Determinant's solving without expansion

[Kartik.]

Solve without expanding the determinant having elements which are listed as follows row-wise = {1^2,2^2,3^2,2^2,3^2,4^2,3^2,4^2,5^2} where , the determinant is of the order 3

I have tried some combinations of operations performed row or column wise that could produce 2 zeros either in one of the rows or columns and then expanding the determinant along that row or column; but is there any simpler way out?

 

[UltrafastPED]

1 4 9
4 9 16
9 16 25

Just put it into upper triangular form ... then the product of the elements along the main diagonal will equal the determinant of the original matrix.

 

[epenguin]

Maybe you are expected to use (a² - b²) = (a - b)(a + b) = (a + b) in this case.

 

[HallsofIvy]

Maybe you are expected to use (a² - b²) = (a - b)(a + b) = (a + b) in this case.

I would think not! In general (a- b)(a+ b) is NOT equal to a+ b.

 

[epenguin]

I would think not! In general (a- b)(a+ b) is NOT equal to a+ b.

Not in general but in this case when subtracting certain rows from the next where (a -b) = 1 .

 

[Kartik.]

1 4 9
4 9 16
9 16 25

Just put it into upper triangular form ... then the product of the elements along the main diagonal will equal the determinant of the original matrix.

How do i do that? Can you please tell me the operations?

 

[Mark44]

How do i do that? Can you please tell me the operations?

No.
Per the Physics Forums rules (https://www.physicsforums.com/showthread.php?t=414380) that you agreed to when you signed up for an account, we don't do your work for you. Here is from the Homework Help Guidelines section of the rules.

On helping with questions: Any and all assistance given to homework assignments or textbook style exercises should be given only after the questioner has shown some effort in solving the problem. If no attempt is made then the questioner should be asked to provide one before any assistance is given. Under no circumstances should complete solutions be provided to a questioner, whether or not an attempt has been made.

Show us what you've done, and we'll steer you in the right direction.

 

[Kartik.]

no.
Per the physics forums rules (https://www.physicsforums.com/showthread.php?t=414380) that you agreed to when you signed up for an account, we don't do your work for you. Here is from the homework help guidelines section of the rules.

Show us what you've done, and we'll steer you in the right direction.

1 4 9
4 9 16
9 16 25
2 times the column 1 - column 2 and taking the negative sign out of the determinant from row 1 and row 2
2 -4 -9
1 -9 -16
2 16 25
row 1 - 2 times row 2 and row 3 - 2 times row 2 and the expanding along column 1
0 -16 -23
1 -9 -16
0 34 57

 

[epenguin]

1 4 9
4 9 16
9 16 25

2 -4 -9
1 -9 -16
2 16 25

0 -16 -23
1 -9 -16
0 34 57

I suggest you modify that post to state what operations you performed, which is not obvious.

Do your operations use in any way the systematic special features of the numbers in the problem or are they just what you would do for any generic 3 X 3 determinant?

 

[LCKurtz]

I don't think any of the above suggestions would count as "not expanding" the determinant. I don't see the trick myself and would certainly like for Kartik to present the solution if this is a classroom problem for which he eventually is given the "answer".

You all probably have seen the trick for expanding the Vandermonde determinant$$
\left | \begin{array}{ccc}
1&a&a^2\\
1&b&b^2\\
1&c&c^2
\end{array}\right|$$where you replace the $a$ by $x$ giving$$
P(x)=\left | \begin{array}{ccc}
1&x&x^2\\
1&b&b^2\\
1&c&c^2
\end{array}\right|$$Then you observe $P(b)=P(c)=0$ so $P(x)=K(x-b)(x-c)$ and $K$ is the coefficient of $x^2$. That can be read from the determinant as $(c-b)$. Therefore $P(x) = (c-b)(x-b)(x-c)$ and the original determinant is $P(a)=(c-b)(a-b)(a-c)$.

I would think they are looking for some trick like that, which I think would count as "not expanding" it. But what trick??

 

[HallsofIvy]

Since you made no attempt to do this yourself it is difficult to say what kind of help you can use (whether you know what "row operations" are) or exactly what you mean by "expansion".

One method I learned, way back in secondary school, for a three by three determinant, is to copy the first two rows again:
$$\begin{vmatrix}1 & 4 & 9 \\ 4 & 9 & 16 \\ 9 & 16 & 25 \end{vmatrix} \begin{vmatrix}1 & 4 \\ 4 & 9 \\ 9 & 16 \end{vmatrix}$$

Now, sum the products of the three "upper left to lower right" diagonals: 1(9)(25)+ 4(16)(9)+ 9(4)(16), sum the products of the three "upper right to lower left" diagonals: 9(9)(9)+ 16(16)(1)+ 25(4)(4). Now subtract the second from the first.

 

[LCKurtz]

Heh heh. You're showing your age Halls. My favorite way for a quick answer too...

 

[epenguin]

Yes I was reminded if the Vandermonde determinant, but this is not the Vandermonde determinant, I haven't looked into an argument of the type of Kurtz', but I have suggested an approach which comes out easily in a couple of lines. Whether it is within the 'without expansion' requirement is not clear, but it is 'mathematical', to the extent it enables solving the more general determinant extending that one in the obvious way - for all order >3 they are all 0 I think.

Karting seems to have done an ordinary general expansion using the numbers without using their particular forms given.

 

[pasmith]

I don't think any of the above suggestions would count as "not expanding" the determinant. I don't see the trick myself and would certainly like for Kartik to present the solution if this is a classroom problem for which he eventually is given the "answer".

You all probably have seen the trick for expanding the Vandermonde determinant

I would think they are looking for some trick like that, which I think would count as "not expanding" it. But what trick??

I obtained the answer by considering
$$P(x,y,z) = \left| \begin{array}{ccc} x^2 & (x+1)^2 & (x+2)^2 \\ y^2 & (y+1)^2 & (y+2)^2 \\ z^2 & (z+1)^2 & (z+2)^2 \end{array}\right|$$
We are looking for $P(1,2,3)$.