3
$\begingroup$

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.?

$\endgroup$
  • 1
    $\begingroup$ Calculus of finite differences! $\endgroup$ – Lord Shark the Unknown Sep 5 '17 at 6:07
  • $\begingroup$ @LordSharktheUnknown I'm not getting your point. Can you explain more? $\endgroup$ – Topo Sep 5 '17 at 6:28
8
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ From where negative sign comes in first step? $\endgroup$ – Topo Sep 5 '17 at 6:51
  • 1
    $\begingroup$ I do not know. It is a mistake. $\endgroup$ – Iuli Sep 5 '17 at 6:58
  • $\begingroup$ Okk...Now it is fine... $\endgroup$ – Topo Sep 5 '17 at 7:06
1
$\begingroup$

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 $(1, 1, 1, 1)$?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.