value of matrix entry, given that determinant is zero

[oldstudent]

Given that the determinant of the following matrix is zero, find the value of w:

. . .\(\displaystyle \L \left\[\begin{array}{ccc}(-5\,-\,w)&6&1 \\-1&(2\,-\,w)&1\\ -8&6&(4\,-\,w)\end{array}\right\]\)

I did cofactor expansion and got:

. . .-6 - ((-16 + 8w) = ((-1(-30 - 6w) + 48) + [(4 - w)(-10 - 3w + w^2) + 6)] = 0

. . .-6 + 16 -8w + 30 + 6w - 48 + [(4 - w)(w^2 - 3w - 4)]

. . .-2w - 8 + [(4 - w)(w - 4)(w + 1)] = 0

But I can't get past this point.

Thoughts?
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Edited by stapel -- Reason for edit: matrix formatting

 

[galactus]

I'll cofactor along the first column.

\(\displaystyle \L\\(-5-w)\begin{vmatrix}2-w&1\\6&4-w\end{vmatrix}-(-1)\begin{vmatrix}6&1\\6&4-w\end{vmatrix}+(-8)\begin{vmatrix}6&1\\2-w&1\end{vmatrix}=0\)

\(\displaystyle \L\\(-5-w)(w^{2}-6w+2)-(-1)(-6(w-3))+(-8)(w+4)=0\)

\(\displaystyle \L\\-w^{3}+w^{2}+14w-24=0\)

\(\displaystyle \L\\(w-3)(w-2)(w+4)=0\)

 

[stapel]

I'm sorry, but I don't follow your expansion...?

The usual method for taking a 3-by-3 determinant produces the following:

. . . . .(-5 - w)(2 - w)(4 - w) + (6)(1)(-8) + (1)(-1)(6)
. . . . . . . - (-8)(2 - w)(1) - (6)(1)(-5 - w) - (4 - w)(-1)(6)

. . . . .= (-5 - w)(8 - 6w + w²) - 48 - 6 + 8(2 - w) - 6(-5 - w) + 6(4 - w)

. . . . .= (-40 + 22w + w² - w³) - 54 + 16 - 8w + 30 + 6w + 24 - 6w

. . . . .= -24 + 14w + w² - w³

Setting this equal to zero, we get:

. . . . .w³ - w² - 14w + 24 = 0

And this factors, having one negative root and two positive ones.

Eliz.

 

[oldstudent]

I'll cofactor along the first column.

\(\displaystyle \L\\(-5-w)\begin{vmatrix}2-w&1\\6&4-w\end{vmatrix}-(-1)\begin{vmatrix}6&1\\6&4-w\end{vmatrix}+(-8)\begin{vmatrix}6&1\\2-w&1\end{vmatrix}=0\)

\(\displaystyle \L\\(-5-w)(w^{2}-6w+2)-(-1)(-6(w-3))+(-8)(w+4)=0\)

\(\displaystyle \L\\-w^{3}+w^{2}+14w-24=0\)

\(\displaystyle \L\\(w-3)(w-2)(w+4)=0\)

Thanks for the help! I had the concept down. but just was making stupid errors...