pure math-matrix

解聯立線性方程, (以此題為例, 為一聯立三元一次方程, in english , simultaneous linear equations in three unknowns)

其中一個可行的方法為guassian elimination

1-k......3.........4
4........2-k.......8
1..........1......4-k

add -4*row3 to row2, -(1-k)*row3 to row 1
0.......k+2.....4-(4-k)(1-k) (<---- = -k^2+5k)
0.......-k-2.....4k-8
1..........1......4-k

add row2 to row 1
0........0.....-k^2+5k+4k-8 (<---=-(k-1)(k-8) )
0.......-k-2.....4k-8
1..........1......4-k

所以第一個可以令system有non-trival solution 的sufficient condtion 是
-(k-1)(k-8)=0
which is ture iff k=1 or k =8


also, look at row2 in the final matrix, espcially the coefficient of y
if -k-2=0 ( whcih is ture iff k=-2)
then it forces z=0,
if z=0, it satisfies row3 simultaneously
ifz=0, it makes row 1 become
x+y=0
then we have nontrival solution
(x,y,z)= (t, -t, 0) where t is some real number

hence the condtion that the system has nontrival solution are
k=1, 8, -2

actually, it is kind of question asking the "eigenvalues of the matrix A" where A in this question equals
1...3...4
4...2...8
1...1...4

let X= transpose of (x1...x2....x3) be a 3 by 1 column matrix
and solving for some real number k such that
AX=kX with non-zero X
then it become (A-kI)X=0, where I is the 3 by 3 identity matrix
then we require the matrix (A-kI) is singular (or non-invertible)
then it is the quickest way to calculate the deteminant of (A-kI)
and forces the quantity to be zero to find the possible value of k

朋友﹐你check清題目了嗎