Determinant Q54: Cofactor Expansion Solution

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Homework Statement



Textbook page with exercises 51-54, Solve for x, showing four 3x3 determinants set equal to zero: 51 has x, 12, 13 / 0, x-1, 23 / 0, 0, x-2; 52 has x, 1, 1 / 1, 1, x / x, 1, x; 53 has 1, 0, x / x^2, 1, 0 / x, 0, 1; 54 has a, b, x-a / x, x+b, x / 0, 1, 1.


Q54.

Homework Equations



Cofactor expansion (along 1st column)

The Attempt at a Solution



Handwritten derivation on lined paper: a 3x3 determinant with rows (a, b, x-a), (x, x+b, x), (0, 1, 1) set equal to zero, expanded by cofactors to a(x+b-x) - x(b-(x-a)), simplifying to x^2 - x(b+a) + ab, with 'Find x?' written beside it. Above is a column vector equation [x, y, z] = [4, 2, -1].
 
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  • Handwritten derivation on lined paper: a 3x3 determinant with rows (a, b, x-a), (x, x+b, x), (0, 1, 1) set equal to zero, expanded by cofactors to a(x+b-x) - x(b-(x-a)), simplifying to x^2 - x(b+a) + ab, with 'Find x?' written beside it. Above is a column vector equation [x, y, z] = [4, 2, -1].
    20180515_125718_1920x1080.jpg
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  • Textbook page showing exercises 51 to 54, 'Solve for x', with four 3 by 3 determinants set equal to zero; exercise 54 has rows (a, b, x - a), (x, x + b, x), (0, 1, 1).
    20180515_123029_1920x1080.jpg
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  • Textbook page with exercises 51-54, Solve for x, showing four 3x3 determinants set equal to zero: 51 has x, 12, 13 / 0, x-1, 23 / 0, 0, x-2; 52 has x, 1, 1 / 1, 1, x / x, 1, x; 53 has 1, 0, x / x^2, 1, 0 / x, 0, 1; 54 has a, b, x-a / x, x+b, x / 0, 1, 1.
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Mark44 said:
What happened to your equation. You're starting with the determinant equalling 0. You lost one side of your equation.
solving the determinant equal to 0
 
azzarooni88 said:
solving the determinant equal to 0
If you're solving an equation, you have to actually have an equation.
In your first line, you start off with "= ..." followed by a bunch of expressions. You seem to have lost one side of your equation.
 
Mark44 said:
If you're solving an equation, you have to actually have an equation.
In your first line, you start off with "= ..." followed by a bunch of expressions. You seem to have lost one side of your equation.
I think my working is logical. I start off with 0=... and just don't continue writing 0 on the LHS as I solve for the RHS. Ultimately I get 0=x2 - x(b+a) +ab
 
azzarooni88 said:
I think my working is logical. I start off with 0=... and just don't continue writing 0 on the LHS as I solve for the RHS. Ultimately I get 0=x2 - x(b+a) +ab

Okay, can you solve that equation for ##x##?