# Determinant of 4 X4

1. Nov 7, 2008

### archita

1. The problem statement, all variables and given/known data
The determinant
a+1 a a a
b b+1 b b
c c c+1 c
d d d d+1
equals =?
option : a) 1+ a + b+ c+ d
b)a + b+ c+ d
c)a.b.c.d
d)1 + a.b.c.d

2. Relevant equations

3. The attempt at a solution
[1] tried using finding the co factors.....but the process becomes very length
[2] tried by row-column operations....as below , but nothing workd....
1 0 0 0 a a a a
0 1 0 0 + b b b b
0 0 1 0 c c c c
0 0 0 1 d d d d

2. Nov 7, 2008

### gabbagabbahey

Hi archita, welcome to PF!

This forum supports LaTeX, which you can use to write equations and matrices in a much nicer format....For example, writing:

[$tex]\begin{pmatrix} a+1 & a & a & a \\ b & b+1 & b & b \\ c & c & c+1 & c \\ d & d & d &d+1 \end{pmatrix}[/$tex]

And deleting the \$ signs gives:

$$\begin{pmatrix} a+1 & a & a & a \\ b & b+1 & b & b \\ c & c & c+1 & c \\ d & d & d &d+1 \end{pmatrix}$$

$$\begin{pmatrix} a+1 & a & a & a \\ b & b+1 & b & b \\ c & c & c+1 & c \\ d & d & d &d+1 \end{pmatrix}=\begin{pmatrix} a & a & a & a \\ b & b & b & b \\ c & c & c & c \\ d & d & d &d \end{pmatrix}+I=I+\begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix} \begin{pmatrix}1 & 1 & 1 &1 \end{pmatrix}$$

Where $I$ is the Identity matrix....use Sylvester's detrminant theorem

Last edited: Nov 7, 2008
3. Nov 12, 2008

### JANm

Then the answer would be one but that is none of the options. shouldn't this Sylvester use his theorem one row at a time?

4. Nov 12, 2008

### gabbagabbahey

Sylvester's theorem does not give an answer of one....For two column vectors $u$ and $v$ of equal dimension, Sylvester's theorem states that :

$$\text{det}(I+uv^T)=1+v^T u$$

In this case you would use $$u=\begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix}$$ and $$v^T=\begin{pmatrix}1 & 1 & 1 &1 \end{pmatrix}$$....surely you do not get one using this?

5. Nov 14, 2008

### JANm

Now I can see that the problem is solved! Clever Sylvester.

6. Nov 14, 2008

### JANm

If I take a=1, b=2, c=3, d=4 and calculate the determinant the old way I get 46

7. Nov 14, 2008

### gabbagabbahey

Then you're calculating the determinant incorrectly....using: a=1, b=2, c=3, d=4 and calculating the determinant "the old way" gives:

$$\begin{vmatrix} 2 & 1 & 1 & 1 \\ 2 & 3 & 2 & 2 \\ 3 & 3 & 4 & 3 \\ 4 & 4 & 4 & 5 \end{vmatrix}=\begin{array}{l} 2[3(4*5-3*4)-2(3*5-3*4)+2(3*4-4*4)]-1[2(4*5-3*4)-2(3*5-3*4)+2(3*4-4*4)] \\ +1[2(3*5-3*4)-3(3*5-3*4)+2(3*4-3*4)]-1[2(3*4-4*4)-3(3*4-4*4)+2(3*4-3*4)] \end{array}$$

$$=2[24-6-8]-[16-6-8]+[6-9+0]-[-8+12+0]=20-2-3-4=11 \neq 46$$

8. Nov 14, 2008

### JANm

You are right my 46 is not just. Excel gave for the four to four matrix indeed 11 too.
So my solution e is not just and one of the four a..d is in this case.

I took (a+1)(b+1)(c+1)(d+1) +3abcd -(a+1)bcd -(b+1)acd -(c+1)abd -(d+1)abc,
a method wich works for three by three matrixes but apparently not for 4 by 4.

9. Nov 21, 2008

### archita

thanks a ton

10. Nov 21, 2008

### HallsofIvy

Staff Emeritus
Since this is a 4 by 4 matrix (even number of rows and columns) its determinant is the same as the determinant of the transpose,
$$\left|\begin{array}{cccc}a+ 1 & b & c & d \\ a & b+ 1 & c & d \\ a & b & c+ 1 & d \\ a & b & c & d+ 1 \end{array}\right\|$$

Swapping two rows multiplies the determinant by -1 so doing that twice gives the same determinant,
$$\left|\begin{array}{cccc}a & b & c & d+1 \\ a & b & c+1 & d \\ a & b+1 & c & d \\ a+1 & b & c & d\end{array}\right|$$

Subtract the first row from each of the other rows,
$$\left|\begin{array}{cccc} a & b & c & d+1 \\ 0 & 0 & 1 & -1 \\ 0 & 1 & 0 & -1 \\ 1 & 0 & 0 & -1\end{array}\right|$$

and, finally, expand that on the first column.

11. Nov 21, 2008

### JANm

So that will be Det(1:4,1:4) = a * det(2:4,2:4) - det(2:4,1:3) ?
I'll check that!...