GO Classes DPP | GATE CS | Linear Algebra | Cayley-Hamilton Theorem

Answer 1

Since the eigenvalues of $A$ are $2$, $3$, and $5$, the characteristic polynomial is

$p(\lambda)=(\lambda-2)(\lambda-3)(\lambda-5)$.

Expanding,

$p(\lambda)=\lambda^3-10\lambda^2+31\lambda-30$.

By the Cayley-Hamilton theorem,

$A^3-10A^2+31A-30I=0$.

Since all eigenvalues are nonzero, $A$ is invertible. Multiplying by $A^{-1}$, we get

$A^2-10A+31I-30A^{-1}=0$.

So,

$30A^{-1}=A^2-10A+31I$,

$\boxed{\therefore A^{-1}=\frac{1}{30}(A^2-10A+31I)}$.

Hence, the correct answer is A.

Answer 2

For a $3 \times 3$ matrix $A$, the characteristic equation is given by:

$$\lambda^3 - S_1\lambda^2 + S_2\lambda - S_3 = 0$$

Where the coefficients are defined as:

• $S_1$ (Sum of Eigenvalues / Trace):

  $$S_1 = \lambda_1 + \lambda_2 + \lambda_3 = \text{tr}(A)$$

• $S_2$ (Sum of Principal Minors):

  $$S_2 = \lambda_1\lambda_2 + \lambda_2\lambda_3 + \lambda_3\lambda_1$$

• $S_3$ (Product of Eigenvalues / Determinant):

  $$S_3 = \lambda_1\lambda_2\lambda_3 = \det(A)$$

• subtituting  given eignen value we  get 

• s1 = 3 + 5 +2 = 10 

• s2 = 2* 3 + 3*5 + 5*2 = 31 

• s3 = 2*3 * 5= 30 

• on simplifiying we get option A .

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