Let the matrix $X$ be $$X = \left [\begin{matrix}8 & -2\-6 & k\end{matrix}\right ].$$

For $X$ to have $0$ as an eigenvalue, what must $k$ be?

Question

Let the matrix $X$ be $$X = \left [\begin{matrix}8 & -2\-6 & k\end{matrix}\right ].$$

For $X$ to have $0$ as an eigenvalue, what must $k$ be?

BlackTom AI · Accepted Answer

We start by recognizing that for a matrix to have 0 as an eigenvalue, its determinant must be zero, since the eigenvalues are the roots of the characteristic polynomial det(X - λI) and plugging λ = 0 gives det(X).

Given X = [[8, -2], [-6, k]], its determinant is det(X) = 8·k − (−2)·(−6) = 8k − 12.

Set this equal to zero to find the condition on k: 8k − 12 = 0.

Solve for k: 8k = 12, so k = 12/8 = 3/2.

Note on the provided information: no answer choices were supplied in the options field, so the derivation stands on det(X) = 0 as the criterion for 0 to be an eigenvalue, leading to k = 3/2.

Let the matrix \(X\) be \[X = \left [\begin{matrix}8 & -2\\-6 & k\end{matrix}\right ].\] For \(X\) to have \(0\) as an eigenvalue, what must \(k\) be?单项选择题

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