[Math] What values of $a$ make this matrix not invertible

linear algebra

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

Best Answer

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 $\mathbb{R}^3$.

Related Solutions

[Math] Different results for row reduction in Matlab

This is already documented in MATLAB.

Roundoff errors may cause this algorithm to compute a different value for the rank than rank, orth and null.

You need to type format long e. After that you can see the difference if you execute N-eye(2) resulting with

N-eye(2)

ans =

   -5.000000000000004e-002    4.500000000000000e-001
    5.000000000000000e-002   -4.500000000000000e-001

Here, the trouble is already visible in the (1,1) element of the matrix. But also

 [1 1]*(N-eye(2))

ans =

   -4.163336342344337e-017    5.551115123125783e-017

gives you the error between seemingly identical elements.

The reason why you get correct results with an additional zero column is (my personal view) due to the term in the tolerance computation for rref given by (max(size(A))*eps *norm(A,inf)). Here, the first term is 3 instead of 2 and that should make the small difference between selecting a rank 1 and rank 2 matrix.

[Math] Invertible Matrix and Linearly Independent Vectors Proof

Let $w_i=Av_i$ for $i=1,\dots,n$. To show linear independence, we must show that $$ c_1w_1+\cdots+c_nw_n=0 \Longrightarrow c_i=0,\quad i=1,\dots n.$$ Since $A$ is invertible, we can left-multiply $c_1w_1+\dots=0$ by $A^{-1}$ to get...

Can you take it from there? Remember that the vectors $v_1,\dots v_n$ are linearly independent.

Best Answer

Related Solutions

[Math] Different results for row reduction in Matlab

[Math] Invertible Matrix and Linearly Independent Vectors Proof

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