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QuestionSolve the system of equations using the Gauss method:
1. x+2y−2t+t=1x + 2y - 2t + t = 1
2. 2x+5y−6z+xt=12x + 5y - 6z + xt = 1
3. x+4y−3t+7t=5x + 4y - 3t + 7t = 5
4. 3x+9y−12t+4t=23x + 9y - 12t + 4t = 2

Studdy Solution

STEP 1

Assumptions1. We have a system of linear equations. . We are asked to solve it using the Gauss method, also known as Gaussian elimination.
3. The system of equations is consistent (has at least one solution).

STEP 2

First, let's rewrite the system of equations in the standard form, where each equation is written as a linear combination of the variables equals a constant.
\begin{align*} x +2y - t &=1 \\ 2x +5y -6z + t &=1 \\ x +4y -4t &=5 \\ x +9y -8t &=2\end{align*}

STEP 3

Now, let's write the system of equations as an augmented matrix. The coefficients of the variables form the columns of the matrix, and the constants on the right side of the equations form the last column.
[12−1125−611−539−82]\begin{bmatrix} 1 &2 & -1 &1 \\ 2 &5 & -6 &1 \\ 1 & & - &5 \\ 3 &9 & -8 &2\end{bmatrix}

STEP 4

We will start by making the first element of the first row a pivot (1), and make all elements below it zero. To do this, we can subtract the first row from the third row, and subtract three times the first row from the fourth row.
[12−112−6102−3403−−1]\begin{bmatrix} 1 &2 & -1 &1 \\ 2 & & -6 &1 \\ 0 &2 & -3 &4 \\ 0 &3 & - & -1\end{bmatrix}

STEP 5

Next, we subtract twice the first row from the second row to make the first element of the second row zero.
[12−1101−4−102−3403−5−1]\begin{bmatrix} 1 &2 & -1 &1 \\ 0 &1 & -4 & -1 \\ 0 &2 & -3 &4 \\ 0 &3 & -5 & -1\end{bmatrix}

STEP 6

Now, we subtract twice the second row from the third row, and three times the second row from the fourth row to make the second elements of the third and fourth rows zero.
[12−1101−4−10056002]\begin{bmatrix} 1 &2 & -1 &1 \\ 0 &1 & -4 & -1 \\ 0 &0 &5 &6 \\ 0 &0 & &2\end{bmatrix}

STEP 7

We subtract1.4 times the third row from the fourth row to make the third element of the fourth row zero.
[12−1101−4−10056000−6.4]\begin{bmatrix} 1 &2 & -1 &1 \\ 0 &1 & -4 & -1 \\ 0 &0 &5 &6 \\ 0 &0 &0 & -6.4\end{bmatrix}

STEP 8

Now, we can solve for the variables by back substitution. From the fourth row, we have 0=−6.40 = -6.4, which is a contradiction. This means the system of equations has no solution.

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