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..

c <- matrix(data = c(1,2,1,2,0,1,2,3,7/4),
nrow=3,
ncol=3,
byrow=TRUE)

det(c)
## [1] 0