# Show that determinant of a symmetric matrix of order $4$ is $0$

Without expanding prove that the determinant of the following matrix is $0$. $$\begin{bmatrix}1^2 & 2^2 & 3^2 & 4^2\\2^2 & 3^2 & 4^2 & 5^2\\3^2 & 4^2 & 5^2 & 6^2\\4^2 & 5^2 & 6^2 & 7^2\end{bmatrix}$$

It is a symmetric matrix. I'm trying to show by operation that any two row or any two column are identical, but I'm unable to do that.

Any hint.?

• Calculus of finite differences! – Lord Shark the Unknown Sep 5 '17 at 6:07
• @LordSharktheUnknown I'm not getting your point. Can you explain more? – Topo Sep 5 '17 at 6:28

Observe that the matrix has the form : $$\begin{bmatrix}P(1) & Q(1) & R(1) & S(1) \\ P(2) & Q(2) & R(2) & S(2) \\ P(3) & Q(3) & R(3) & S(3) \\ P(4) & Q(4) & R(4) & S(4)\end{bmatrix}$$ where $P(X)=X^2$, $Q(X)=(X+1)^2$, $R(X)=(X+2)^2$, and $S(X)=(X+3)^2$.

However, the set of polynomials with degree less than or equal to 2 is a 3-dimensional vector space. Thus, the familly $(P,Q,R,S)$ must be linearly dependent and this implies that one of the column of the matrix can be expressed with the three others. Hence, the matrix cannot be invertible.

For each $i$, $a_{i,j}$ is a quadratic polynomial in $j$, i.e. of the form $\alpha j^2+\beta j+\gamma$.

$$\begin{vmatrix}1^2 & 2^2 & 3^2 & 4^2\\2^2 & 3^2 & 4^2 & 5^2\\3^2 & 4^2 & 5^2 & 6^2\\4^2 & 5^2 & 6^2 & 7^2\end{vmatrix}=\begin{vmatrix}1^2 & 2^2 & 3^2 & 4^2-1^2\\2^2 & 3^2 & 4^2 & 5^2-2^2\\3^2 & 4^2 & 5^2 & 6^2-3^2\\4^2 & 5^2 & 6^2 & 7^2-4^2\end{vmatrix}=\begin{vmatrix}1^2 & 2^2 & 3^2 & 5 \cdot 3\\2^2 & 3^2 & 4^2 & 7 \cdot3\\3^2 & 4^2 & 5^2 & 9 \cdot 3\\4^2 & 5^2 & 6^2 & 11 \cdot3\end{vmatrix}=3\begin{vmatrix}1^2 & 2^2 & 3^2 & 5 \\2^2 & 3^2 & 4^2 & 7 \\3^2 & 4^2 & 5^2 & 9 \\4^2 & 5^2 & 6^2 & 11 \end{vmatrix}=3\begin{vmatrix}1^2 & 2^2 & 3^2-2^2 & 5 \\2^2 & 3^2 & 4^2-3^2 & 7 \\3^2 & 4^2 & 5^2-4^2 & 9 \\4^2 & 5^2 & 6^2-5^2 & 11 \end{vmatrix}=3\begin{vmatrix}1^2 & 2^2 & 1 \cdot 5 & 5 \\2^2 & 3^2 & 1 \cdot7 & 7 \\3^2 & 4^2 & 1 \cdot 9 & 9 \\4^2 & 5^2 & 1 \cdot 11 & 11 \end{vmatrix}=3\begin{vmatrix}1^2 & 2^2 & 5 & 5 \\2^2 & 3^2 & 7 & 7 \\3^2 & 4^2 & 9 & 9 \\4^2 & 5^2 & 11 & 11 \end{vmatrix}=0.$$

• From where negative sign comes in first step? – Topo Sep 5 '17 at 6:51
• I do not know. It is a mistake. – Iuli Sep 5 '17 at 6:58
• Okk...Now it is fine... – Topo Sep 5 '17 at 7:06

You can try applying your matrix to some specific vectors. For example, if $v = (-1, 1, 0, 0)$, then $Mv = (3, 5, 7, 9)$. If $w = (0, 0, -1, 1)$, then $Mw = (7, 9, 11, 13)$. This means that $M(w - v) = Mw - Mv = (4, 4, 4, 4)$.

Can you find another vector $z$, linearly independent to $w - v$, such that $Mz$ is proportional to $(1, 1, 1, 1)$?