6.2.8 {{k,4},{3,k}}=k what is k

[karush]

What positive real value of \textbf{k}, is the determinant of the Matrix
$\begin{bmatrix}
k & 4\\3 & k
\end{bmatrix}$
equal to $k$?
a.3 b.4 c.12 d. $\sqrt{12}$ e. none $k^2-12=k\implies k^2-k-12=0\implies k^2-k-12=0\implies (k+3)(k-4)=0$
$k=\boxed{4}$

ok basically I think most could just eyeball this and get it
but when I tried to ck it in W|A it froze,,,

W|A

 

[skeeter]

https://www.wolframalpha.com/input?i2d=true&i=det{{k,4},{3,k}}=k

 

[karush]

got it :unsure:

 

[HOI]

Yes, what you have done is exactly correct!
\(\displaystyle \left|\begin{array}{cc} k & 4\\ 3 & k\end{array}\right|= k^2- 12= k\).
\(\displaystyle k^2- k- 12= (k- 4)(k+ 3)= 0\).

k= -3 and k= 4. Since the question asks for the "positive" solution, the answer is "4". Well done!

I would not use Wolfram alpha or any tool to check- just put k= 4 back in the original equation:
\(\displaystyle \left|\begin{array}{cc} k & 4\\ 3 & k\end{array}\right|= \left|\begin{array}{cc} 4 & 4\\ 3 & 4\end{array}\right|= 4^2- 12= 16- 12= 4= k\).