Determinants with arithmetic progressions as columns

Question

Prove that determinants of the following form all vanish: $$\det \begin{bmatrix} x-3 & x-4 & x-a \\ x-2 & x-3 & x-b \\ x-1 & x-2 & x-c\end{bmatrix} = 0$$ Here $a$, $b$, $c$ are three consecutive terms in an arithmetic progression.

Answer 1

Considering the property $$\lambda \cdot \begin{vmatrix} a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3 \end{vmatrix}=\begin{vmatrix} a_1 & a_2 & a_3\\ \lambda b_1 & \lambda b_2 & \lambda b_3\\ c_1 & c_2 & c_3 \end{vmatrix} ,$$ we have: $$\begin{vmatrix} x - 3 & x-4 & x-a \\ x-2 & x-3 & x-b\\ x-1 & x-2 & x-c \end{vmatrix}= \frac 12\cdot \begin{vmatrix} x - 3 & x-4 & x-a \\ 2x-4 & 2x-6 & 2x-2b\\ x-1 & x-2 & x-c \end{vmatrix} $$

Also, we take advantage of the property which declares that adding multiples of any rows to a specific one the determinant remains invariant. We add to the second row the first row multiplied by $(-1)$ and the third row multiplied by $(-1)$. Thus, we have: $$\begin{vmatrix} x-3 & x-4 & x-a \\ 0 & 0 &a+c - 2b \\ x-1 & x-2 & x-c \end{vmatrix}. $$ However, $a +c =2b$, because $a,b,c$ are consecutive terms of an arithmetic progression. That means that we have a zero row, which implies that the determinant is zero!

Answer 2

$$\det \begin{bmatrix} x-3 & x-4 & x-a \\ x-2 & x-3 & x-b \\ x-1 & x-2 & x-c\end{bmatrix} = \\ (x-3) \begin{vmatrix}x-3 &x-b \\x-2 & x-c \end{vmatrix} - (x-4)\begin{vmatrix}x-2 &x-b \\x-1 & x-c \end{vmatrix} + (x-a) \begin{vmatrix}x-2 &x-3 \\x-1 & x-2 \end{vmatrix}$$

If a, b and c are consecutive terms of an arithmetic progression,

$a = a,\ b = a + d, \ c = a + 2d $

Solving the last 2 equations gives: $d = c-b$, $b- a = d$

Hence, $b-a = c-b$ $\implies$ $a+c = 2b$

Now try and expand $$\det \begin{bmatrix} x-3 & x-4 & x-a \\ x-2 & x-3 & x-b \\ x-1 & x-2 & x-c\end{bmatrix}$$ and use the fact that $a+c = 2b$ and see what you get

You have to make some effort here I'm afraid.

Answer 3

Adding to a row another row multiplied by any number doesn't change the determinant: \begin{align} \begin{bmatrix} x-3 & x-4 & x-a \\ x-2 & x-3 & x-b \\ x-1 & x-2 & x-c \end{bmatrix} &\to \begin{bmatrix} -1 & -1 & b-a \\ x-2 & x-3 & x-b \\ x-1 & x-2 & x-c \end{bmatrix} &&R_1\gets R_1-R_2 \\[6px]&\to \begin{bmatrix} -1 & -1 & b-a \\ 0 & -1 & x-b+(x-2)(b-a) \\ x-1 & x-2 & x-c \end{bmatrix} &&R_2\gets R_2+(x-2)R_1 \\[6px]&\to \begin{bmatrix} -1 & -1 & b-a \\ 0 & -1 & x-b+(x-2)(b-a) \\ 0 & -1 & x-c+(x-1)(b-a) \end{bmatrix} &&R_3\gets R_3+(x-1)R_1 \\[6px]&\to \begin{bmatrix} -1 & -1 & b-a \\ 0 & -1 & x-b+(x-2)(b-a) \\ 0 & 0 & 2b-a-c \end{bmatrix} &&R_3\gets R_3-R_2 \end{align} When is the determinant $0$?

Answer 4

$(1,-2,1)$ is a left eigenvector associated with a zero eigenvalue, so the determinant is trivially zero since that matrix is not invertible.