Question

find k if the following matrices are singular
[k-1 2 3]
[3 1 2]
[1 -2 4 ]​

Answer 1

k=$\frac{49}{8}$

Step-by-step explanation:

1. Let A=$\left[\begin{array}{ccc}k-1&2&3\\3&1&2\\1&-2&4\end{array}\right]$

$\mid A\mid=$$\begin{vmatrix}k-1&2&3\\3&1&2\\1&-2&4\end{vmatrix}$

When matrix is singular then det A=0

$k-1(4(1)+2(2))-2(3(4)-2(1))+3(3(-2)-1)=0$

$(k-1)(8)-2(10)+3(-6-1)=0$

$8(k-1)-20-21=0$

$8(k-1)-41=0$

$8(k-1)=41$

$k-1=\frac{41}{8}$

$k-1=\frac{41}{8}+1=\frac{41+8}{8}=\frac{49}{8}$

The value of k=$\frac{49}{8}$

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