GATE 2018 EE – Question 54
Let $A=\begin{bmatrix}1&0&-1\\-1&2&0\\0&0&-2\end{bmatrix}$ and $B=A^3-A^2-4A+5I$, where $I$ is the $3\times3$ identity matrix. The determinant of $B$ is ________ (up to 1 decimal place).
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Correct answer: 1
Explanation
$A$ has eigenvalues $1,2,-2$ (it is block triangular). The eigenvalues of $B$ are $\lambda^3-\lambda^2-4\lambda+5$, which equal $1$ at $\lambda=1$, $1$ at $\lambda=2$ and $-8-4+8+5=1$ at $\lambda=-2$. So $\det B=1$.