# Determinant

#### Monster32432421

show that det
( 1+x , 2 , 3 , 4 )
( 1 , 2+x , 3 , 4 )
( 1 , 2 , 3+x , 4 )
( 1 , 2 , 3 , 4+x)
= x^3(x+10)

note that it is one whole matrix.

thansk

#### Swlabr

show that det
( 1+x , 2 , 3 , 4 )
( 1 , 2+x , 3 , 4 )
( 1 , 2 , 3+x , 4 )
( 1 , 2 , 3 , 4+x)
= x^3(x+10)

note that it is one whole matrix.

thansk
Where is it you are stuck on this problem? What have you already tried?

#### Monster32432421

Where is it you are stuck on this problem? What have you already tried?
I just dont know how to start...

#### dwsmith

MHF Hall of Honor
I just dont know how to start...
The determinant of an nxn matrix A, denoted det(A), is a scalar associated with the matrix A that is defined inductively as
$$\displaystyle det(A)= \begin{cases} a_{11}, & \mbox{if }n=1 \\ a_{11}A_{11}+a_{12}A_{12}+\dots+a_{1n}A_{1n}, & \mbox{if }n>1 \end{cases}$$ where $$\displaystyle A_{1j}=(-1)^{1+j}det(M_{1j}),\ j=1,...,n$$ are the cofactors associated with the entries in the first row of A.

Leon, S. (2010). Linear algebra with applications. Upper Saddle River, NJ: Pearson.

Monster32432421

#### Monster32432421

The determinant of an nxn matrix A, denoted det(A), is a scalar associated with the matrix A that is defined inductively as
$$\displaystyle det(A)= \begin{cases} a_{11}, & \mbox{if }n=1 \\ a_{11}A_{11}+a_{12}A_{12}+\dots+a_{1n}A_{1n}, & \mbox{if }n>1 \end{cases}$$ where $$\displaystyle A_{1j}=(-1)^{1+j}det(M_{1j}),\ j=1,...,n$$ are the cofactors associated with the entries in the first row of A.

Leon, S. (2010). Linear algebra with applications. Upper Saddle River, NJ: Pearson.
I dont really get what you are saying.. cause normally i use something along the lines of gaussian elimination.. and then diagonally times the numbers together

#### roninpro

You can use Gaussian elimination on that matrix, if you want.

I personally would use expansion by minors, which is what dwsmith is suggesting.

#### dwsmith

MHF Hall of Honor
$$\displaystyle det\left(\begin{bmatrix} 1-x & 2 & 3 & 4\\ 1 & 2-x & 3 & 4\\ 1 & 2 & 3-x & 4\\ 1 & 2 & 3 & 4-x \end{bmatrix}\right)=(1-x)(-1)^{1+j}det(M_{11})$$$$\displaystyle =(1-x)(-1)^{1+1}\begin{vmatrix} 2-x & 3 & 4\\ 2 & 3-x & 4\\ 2 & 3 & 4-x \end{vmatrix}=(1-x)(-1)^2\left[(2-x)(-1)^{1+2}\begin{vmatrix} 3-x & 4\\ 3 & 4-x \end{vmatrix}\right]$$

This is just the first term expanded. You still need to do 3 more terms.

Last edited:
Monster32432421

#### Monster32432421

$$\displaystyle (2)(-1)^{1+2}\begin{vmatrix} 1 & 3 & 4\\ 1 & 3-x & 4\\ 1 & 3 & 4-x \end{vmatrix}$$

would that be the next one?

#### dwsmith

MHF Hall of Honor
$$\displaystyle (2)(-1)^{1+2}\begin{vmatrix} 1 & 3 & 4\\ 1 & 3-x & 4\\ 1 & 3 & 4-x \end{vmatrix}$$

would that be the next one?
Yes

Last edited:
Monster32432421

#### simplependulum

MHF Hall of Honor
Let $$\displaystyle D(a_1,a_2,...,a_n) =$$

$$\displaystyle det \begin{bmatrix} a_1 + x & a_2&... & a_n\\ a_1 & a_2 +x &... & a_n\\ ... & ... & ...& a_n\\ a_1 & a_2 & ...& a_n+x \end{bmatrix}$$

and consider the system of equations :

$$\displaystyle (a_1 + x )t_1 + a_2t_2 + ... + a_nt_n = 0$$
$$\displaystyle a_1t_1 + (a_2+x)t_2 + ... + a_n t_n = 0$$
$$\displaystyle ................................................$$
$$\displaystyle a_1t_1 + a_2t_2 + ... + (a_{n-1}+x)t_{n-1} + a_nt_n = 0$$
$$\displaystyle a_1t_1 + a_2t_2 + ... + (a_n+x)t_n = 1$$

Let $$\displaystyle Y = a_1t_1 + a_2t_2 + ... + a_nt_n$$ and we have

$$\displaystyle Y + xt_1 = 0 ~~...(1)$$
$$\displaystyle Y + xt_2 = 0 ~~...(2)$$
$$\displaystyle ..............$$
$$\displaystyle Y + xt_{n-1} = 0 ~~...(n-1)$$
$$\displaystyle Y + xt_n = 1 ~~...(n)$$

From $$\displaystyle (1),(2),...,(n-1)$$ we obtain

$$\displaystyle t_1 = t_2 = ... = t_{n-1} = - \frac{Y}{x}$$

Consider $$\displaystyle Y = a_1t_1 + a_2t_2 + ... + a_{n-1} t_{n-1} + a_nt_n$$ $$\displaystyle = (a_1 + a_2 + ... + a_{n-1} ) \left( - \frac{Y}{x} \right) + a_nt_n$$ .

Hence we have these two equations with variants $$\displaystyle t_n$$ and $$\displaystyle Y$$

By eliminating $$\displaystyle Y$$ we obtain

$$\displaystyle t_n = \frac{ x + \sum_{i=1}^{n-1}a_i }{x(x + \sum_{i=1}^n a_i )}$$

Also by cramer's rule

$$\displaystyle t_n = \frac{ D(a_1,a_2,...,a_{n-1})}{ D(a_1,a_2,...,a_n)}$$

Therefore , $$\displaystyle \frac{ x + \sum_{i=1}^{n-1}a_i }{x(x + \sum_{i=1}^n a_i )} = \frac{ D(a_1,a_2,...,a_{n-1})}{ D(a_1,a_2,...,a_n)}$$

We have $$\displaystyle D(a_1) = x+ a_1$$ then $$\displaystyle D(a_1,a_2) = \frac{x(x+a_1+a_2)}{x+ a_1 } (x+a_1 ) = x(x+ a_1 + a_2)$$ so we deduce that

$$\displaystyle D(a_1,a_2,...,a_n) = x^{n-1} ( x + \sum_{i=1}^n a_i )$$