F (x) = determinant first line x 1 2 3 second line 3 x 1 2 third line 2 3 x 1 fourth line 1 2 3 x find f (4), please explain it in detail

F (x) = determinant first line x 1 2 3 second line 3 x 1 2 third line 2 3 x 1 fourth line 1 2 3 x find f (4), please explain it in detail

Add column 2.3.4 to column 1, and the first column becomes 6 + X
And then we put forward 6 + X of the first column, the first column becomes 1, and now the determinant becomes 1
1 1 2 3
1 x 1 2
1 3 x 1
1 2 3 x = 4
Calculate the value of determinant 1 2 3
Two lines minus one line = 0 3 - 1 - 1
Three lines minus one line = 0.22 - 2
Four lines minus one line = 0.1
To trigonometric determinants
0 1 - 1 / 3 - 1 / 3 put forward 3
0 04 / 3 - 2 / 3 put forward 2
0 0 4/3 4/3
|A|=1*1*4/3*2=8/3
f(4)=(6+4)*3*2*8/3=160

If the elements in the first row of a fourth-order determinant are - 1,7,3 and - 4 in turn, and the cofactors in the third row are 7, x, 9 and 2 respectively, then x = ----

The inference of the determinant expansion theorem is that the sum of the products of the elements in one row and the algebraic cofactors of the elements in another row is equal to 0
So there are
(-1) * (-1)^(1+1)*7 + 7* (-1)^(1+2) * x + 3* (-1)^(1+3)*9 + (-4)*(-1)^(1+4)*2
= 7 - 7x +27 + 8
= 42 - 7x
= 0
So x = 6

The first row of determinant LG ^ 2x 2 4, the second row 2lgx 1 1, the third row 0 1 3 is less than or equal to 6
lg^2 x - 2lgx-3≤0

lg^2 x - 2lgx-3≤0
(lgx-3)(lgx+1)≤0
-1≤lgx≤3
==>0.1≤x≤1000