text{ Therefore, determinant of } (A^{-1})^T \text{ = } 1/8 = 0.125$$