Determinant

Mar 2010
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0
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
 
May 2009
1,176
412
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?
 

dwsmith

MHF Hall of Honor
Mar 2010
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Florida
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.
 
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Mar 2010
45
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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
 
Nov 2009
485
184
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
Mar 2010
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Florida
\(\displaystyle det\left(\begin{bmatrix}
1-x & 2 & 3 & 4\\
1 & 2-x & 3 & 4\\
1 & 2 & 3-x & 4\\
1 & 2 & 3 & 4-x
\end{bmatrix}\right)=(1-x)(-1)^{1+j}det(M_{11})\)\(\displaystyle =(1-x)(-1)^{1+1}\begin{vmatrix}
2-x & 3 & 4\\
2 & 3-x & 4\\
2 & 3 & 4-x
\end{vmatrix}=(1-x)(-1)^2\left[(2-x)(-1)^{1+2}\begin{vmatrix}
3-x & 4\\
3 & 4-x
\end{vmatrix}\right]\)

This is just the first term expanded. You still need to do 3 more terms.
 
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Mar 2010
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\(\displaystyle (2)(-1)^{1+2}\begin{vmatrix}
1 & 3 & 4\\
1 & 3-x & 4\\
1 & 3 & 4-x
\end{vmatrix}\)

would that be the next one?
 

simplependulum

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Jan 2009
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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

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