GATE Breaker

If the matrix $A = \begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & -1 & 0 & -1 \\ 0 & 0 & i & i \\ 0 & 0 & 0 & -i \end{bmatrix}$, then find the value of $A^4$ using Cayley-Hamilton theorem.  (GATE 1993)  Answer: $A^4 = I$.  Explanation:  Since the given matrix $A$ is an upper triangular matrix, the eigenvalues of $A$ are the main diagonal values.  Therefore, the eigenvalues of $A$ are $1$, $-1$, $i$, $-i$.  Hence, the characteristic equation of of $A$ is  $(\lambda - 1)(\lambda + 1)(\lambda - i)(\lambda + i) = 0$   $\Rightarrow  (\lambda^2 - 1)(\lambda^2 + 1) = 0$  $\Rightarrow  (\lambda^4 - 1) = 0$  Therefore, by Calley-Hamilton theorem, we have $A^4 - I = 0$, i.e., $A^4 = I$. 

Suppose that the eigenvalues of matrix $A$ are 1, 2, 4. Find the determinant of $(A^{-1})^T$.  Answer: 0.125  Explanation:  The eigenvalues of matrix $A$ are 1, 2, 4.  Therefore, determinant of $A$, $|A|$ = product of the  eigenvalues of $A$ = 8.  Now, $|A^{-1}| = \frac{1}{|A|} = \frac{1}{8}$.  Hence, $|(A^{-1})^T| = |A^{-1}| = \frac{1}{8} = 0.125$. 

Two eigenvalues of a $3 \times 3$ real matrix $P$ are $(2 + \sqrt{-1})$ and $3$. Find the determinant of $P$.  (GATE 2016)  Answer: 15  Explanation:  Here, $P$ is a $3 \times 3$ real matrix.  The given eigenvalues are $2 + i$ and $3$.  Since $P$ is a real matrix, the third eigenvalue must be $2 - i$.  We know that the determinant of a real matrix is equal to the product of its eigenvalues.  Therefore, the determinant of $P$ = product of eigenvalues = $(2 + i) \times 3 \times (2 - i)$ = $15$. 

Find the determinant of the matrix \[\begin{bmatrix} 6 & -8 & 1 & 1\\ 0 & 2 & 4 & 6\\ 0 & 0 & 4 & 8\\ 0 & 0 & 0 & -1 \end{bmatrix}\].  (A) 11, (B) -48, (C) 0, (D) -24 (GATE 1997) Answer: (B) -48  Explanation:  The given matrix is an upper triangular matrix.  We know that the determinant of an upper triangular matrix is the product of the main diagonal elements.  Therefore, the determinant of the given matrix is $= 6 \times 2 \times 4 \times (-1) = -48$.   Previous Post Next Post