MHB 6.2.8 {{k,4},{3,k}}=k what is k

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The discussion centers on finding the positive real value of k for which the determinant of the matrix [[k, 4], [3, k]] equals k. The determinant is calculated as k^2 - 12, leading to the equation k^2 - k - 12 = 0. Factoring this equation gives (k - 4)(k + 3) = 0, resulting in k = 4 and k = -3. Since only the positive solution is relevant, k is determined to be 4. Verification of the solution confirms that substituting k back into the original determinant equation holds true.
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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
 
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got it :unsure:
 
Yes, what you have done is exactly correct!
$$\left|\begin{array}{cc} k & 4\\ 3 & k\end{array}\right|= k^2- 12= k$$.
[math]k^2- k- 12= (k- 4)(k+ 3)= 0[/math].

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:
$$\left|\begin{array}{cc} k & 4\\ 3 & k\end{array}\right|= \left|\begin{array}{cc} 4 & 4\\ 3 & 4\end{array}\right|= 4^2- 12= 16- 12= 4= k$$.
 
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