1. n by n matrix

can someone please give me an example of an n x n matrix, where the last column is the sum of the previous n-1 columns?
i have to prove whether it is invertible or not

thanks

2. Is there something preventing you from doing this on your own?

3. i'm not sure whether the example i made up was correct

a11 a12 a13 | (a13+ a23+ a33)
a21 a22 a23 | (a12 + a22 +a32)
a31 a32 a33 | (a11 + a21 +a31)

is this correct?

4. That's not an $n\times n$ matrix, though.

5. woops. sorry i didn't see that

a11 a12 (a12:a32)
a21 a22 (a11:a31)
a31 a32 (...)

i get stuck when it comes to a33.

6. Originally Posted by Kbotz
i'm not sure whether the example i made up was correct

a11 a12 a13 | (a13+ a23+ a33)
a21 a22 a23 | (a12 + a22 +a32)
a31 a32 a33 | (a11 + a21 +a31)

is this correct?
i think you should add an extra row

7. Originally Posted by Kbotz
can someone please give me an example of an n x n matrix, where the last column is the sum of the previous n-1 columns?
i have to prove whether it is invertible or not

thanks
To disprove something, all you need is an example. Since there aren't many stipulations on this matrix, I can think of many singular nxn matrices that meet your criteria.

8. Originally Posted by Dgphru
i think you should add an extra row
Adding an extra row makes the matrix 4x3 which isn't nxn

9. Assume this is nxn

$\begin{bmatrix}
a & a & b & \dots & \sum_{x=1}^{n}a_{1x}\\
a & a & c & & \\
a & a & d & & \vdots\\
a & a & e & \ddots & \\
a & a & f & & \sum_{x=1}^{n}a_{nx}
\end{bmatrix}$

This matrix isn't invertible since column 1 and 2 are lin. dep.

10. Originally Posted by dwsmith
Assume this is nxn

$\begin{bmatrix}
a & a & b & \dots & \sum_{x=1}^{n}a_{1x}\\
a & a & c & & \\
a & a & d & & \vdots\\
a & a & e & \ddots & \\
a & a & f & & \sum_{y=1}^{n}\sum_{x=1}^{n}a_{yx}
\end{bmatrix}$

This matrix isn't invertible since column 1 and 2 are lin. dep.

are b,c,d...f the sum of the previous columns?

11. If they have to be, then yes. By last column, do you mean every column needs to be the sum? I thought you mean column n is the sum of all the columns.

It doesn't change anything then because column 2 is the sum of column 1 then; thus, column 1 and 2 are lin. ind.

12. sorry, my initial question was a little vague.

Let B be an n by n matrix. suppose its last column is the sum of the previous n-1 columns. is it possible for B to be invertible? why or why not?

13. i found it quite difficult to provide an example since for a n by n matrix, the right hand bottom corner of the matrix (a(mn)) will not have a value.

14. Originally Posted by Kbotz
sorry, my initial question was a little vague.

Let B be an n by n matrix. suppose its last column is the sum of the previous n-1 columns. is it possible for B to be invertible? why or why not?
What do you mean the last column is the sum of previous?

Does that mean the nth column is the sum or does that mean column two is the sum of column 1, column 3 is the sum of column 1 and 2....? If the second is the case, you can make the matrix columns the Fibonacci numbers.
1, 1, 2, 3, 5, 8, ....
*note $v_1$ means a $nx1$ column vector of the nxn matrix
If you make $v_1$ a column vector of all 1s, then $v_2$= $v_1$, $v_3$= $2v_1$, $v_4$= $5v_1$,....

15. i'm hoping it's the second case, but i'm not too sure since it says n-1.
the question itself is quite vague.
sorry for the trouble

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