Math Problem Solver

The system of equations has one solution.\[t = \frac{1}{3}\] \[x = \frac{1}{6}\] \[y = \frac{1}{6}\] \[z = \frac{1}{3}\]

Solve by substitution:

Solve \(x = \frac{z}{2}\) for \(x\). \[x = \frac{z}{2}\]

Substitute \(\frac{z}{2}\) for \(x\) in each of the remaining equations and simplify. $$\begin{aligned}z &= \frac{t}{2} + \frac{x}{2} + \frac{y}{2} \\ z &= \frac{t}{2} + \frac{\left(\frac{z}{2}\right)}{2} + \frac{y}{2} \\ 2 t &= - 2 y + 3 z \end{aligned}$$$$\begin{aligned}t &= \frac{t}{2} + \frac{x}{2} + \frac{y}{2} \\ t &= \frac{t}{2} + \frac{\left(\frac{z}{2}\right)}{2} + \frac{y}{2} \\ 2 t &= 2 y + z \end{aligned}$$$$\begin{aligned}t + x + y + z &= 1 \\ t + \left(\frac{z}{2}\right) + y + z &= 1 \\ t + y + \frac{3 z}{2} &= 1 \end{aligned}$$

Solve \(2 t = - 2 y + 3 z\) for \(y\). \[y = - t + \frac{3 z}{2}\]

Substitute \(- t + \frac{3 z}{2}\) for \(y\) in each of the remaining equations and simplify. $$\begin{aligned}2 t &= 2 y + z \\ 2 t &= 2 \left(- t + \frac{3 z}{2}\right) + z \\ t &= z \end{aligned}$$$$\begin{aligned}t + y + \frac{3 z}{2} &= 1 \\ t + \left(- t + \frac{3 z}{2}\right) + \frac{3 z}{2} &= 1 \\ z &= \frac{1}{3} \end{aligned}$$

Solve \(t = z\) for \(z\). \[z = t\]

Substitute \(t\) for \(z\) in \(z = \frac{1}{3}\) and simplify. $$\begin{aligned}z &= \frac{1}{3} \\ \left(t\right) &= \frac{1}{3} \\ t &= \frac{1}{3} \end{aligned}$$

Use substitution of the numerical value of \(t\) to get the values of \(x, y\) and \( z\). $$\begin{aligned}z &= t \\ z &= 1 \cdot \frac{1}{3} \\ z &= \frac{1}{3}\end{aligned}$$$$\begin{aligned}x &= \frac{z}{2} \\ x &= 1 \cdot \frac{1}{3} \cdot \frac{1}{2} \\ x &= \frac{1}{6}\end{aligned}$$

Solve by Gauss-Jordan Elimination:

Begin by writing the augmented matrix of the system of equations. $$\begin{bmatrix}0 & 1 & 0 & - \frac{1}{2} &\bigm |& 0\\0 & 0 & 1 & - \frac{1}{2} &\bigm |& 0\\- \frac{1}{2} & - \frac{1}{2} & - \frac{1}{2} & 1 &\bigm |& 0\\\frac{1}{2} & - \frac{1}{2} & - \frac{1}{2} & 0 &\bigm |& 0\\1 & 1 & 1 & 1 &\bigm |& 1\end{bmatrix}$$

Then use a series of elementary row operations to convert the matrix into reduced-row echelon form. The three elementary row operations are:

  1. Swap the positions of any two rows.

  2. Multiply any row by a nonzero scalar.

  3. Multiply a row by a nonzero scalar and add it to any other row.

First, switch the rows in the matrix such that the row with the leftmost non-zero entry with the greatest magnitude is at the top.

$$\begin{bmatrix}1 & 1 & 1 & 1 &\bigm |& 1\\0 & 0 & 1 & - \frac{1}{2} &\bigm |& 0\\- \frac{1}{2} & - \frac{1}{2} & - \frac{1}{2} & 1 &\bigm |& 0\\\frac{1}{2} & - \frac{1}{2} & - \frac{1}{2} & 0 &\bigm |& 0\\0 & 1 & 0 & - \frac{1}{2} &\bigm |& 0\end{bmatrix}$$

