Gaussian Elimination, Question Check.

Question

I'm going through my practice problems, and just want to know if I am doing this right:

$$ 2x_1 - 3x_2 = -2$$ $$ 2x_1 + x_2 = 1$$ $$ 3x_1 + 2x_2 = 1$$

And this is my solution: $$ \begin{align} \begin{bmatrix} 2 & 3 & -2\\ 2 & 1 & 1\\ 3 & 2 & 1 \end{bmatrix} \end{align} $$

$$ \begin{align} \begin{bmatrix} 1 & -3/2 & -1\\ 2 & 1 & 1\\ 3 & 2 & 1 \end{bmatrix} \end{align} $$

$$ \begin{align} \begin{bmatrix} 1 & -3/2 & -1\\ 0 & -1 & -1\\ 3 & 2 & 1 \end{bmatrix} \end{align} $$

$$ \begin{align} \begin{bmatrix} 1 & -3/2 & -1\\ 0 & -1 & -1\\ 0 & -1 & -2 \end{bmatrix} \end{align} $$

$$ \begin{align} \begin{bmatrix} 1 & -3/2 & -1\\ 0 & 1 & 1\\ 0 & -1 & -2 \end{bmatrix} \end{align} $$

$$ \begin{align} \begin{bmatrix} 1 & -3/2 & -1\\ 0 & 1 & 1\\ 0 & 0 & -1 \end{bmatrix} \end{align} $$

Can anyone tell me if I did this right, or if I did a mistake where? Thanks.

Answer 1

In your second matrix, assuming you meant to divide by 2 in the first row, you incorrectly negated the second entry. If you meant to divide by $-2$, then the first entry should be negated, and the third entry should be positive.

In your third matrix, when you doubled row 1 and added to row 2, your entry in the middle row, middle column should be $-3 + 1 =-2$. I'm afraid then, that the error will "trickle down" in later reductions using that row.

Likewise, in the fourth matrix, when you tripled row 1 and added to row 3, your middle value in the bottom row should be $-\frac 92 + 2 = -\frac 52$.

$$ \begin{align} \begin{bmatrix} 2 & 3 & -2\\ 2 & 1 & 1\\ 3 & 2 & 1 \end{bmatrix} \end{align} $$

Subtract row 1 from row 2: $$ \begin{align} \begin{bmatrix} 2 & 3 & -2\\ 0 & -2 & 3\\ 3 & 2 & 1 \end{bmatrix} \end{align} $$

1/2 row 1:

$$ \begin{align} \begin{bmatrix} 1 & 3/2 & -1\\ 0 & -2 & 3\\ 3 & 2 & 1 \end{bmatrix} \end{align} $$ Add $-3 R1$ to row 3 $$ \begin{align} \begin{bmatrix} 1 & 3/2 & -1\\ 0 & -2 & 3\\ 0 & -5/2 & 4 \end{bmatrix} \end{align} $$ -1/2 Row 2, 2 times Row 3: $$ \begin{align} \begin{bmatrix} 1 & 3/2 & -1\\ 0 & 1 & -3/2\\ 0 & -5 & 8 \end{bmatrix} \end{align} $$ 5 Row 2 + row 3 = row 3 $$ \begin{align} \begin{bmatrix} 1 & 3/2 & -1\\ 0 & 1 & 1\\ 0 & 0 & 1/2 \end{bmatrix} \end{align} $$ We can now multiply the last row by 2: $$ \begin{align} \begin{bmatrix} 1 & 3/2 & -1\\ 0 & 1 & 1\\ 0 & 0 & 1 \end{bmatrix} \end{align} $$

Answer 2

first consider $2x_1 - 3x_2 = -2$ and $2x_1 + x_2 = 1$ your solution is $x_1=\frac{1}{8} , x_2=\frac{3}{4}$

now consider : $2x_1 + x_2 = 1$ and $3x_1 + 2x_2 = 1$ your solution is $x_1=1,x_2=-1$

so your system does not an answer

and 3 th row of your last matrix \begin{align} $\begin{bmatrix} 1 & -3/2 & -1\\ 0 & 1 & 1\\ 0 & 0 & -1 \end{bmatrix} \end{align}$ say that your matrix has not answer too

($ 0x_1+0x_2=1 $ is impossible)