Find all values of $x$ where the determinant equals 0

Question

$$\det\begin{bmatrix}x & x & x & x & x\\a & x & x & x & x\\a & a & x & x & x\\ a &a&a& x& x\\a&a&a&a&x\end{bmatrix} = 0$$

Having this matrix I am trying to find all the values of $x$ that fulfill the above equation. I tried it with the long way of computing the cofactors but that just takes too much time. Is there another more efficient way?

Answer 1

Clearly $x=0$ is one of those values. Now, if $x\ne 0$, then

\begin{align} \det\begin{pmatrix}x & x & x & x & x\\a & x & x & x & x\\a & a & x & x & x\\ a &a&a& x& x\\a&a&a&a&x\end{pmatrix} &=x\det\begin{pmatrix}1 & 1 & 1 & 1 & 1\\a & x & x & x & x\\a & a & x & x & x\\ a &a&a& x& x\\a&a&a&a&x\end{pmatrix} \\ &=x\det\begin{pmatrix}1 & 1 & 1 & 1 & 1\\0 & x-a & x-a & x-a & x-a\\0 & 0 & x-a & x-a & x-a\\ 0 &0&0& x-a& x-a\\0&0&0&0&x-a\end{pmatrix} \\ &=x(x-a)^4 \end{align}

Answer 2

Perform: $C_1-C_2, C_2-C_3,C_3-C_4,C_4-C_5$ (column operations). These preserve the determinant, so

$$\begin{vmatrix}x & x & x & x & x\\a & x & x & x & x\\a & a & x & x & x\\ a &a&a& x& x\\a&a&a&a&x\end{vmatrix} = \begin{vmatrix}0 & 0 & 0 & 0 & x\\a-x & 0 & 0 & 0 & x\\0 & a-x & 0 & 0 & x\\ 0 &0&a-x& 0& x\\0&0&0&a-x&x\end{vmatrix}$$

Now you can expand along any row (for example first row) and use the idea of upper/lower triangular (or better yet diagonal) matrix. To get $$\det=x\begin{vmatrix}a-x & 0 & 0 & 0 \\0 & a-x & 0 & 0\\ 0 &0&a-x& 0\\0&0&0&a-x\end{vmatrix}=x(a-x)^4.$$

So $\det=0$ for $x=0,a$ only.

Answer 3

Expanding my comment, a reasonable approach might be to reduce the problem to something a little more tractable via elementary row operations. Suppose that $A$ is a matrix.

• Let $A_\text{swap}$ be the matrix obtained by swapping two rows of $A$. Then $\det(A_{\text{swap}}) = -\det(A)$.
• Let $A_{\text{scale}}$ be the matrix obtained by scaling one row of $A$ by some constant $C$. Then $\det(A_{\text{scale}}) = \frac{1}{C} \det(A)$.
• Let $A_{\text{add}}$ be the matrix obtained by adding a row of $A$ to a different row of $A$. Then $\det(A_{\text{add}}) = \det(A)$.

So, if we use elementary row operations to reduce your matrix to something more tractable (and we keep track of those operations), we are in business. So, try the following: \begin{align} A = \begin{pmatrix} x & x & x & x & x \\ a & x & x & x & x \\ a & a & x & x & x \\ a & a & a & x & x \\ a & a & a & a & x \end{pmatrix} &\implies \begin{pmatrix} x & x & x & x & x \\ -a & -x & -x & -x & -x \\ a & a & x & x & x \\ -a & -a & -a & -x & -x \\ a & a & a & a & x \end{pmatrix} \\ &\implies \begin{pmatrix} x-a & 0 & 0 & 0 & 0 \\ 0 & a-x & 0 & 0 & 0 \\ 0 & 0 & x-a & 0 & 0 \\ 0 & 0 & 0 & a-x & 0 \\ a & a & a & a & x \end{pmatrix} =: A'. \end{align} In the first reduction, we multiply the second and third rows by $-1$. This has no net effect on the determinant, since it amounts to multiplying the determinant by $-1$ twice. In the last step, we are just adding row $j+1$ to row $j$ for each $j=1,2,3,4$; this also doesn't change the determinant. In other words, we have $$ \det(A) = \det(A'). $$ But $A'$ is a lower triangular matrix, so the determinant is the product of the terms on the diagonal. In other words $$ \det(A') = (x-a)(a-x)(x-a)(a-x)x = x(x-a)^4. $$ This is zero only when $x \in \{0, a\}$.

Answer 4

Consider the $n \times n$ generalization of this.
Write your matrix as $A = a {\bf 1 1}^T + (x-a) U = (x-a)(a (x-a)^{-1} \bf{1 1}^T U^{-1} + I) U$, where $U$ is the matrix with all $1$'s on and above the diagonal and $0$'s elsewhere, and $\bf 1$ the column vector of all $1$'s. By Sylvester's determinant identity and the fact that $\det(U)=1$, $$\det(A) = (x-a)^n \det(I + a (x-a)^{-1} {\bf 1}^T U^{-1} {\bf 1})$$ Now $U^{-1} \bf 1$ is the column vector with last entry $1$ and all others $0$, so $I + a(x-a)^{-1} {\bf 1}^T U^{-1} {\bf 1} = 1 + a (x-a)^{-1} = x/(x-a)$ (a $1 \times 1$ matrix, whose determinant is its entry). Thus $$\det(A) = x (x-a)^{n-1}$$