The leading term of row \(1\) is already \(1\) so this row does not need to be multiplied by a scalar.

$$\begin{bmatrix}1 & 1 & 1 & 1 &\bigm |& 1\\0 & 0 & 1 & - \frac{1}{2} &\bigm |& 0\\- \frac{1}{2} & - \frac{1}{2} & - \frac{1}{2} & 1 &\bigm |& 0\\\frac{1}{2} & - \frac{1}{2} & - \frac{1}{2} & 0 &\bigm |& 0\\0 & 1 & 0 & - \frac{1}{2} &\bigm |& 0\end{bmatrix}$$

Multiply row \(1\) by scalar \(\frac{1}{2}\) and add it to row \(3\).

$$\begin{bmatrix}1 & 1 & 1 & 1 &\bigm |& 1\\0 & 0 & 1 & - \frac{1}{2} &\bigm |& 0\\0 & 0 & 0 & \frac{3}{2} &\bigm |& \frac{1}{2}\\\frac{1}{2} & - \frac{1}{2} & - \frac{1}{2} & 0 &\bigm |& 0\\0 & 1 & 0 & - \frac{1}{2} &\bigm |& 0\end{bmatrix}$$

Multiply row \(1\) by scalar \(- \frac{1}{2}\) and add it to row \(4\).

$$\begin{bmatrix}1 & 1 & 1 & 1 &\bigm |& 1\\0 & 0 & 1 & - \frac{1}{2} &\bigm |& 0\\0 & 0 & 0 & \frac{3}{2} &\bigm |& \frac{1}{2}\\0 & -1 & -1 & - \frac{1}{2} &\bigm |& - \frac{1}{2}\\0 & 1 & 0 & - \frac{1}{2} &\bigm |& 0\end{bmatrix}$$

Switch the rows in the matrix such that the row with the next leftmost non-zero entry with the next greatest magnitude is the next row from the top.

$$\begin{bmatrix}1 & 1 & 1 & 1 &\bigm |& 1\\0 & -1 & -1 & - \frac{1}{2} &\bigm |& - \frac{1}{2}\\0 & 0 & 0 & \frac{3}{2} &\bigm |& \frac{1}{2}\\0 & 0 & 1 & - \frac{1}{2} &\bigm |& 0\\0 & 1 & 0 & - \frac{1}{2} &\bigm |& 0\end{bmatrix}$$

Multiply row \(2\) by scalar \(-1\) to make the leading term \(1\).

$$\begin{bmatrix}1 & 1 & 1 & 1 &\bigm |& 1\\0 & 1 & 1 & \frac{1}{2} &\bigm |& \frac{1}{2}\\0 & 0 & 0 & \frac{3}{2} &\bigm |& \frac{1}{2}\\0 & 0 & 1 & - \frac{1}{2} &\bigm |& 0\\0 & 1 & 0 & - \frac{1}{2} &\bigm |& 0\end{bmatrix}$$

Multiply row \(2\) by scalar \(-1\) and add it to row \(1\).

$$\begin{bmatrix}1 & 0 & 0 & \frac{1}{2} &\bigm |& \frac{1}{2}\\0 & 1 & 1 & \frac{1}{2} &\bigm |& \frac{1}{2}\\0 & 0 & 0 & \frac{3}{2} &\bigm |& \frac{1}{2}\\0 & 0 & 1 & - \frac{1}{2} &\bigm |& 0\\0 & 1 & 0 & - \frac{1}{2} &\bigm |& 0\end{bmatrix}$$

Multiply row \(2\) by scalar \(-1\) and add it to row \(5\).

