Determinant Q54: Cofactor Expansion Solution

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The discussion centers on solving a determinant using cofactor expansion along the first column, where a participant expresses confusion about losing one side of the equation while attempting to set the determinant equal to zero. There is a repeated emphasis on the necessity of maintaining both sides of the equation throughout the solution process. Another participant defends their approach, arguing that they logically start with 0 on one side and focus on the right-hand side. The conversation highlights the importance of clear equation representation in mathematical problem-solving. Ultimately, the discussion leads to a request for solving the resulting equation for x.
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Homework Statement



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Q54.

Homework Equations



Cofactor expansion (along 1st column)

The Attempt at a Solution



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azzarooni88 said:

Homework Statement



View attachment 225748

Homework Equations



Cofactor expansion (along 1st column)

The Attempt at a Solution



View attachment 225746
What happened to your equation. You're starting with the determinant equalling 0. You lost one side of your equation.
 
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##?
 

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