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I came across this exercise:

Use Cramer's rule to find the values of k for which the system of equations below has a unique solution:

$$x - 2y + k = k + 4$$

$$(k-1)x + y + z = k-3$$

$$-kx + 2y - 2z = -10 $$

So my question is how is Cramer's rule relevant here? An obvious way to find the values of k for which the system has a solution would be finding the values of k for which the determinant of the coefficients matrix is $\neq 0$. Are the authors of the exercise expecting an answer along these lines?

Using Cramer's rule to find the solution of system we have:

$x = \frac{D_x}{D}, y = \frac{D_y}{D}, z = \frac{D_z}{D}$

In order for the above solution to exist, $D \neq 0$ etc. etc. etc. and we find the values of k for which $D \neq 0$

Am I missing something?

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1 Answer 1

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I do not think you are missing something.

It suffices to find out the values of $k$ for which the determinant of the coefficient matrix is not zero.

More precisely, one has the following equation to solve: \begin{align*} \det(A) & = \begin{vmatrix} 1 & -2 & 0\\ k - 1 & 1 & 1\\ -k & 2 & -2 \end{vmatrix} \neq 0 \end{align*}

Can you take it from here?

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  • $\begingroup$ Thanks for your response. I know how to solve the exercise. I just don't see how Cramer's rule is relevant $\endgroup$
    – Eva K.
    Mar 7, 2022 at 1:47
  • $\begingroup$ It depends on what you call Cramer's rule. By the context of the exercise, I suppose it is enough to determine the above-mentioned values of $k$ which do not vanish the determinant. $\endgroup$ Mar 7, 2022 at 1:48
  • $\begingroup$ @VakiPitsi See Cramer's Rule and note that the pertinent denominator is not allowed to equal $(0)$. $\endgroup$ Mar 7, 2022 at 1:55
  • $\begingroup$ Thank you both just as I suspected $\endgroup$
    – Eva K.
    Mar 7, 2022 at 2:57
  • $\begingroup$ @VakiPitsi you can upvote/choose my answer if you think it deserves. $\endgroup$ Mar 13, 2022 at 22:09

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