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$$ \begin{bmatrix} 3 & 2 & -2 \\ 1 & 1 & -1 \\ -1 & 2 & k \\ \end{bmatrix} $$

Find the values k for which rank(A) = 3 From what I have tried myself, I have brought it down to row-echelon form where it looks something like this:

$$ \begin{bmatrix} 1 & 1 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & k+2 \\ \end{bmatrix} $$

Through various row operations, but I seem to be stuck on whether or not I'm doing the right thing. Thanks!

To start: $$ \begin{bmatrix} 3 & 2 & -2 \\ 1 & 1 & -1 \\ -1 & 2 & k \\ \end{bmatrix} $$ Interchanging Rows 1 and 2.

$$ \begin{bmatrix} 1 & 1 & -1 \\ 3 & 2 & -2 \\ -1 & 2 & k \\ \end{bmatrix} $$

then, Row 2 minus 3 times Row 1 and Row 3 plus Row 1.

$$ \begin{bmatrix} 1 & 1 & -1 \\ 0 & -1 & 1 \\ 0 & 3 & k-1 \\ \end{bmatrix} $$

then, multiply Row 2 by (-1)

$$ \begin{bmatrix} 1 & 1 & -1 \\ 0 & 1 & -1 \\ 0 & 3 & k-1 \\ \end{bmatrix} $$

Finally, Row 3 minus 3 times Row 2.

$$ \begin{bmatrix} 1 & 1 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & k+2 \\ \end{bmatrix} $$

Siccamende
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  • I have not verified that the final form is correct, but what you have described is a good way to solve the problem. Reduce the matrix so that you either obtain the reduced row echelon form, or a form that is close enough (so that you can read off the rank). From the second matrix you have (which I have not verified is correct) you could conclude that the rank of the matrix is 3 if and only if $k\neq -2$. – Dave Nov 07 '17 at 03:22
  • Oh no. That's bad and embarrassing. – Siccamende Nov 07 '17 at 03:42
  • I also keep trying doing the same legal row operations and I keep getting the same result. (I'm interchanging rows 1 and 2, then Row 2 - 3Row 1, then Row 3 + Row 1, then Row - 3Row 2). – Siccamende Nov 07 '17 at 03:53
  • Let me write the all my steps, hold on. Because its the second row minus 3 times the first row. – Siccamende Nov 07 '17 at 03:59
  • I have added the changes and showed my steps. – Siccamende Nov 07 '17 at 04:07

3 Answers3

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Hint:- Rank$=3$ iff $\det(A)\ne 0$

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Assuming that you row reduced correctly, you should just check what happens when $k+2=0$, and when $k+2 \neq 0$, recalling that $rk=3 \iff det(A) \neq 0$.

A nice way to see this is by checking what happens to the determinant under row operations and checking that the determinant of the original matrix has a zero if and only if the reduced matrix does.

Note that the determinant of your row reduced matrix is $k+2$.

Andres Mejia
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$\begin{align*} \det(A)&=3(k+2)-2(k-1)-2(2+1)\\ &=k+2 \end{align*}$

So $\det(A)=0 \iff k=-2$.

Laars Helenius
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