Answer to Question #230373 in Linear Algebra for moe

Question

A system of linear equations has the augmented matrix

$\left(\begin{array}{cccc}1&1&k&:&1\\-1&1&2k&:&-3\\k&1&1&: &k+1\end{array}\right)$

where k

k is a constant. What is the value of k such that the system above has no solution?

1. $k=-1$
2. $k=-2$
3. $k=0$
4. $k=1$

Answer 1

Solution:

For no solution, $\det \begin{pmatrix}1&1&k\\ \:-1&1&2k\\ \:k&1&1\end{pmatrix}=0$

$\Rightarrow 0=1\cdot \det \begin{pmatrix}1&2k\\ 1&1\end{pmatrix}-1\cdot \det \begin{pmatrix}-1&2k\\ k&1\end{pmatrix}+k\det \begin{pmatrix}-1&1\\ k&1\end{pmatrix}$

$\Rightarrow 0=1\cdot \left(1-2k\right)-1\cdot \left(-1-2k^2\right)+k\left(-1-k\right) \\\Rightarrow 0=k^2-3k+2 \\\Rightarrow k^2-2k-k+2=0 \\\Rightarrow (k-2)(k-1)=0 \\\Rightarrow k=2,k=1$

Thus, option 4 is correct.