Chapter D, Section DM, Question M11

Find a value of k so that the matrix

\[ A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 0 & 1 \\ 2 & 3 & k \end{bmatrix} \] has \[det(A) = 0 \] or explain why it is not possible.

Solution

\[|A| = 1\begin{bmatrix} 0 & 1 \\ 3 & k \end{bmatrix} -2 \begin{bmatrix} 2 & 1 \\ 2 & k \end{bmatrix} + 1 \begin{bmatrix} 2 & 0 \\ 2 & 3 \end{bmatrix}\]

\[|A| = 1((0.k) - (3 . 1)) - 2((2.k) - (2.1)) + 1 ((2.3) - (2.0)) \] \[|A| = -3 - 4k + 4 + 6 \] \[|A| = 7 - 4k \] Hence.. \[ k = 7/4 \]

Verification..

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