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so I got this task from my professor and wanted to ask for help I have this following matrices

(a)

$$A = \begin{pmatrix} -3 & -11 & -11 & 45 \\ 1 & 11 & 10 & -83 \\ 1 & -6 & -5 & 81 \\ 0 & -3 & -3 & 42 \end{pmatrix}$$

I did this one with the Laplace expansion stuff $4 \times 4$ and got $42$, though I don't know if it's right yet.

Now here comes the real problem

(b)

$$ B = \begin{pmatrix} 1+a_1 & a_2 & \dots & a_n \\ a_1 & 1+a_2 & \dots & a_n \\ \dots & \dots & \dots & \dots \\ a_1 & a_2 & \dots & 1+a_n \end{pmatrix}$$

with $a_1, \dots , a_n$ are elements of $\mathbb{R}$.

So how do I do this one ? also with the Laplace expansion ? What kind of value should come out ? a something ? If there were more numbers I could solve it more easily and I don't know how big it is.. like $4\times 4$ or something.. but it's $n\times n$ so how am I doing this one

And at last

(c) $$C = (c_{ij})$$ with $c_{ij} = 0$ if $i = j$ $1$ if $i != j$ for $1<i,j<n$

I have the same problem with this one (similiar to b)

It would be nice if someone could give me hints/a solution/etc Thanks for reading

Trajan
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MenMan
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  • There's probably a clever solution to $b$. But I would subtract row 2 from row 1 before attempting cofactor expansion to see a quick path of reduction – David P Jan 14 '15 at 22:40
  • You mean substracting 1+a1, a1, a1.. a1 - a2, 1+a2, a2.. a2 ? – MenMan Jan 14 '15 at 23:05
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    About $(b)$, look up Matrix determinant lemma. In particular, this is the special case $$\det(I + uv^T) = 1 + v^T u$$ with $u^T = (1,1,\ldots)$ and $v^T = (a_1,a_2,\ldots,a_n)$. – achille hui Jan 14 '15 at 23:24
  • I don't understand how this can help me after reading the wikipedia article – MenMan Jan 15 '15 at 14:43
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    $$ \begin{align} \det\begin{bmatrix} 1+a_1 & a_2 & \dots & a_n \ a_1 & 1+a_2 & \dots & a_n \ \vdots & \vdots & \ddots & \vdots \ a_1 & a_2 & \dots & 1+a_n \end{bmatrix} = & \det\left(I_n + \begin{bmatrix}1 \ 1 \ \vdots \ 1\end{bmatrix} \otimes \begin{bmatrix}a_1 & a_2 & \ldots & a_n\end{bmatrix} \right)\ = & 1 + \begin{bmatrix}a_1 & a_2 & \ldots & a_n\end{bmatrix}\cdot \begin{bmatrix}1 \ 1 \ \vdots \ 1\end{bmatrix}\ = & 1 + ( a_1 \cdot 1 + a_2 \cdot 1 + \cdots + a_n \cdot 1 )\ = & 1 + a_1 + a_2 + \cdots + a_n\ \end{align} $$ – achille hui Jan 15 '15 at 18:36

1 Answers1

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What I meant for $b$ was:

$$\left|\begin{matrix}1+a & b & c \\ a & 1+b & c \\a & b & 1+c\end{matrix}\right|=\left|\begin{matrix}1 & -1 & 0 \\ a & 1+b & c \\a & b & 1+c \end{matrix}\right|$$

$$=\left|\begin{matrix}1+b & c \\ b & 1+c\end{matrix}\right|+\left|\begin{matrix}a & c \\ a & 1+c\end{matrix}\right|$$

$$=\left|\begin{matrix}1 & -1 \\ b & 1+c\end{matrix}\right|+a\left|\begin{matrix}1 & c \\ 1 & 1+c\end{matrix}\right|$$

$$=([1+c]+[b])+a([1+c]-[c])=a+b+c+1$$

If you know induction, it will be a very quick proof (if you can see the general solution from the above). Achille Hui's comment gives a much faster route to the solution.

Now for question $c$, it is almost a special case of $(b)$. Notice if we choose $a_i=-1$ for all $i$ then the matrix becomes

$$\left(\begin{matrix}0 & -1 & \cdots & -1\\ -1 & 0 & \cdots & -1\\ \vdots & \vdots & \ddots & -1\\ -1 & -1 & \cdots & 0\end{matrix}\right)$$

From the result in $(b)$ the determinant of this matrix is $1-n$. The matrix in $(c)$ is the negative of this one. So factoring out a $(-1)$ from each row will finish it off.

David P
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  • Thanks for alot for your help but I think I can't choose ai = -1 it can only be 0 or 1.. according to my understanding.. besides that i, j lie between 1 and n anyway.. so it's not possible for it to be -1 in my opinion unless I have a thinking error.. – MenMan Jan 15 '15 at 00:08
  • While when I looked at the B of Achilles I kinda got the idea but didn't understand it fully yet.. so I will try to think of an own idea I can understand better.. with your ideas as support – MenMan Jan 15 '15 at 00:09
  • Take a minute and digest what I wrote. I did not specifically refer to the given matrix. But the matrix in the question is the negative of the one I gave – David P Jan 15 '15 at 00:10
  • And in your solution I can see that you used the Laplace expansion I guess but I don't understand the values 1 - 1 and 0 yet..how you got them I mean.. or I'm just too tired atm to do math since I've been ill for a while – MenMan Jan 15 '15 at 00:11
  • Adding a multiple of one row to another does not affect the determinant. – David P Jan 15 '15 at 00:14
  • Alright I kinda did the (c) before the b – MenMan Jan 15 '15 at 20:27
  • All main diagonals are 0 there.. the rest is 1.. I think the determinant is 0 because it's a skew symmetric matrix.. are my thoughts right ? – MenMan Jan 15 '15 at 20:28
  • No, the determinant is $(-1)^n(1-n)$ as I tried to explain – David P Jan 15 '15 at 22:03
  • Ok I proved the (c) of the other task.. now I tried to comprehend what you did here.. but aren't you missing by the cofactor expansion the case: a 1+b, a +b ? and how do you get the a in front of the other one ? I thought of a * 1 = a1 and a 1 first for the first row but what about the c then.. can you explain me that ? – MenMan Jan 17 '15 at 00:04
  • I think your past teachers have done you a disservice on how to find determinants. This is a topic for an entire lecture at a university. You can factor constants out of rows or columns by multiplying the determinant by that constant. See page 4 of this noteset: http://www.math.pitt.edu/~annav/0290H/row_reduction.pdf – David P Jan 17 '15 at 00:45