Solving systems of linear equations with an unknown 'a' using matrices and elementary row operations

velitshh

velitshh

Answered question

2022-05-28

Solving systems of linear equations with an unknown 'a' using matrices and elementary row operations
Came across this one the other day... while I can narrow 'a' down I can't seem to find an exact/ optimised figure. For example 'a' cannot equal 1/3, 'a' must be less than 0.5...
Anyway, here's the problem. I've got a 3x3 matrix by a 3x1 which equates to a 3x1.
{0.6, 0.2, a},{0.4, 0.3, a},{0.0, 0.5, 1-2a}.{x, y, z} = {1360, 1260, 2000}

Answer & Explanation

asafand2c

asafand2c

Beginner2022-05-29Added 11 answers

Hint: do Gaussian Elimination or Row-Reduced-Echelon-Form (RREF). Note that I changed all the decimals to fractions to get exact results. You will end up with the identity matrix on the left of your augmented matrix, with:
z = 940 3 a 1
y = 40 ( 206 a 53 ) 3 a 1
x = 20 ( 281 a 78 ) 3 a 1
From this you can see that a 1 3 , otherwise a is a "free variable" that you can make anything you like.
You effectively found the inverse of the matrix on the LHS times the column vector on the RHS.

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