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$A=\begin{vmatrix} s&s&s &\cdots & s&s\\ s&1&s &\cdots & s&s\\ s&s&2 &\cdots & s&s\\\vdots & \vdots&\vdots&\ddots&\vdots&\vdots&\\ s&s&s &\cdots & n-1&s\\ s&s&s &\cdots & s&n\\ \end{vmatrix}$

a) calculate $\det A$ when $s =n+1$ Prove that in that case is invertible b) It may happened that $s=4n$ and $\det A=26\times 3^5$

in a) $\det A = -1$ right? how to prove the invertile?

b) plz some hint

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b) Is $s=4n$, the determinant probably depends on $n$. – 1015 Feb 1 '13 at 17:08
up vote 0 down vote accepted

Adding a multiple of one row to another row in a square matrix does not change the determinant (link). Hence, by subtracting the first row from all the others, we find $$A=\begin{pmatrix} s&s&s &\cdots & s&s\\ s&1&s &\cdots & s&s\\ s&s&2 &\cdots & s&s\\ \vdots & \vdots&\vdots&\ddots&\vdots&\vdots&\\ s&s&s &\cdots & n-1&s\\ s&s&s &\cdots & s&n\\ \end{pmatrix}$$ and $$B=\begin{pmatrix} s&s&s &\cdots & s&s\\ 0 & 1-s & 0 & \cdots & 0 & 0 \\ 0 & 0 & 2-s & \cdots & 0 & 0 \\ \vdots & \vdots&\vdots&\ddots&\vdots&\vdots&\\ 0 & 0 & 0 & \cdots & n-1-s & 0 \\ 0 & 0 & 0 & \cdots & 0 & n-s \\ \end{pmatrix}$$ have the same determinant. The determinant of $B$ can be "read off" using the Leibniz Formula; the only non-zero contribution to the sum comes from the identity permutation. Hence $$\det(A)=\det(B)=s\prod_{i=1}^n (i-s).$$ In the case when $s=n+1$, we have \begin{align*} \det(A) &= (n+1) \times -n \times -(n-1) \times \cdots \times -1 \\ &= (-1)^{n-1} (n+1)!. \end{align*} Since $\det(A) \neq 0$ the matrix $A$ is invertible.

In the case when $s=4n$, we have \begin{align*} \det(A) &= 4n \times (1-4n) \times (2-4n) \times \cdots \times (n-4n). \end{align*} For no value of $n$ does this equal $26 \times 3^5$.

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Hint: One way of finding the determinant is to do row operations and track how this changes the determinant. Start by subtracting the first row from all the others. You should be able to come up with a formula for the determinant in terms of $s$ and $n$.

Also for part (a) remember the following theorem:

  • A square matrix is invertible if and only if it's determinant is invertible.
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I recived triangle determinant and on the diagonal have $s*(1-s)(2-s)...(n-s)$ is it correct? – aiki93 Feb 1 '13 at 17:33
Yes, that's correct. – Jim Feb 1 '13 at 18:32
b) it is depending on the odd n $\pm \frac{(4n)!}{(3n-1)!}$ and how to prove that its $\neq 26*3^5$ ?? – aiki93 Feb 1 '13 at 18:40

If $\det A \ne 0$ the A is invertible, you can construct the inverse by Cramer formulae.

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yaa i forgotten about it, thanks you – aiki93 Feb 1 '13 at 17:13

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