Question

solve 5x+5y+2z=12 and2x+4y+5z=2 and 39x+43y+45z=74 by guass elimination method​

Answer 1

Step-by-step explanation:

You can write this as a linear algebra equation Az⃗ =c⃗ :

⎛⎝⎜523934432545⎞⎠⎟⎛⎝⎜xyz⎞⎠⎟=⎛⎝⎜122c⎞⎠⎟

Then detA=5(4×45−5×43)−3(2×45−5×39)+2(2×43−4×39)=0

So the matrix does not have an inverse.

Translating A to row echelon form (basically, using row operations to eliminate the leading coefficient from the 2nd and 3rd rows, and repeating for the 3rd row) we end up with:

⎛⎝⎜500320230⎞⎠⎟⎛⎝⎜xyz⎞⎠⎟=⎛⎝⎜12−25c−370⎞⎠⎟

You might get a different result with different row order or scaling factors, but the main point is that the rank of the matrix is only 2, so there are no solutions unless 5c−370=0 or c=74 at which point, there are an infinite number of solutions, given by the equations of the first two rows:

5x+3y+2z=12

2y+3z=−2

We can write both x and y in terms of z . From the second equation, we get:

y=−1−32z

And substituting this into the first equation, we get:

2x−z=6

And checking the solution in the question: If y=2,z=−2⟹x=2 , which we can check is a solution to the equation, but clearly y can have values other than 2 - for example, (x,y,z)=(5,−7,4) also satisfies the equations.