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So I'm given this matrix:

$$\left(\begin{array}{c} a & 1 & 1 \\ 1 & a & 1 \\ 1 & 1 & a\end{array}\right)$$

and am told to find the values of a which make it not invertible.

I know that $a = 0$ means our matrix is invertible (since the column vectors span $\mathbb{R}^3$) but I'm not sure how to go about finding all values of $a$ which make the matrix not invertible.

My thought was to row reduce it and find values of $a$ for which rref isn't the identity. The row reduction with $a$ instead of numbers is tripping me up.

Any help?

Thanks, Mariogs

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Calculate the determinant of A $$\det(A)=a(a^2-1)-1(a-1)+1(1-a)=a^3-a-a+1+1-a=a^3-3a+2$$ The matrix is not invertible when det(A) is equal to zero.

You can solve or guess the solution the obvious solutions which are $a=1$ (double root) and $a=-2$. As stated in the comments, the value $a=1$ is obvious already before that point, since it is the value that makes all columns of the matrix equal (and thus linearly dependent).

So, for every other value of $a$, i.e for $a\in \mathbb{R}-\{1,-2\}$ the matrix is invertible, which means that in this case the column vectors are linearly independent and therefore they span $\mathbb{R}^3$.

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  • $\begingroup$ @BabyDragon True, did not see it. +1. $\endgroup$ – Jimmy R. Feb 27 '14 at 2:59
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Begin by noting that $\begin{pmatrix}a&1&1\\1&a&1\\1&1&a\end{pmatrix}=-\begin{pmatrix}-a&-1&-1\\-1&-a&-1\\-1&-1&-a\end{pmatrix}=-((1-a)I-J)$, where $J=\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}$.

Let's compute the characteristic polynomial of $J$. For this, we note that $J$ has rank $1$, so the null space has dimension $2$, which means that the equation $Jx=0$ has at least two independent solutions, that is, $0$ is an eigenvalue with multiplicity at least $2$. On the other hand, clearly $u=\begin{pmatrix}1\\1\\1\end{pmatrix}$ is an eigenvector (since $Jx$ is a multiple of $u$ for any $x$), and a direct computation gives us that $Ju=3u$. It follows that $3$ is also an eigenvalue of $J$, and since $J$ can have no more than $3$ eigenvalues, then they are $0$, $0$, and $3$. The characteristic polynomial of $J$ is therefore $\det(xI-J)=x^2(x-3)$.

Finally, let $x=1-a$. We see that $$\det((1-a)I-J)=(1-a)^2(1-a-3)=-(1-a)^2(a+2).$$ The matrix we are interested in, $-((1-a)I-J)$, is invertible iff its determinant is nonzero. That means that $a$ must be different from $1$ and from $-2$, and these are the only restrictions.

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Here are some hints:

  1. What do you know about determinants and invertibility?

  2. Alternatively, row reduce the matrix. Since you've already noted that $a = 0$ makes the matrix invertible, you needn't worry about this case when row reducing.

For row reduction, I'll do the first step so you can see how it's done. Let's have $R_2 \rightarrow R_1-aR_2$ (i.e. in place of row 2 put row 1 - $a$ times row 2). Then the matrix becomes

$$\left(\begin{array}{c} a & 1 & 1 \\ 0 & 1-a^2 & 1-a \\ 1 & 1 & a\end{array}\right).$$

Doing the same thing for row 3 (that is row 3 goes to row 1 minus $a$ times row 3) gives

$$\left(\begin{array}{c} a & 1 & 1 \\ 0 & 1-a^2 & 1-a \\ 0 & 1-a & 1-a^2\end{array}\right).$$

We want to eliminate downwards again. Before doing so, note that if $a=1$, our second row becomes $0$, meaning that for $a=1$, our matrix is not invertible (since it is row equivalent to a matrix with a row of zeroes). With that said, let's replace row 3 by row 3 times $1+a$ minus row 2. Then our matrix becomes

$$\left(\begin{array}{c} a & 1 & 1 \\ 0 & 1-a^2 & 1-a \\ 0 & 0 & 1+a-a^2-a^3-1+a\end{array}\right).$$

The lower right element can be rewritten as $2a - a^2 -a^3.$ For our matrix to not be invertible, we need this element to be $0$ (so we have a row of zeroes). But the zeroes of this are $a = 0,1,-2$. We've already ruled out $a=0$ as a solution (well you did that) and accounted for $a=1$ as a value that makes the matrix noninvertible so the only other solution is $a=-2$.

So for $a=1,-2$ the matrix is not invertible.

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  • $\begingroup$ 1. basically nothing. 2. i'm not totally sure what your notation re: R2-->R1 - aR2 means. What elementary row operation are you performing? Thanks again for the help! $\endgroup$ – bclayman Feb 27 '14 at 14:59
  • $\begingroup$ (please see above) $\endgroup$ – bclayman Feb 27 '14 at 17:49
  • $\begingroup$ I was adding adding $-a$ times row 2 to row 1 (in place of row 2). That's what that notation means. I'm going to edit my post to work out the solution in more detail. $\endgroup$ – Cameron Williams Feb 27 '14 at 19:11
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For values a=1 and -2, matrix is singular. Hence non-invertible.

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    $\begingroup$ It is highly advised not to give answers like this. It robs the student of a learning experience. $\endgroup$ – Cameron Williams Feb 27 '14 at 2:47
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A bit of reducing: $$\left(\begin{array}{c} a & 1 & 1 \\ 1 & a & 1 \\ 1 & 1 & a\end{array}\right),$$ subtract the third row from the other two: $$\left(\begin{array}{c} a-1 & 0 & 1-a \\ 0 & a-1 & 1-a \\ 1 & 1 & a\end{array}\right),$$ then add the first two columns to the third one: $$\left(\begin{array}{c} a-1 & 0 & 0 \\ 0 & a-1 & 0 \\ 1 & 1 & a+2\end{array}\right).$$

Now you can easily obtain the determinant.

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In this answer it is shown that $$ \det(\lambda I_n-AB)=\lambda^{n-m}\det(\lambda I_m-BA)\tag{1} $$ We can write your matrix as $$ \begin{pmatrix}a&1&1\\1&a&1\\1&1&a\end{pmatrix} =(a-1)I_3+\begin{pmatrix}1\\1\\1\end{pmatrix}\begin{pmatrix}1&1&1\end{pmatrix}\tag{2} $$ Applying $(1)$ to $(2)$, with $\lambda=a-1$, yields $$ \begin{align} \det\begin{pmatrix}a&1&1\\1&a&1\\1&1&a\end{pmatrix} &=(a-1)^{3-1}\det\left((a-1)I_1+\begin{pmatrix}1&1&1\end{pmatrix}\begin{pmatrix}1\\1\\1\end{pmatrix}\right)\\ &=(a-1)^2\det((a+2)I_1)\\[9pt] &=(a-1)^2(a+2)\tag{3} \end{align} $$

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