$$\begin{bmatrix}1 & 0 & 0 & \frac{1}{2} &\bigm |& \frac{1}{2}\\0 & 1 & 1 & \frac{1}{2} &\bigm |& \frac{1}{2}\\0 & 0 & 0 & \frac{3}{2} &\bigm |& \frac{1}{2}\\0 & 0 & 1 & - \frac{1}{2} &\bigm |& 0\\0 & 0 & -1 & -1 &\bigm |& - \frac{1}{2}\end{bmatrix}$$

Switch the rows in the matrix such that the row with the next leftmost non-zero entry with the next greatest magnitude is the next row from the top.

$$\begin{bmatrix}1 & 0 & 0 & \frac{1}{2} &\bigm |& \frac{1}{2}\\0 & 1 & 1 & \frac{1}{2} &\bigm |& \frac{1}{2}\\0 & 0 & 1 & - \frac{1}{2} &\bigm |& 0\\0 & 0 & 0 & \frac{3}{2} &\bigm |& \frac{1}{2}\\0 & 0 & -1 & -1 &\bigm |& - \frac{1}{2}\end{bmatrix}$$

The leading term of row \(3\) is already \(1\) so this row does not need to be multiplied by a scalar.

$$\begin{bmatrix}1 & 0 & 0 & \frac{1}{2} &\bigm |& \frac{1}{2}\\0 & 1 & 1 & \frac{1}{2} &\bigm |& \frac{1}{2}\\0 & 0 & 1 & - \frac{1}{2} &\bigm |& 0\\0 & 0 & 0 & \frac{3}{2} &\bigm |& \frac{1}{2}\\0 & 0 & -1 & -1 &\bigm |& - \frac{1}{2}\end{bmatrix}$$

Multiply row \(3\) by scalar \(-1\) and add it to row \(2\).

$$\begin{bmatrix}1 & 0 & 0 & \frac{1}{2} &\bigm |& \frac{1}{2}\\0 & 1 & 0 & 1 &\bigm |& \frac{1}{2}\\0 & 0 & 1 & - \frac{1}{2} &\bigm |& 0\\0 & 0 & 0 & \frac{3}{2} &\bigm |& \frac{1}{2}\\0 & 0 & -1 & -1 &\bigm |& - \frac{1}{2}\end{bmatrix}$$

Multiply row \(3\) by scalar \(1\) and add it to row \(5\).

$$\begin{bmatrix}1 & 0 & 0 & \frac{1}{2} &\bigm |& \frac{1}{2}\\0 & 1 & 0 & 1 &\bigm |& \frac{1}{2}\\0 & 0 & 1 & - \frac{1}{2} &\bigm |& 0\\0 & 0 & 0 & \frac{3}{2} &\bigm |& \frac{1}{2}\\0 & 0 & 0 & - \frac{3}{2} &\bigm |& - \frac{1}{2}\end{bmatrix}$$

Switch the rows in the matrix such that the row with the next leftmost non-zero entry with the next greatest magnitude is the next row from the top.

$$\begin{bmatrix}1 & 0 & 0 & \frac{1}{2} &\bigm |& \frac{1}{2}\\0 & 1 & 0 & 1 &\bigm |& \frac{1}{2}\\0 & 0 & 1 & - \frac{1}{2} &\bigm |& 0\\0 & 0 & 0 & \frac{3}{2} &\bigm |& \frac{1}{2}\\0 & 0 & 0 & - \frac{3}{2} &\bigm |& - \frac{1}{2}\end{bmatrix}$$

Multiply row \(4\) by scalar \(\frac{2}{3}\) to make the leading term \(1\).

$$\begin{bmatrix}1 & 0 & 0 & \frac{1}{2} &\bigm |& \frac{1}{2}\\0 & 1 & 0 & 1 &\bigm |& \frac{1}{2}\\0 & 0 & 1 & - \frac{1}{2} &\bigm |& 0\\0 & 0 & 0 & 1 &\bigm |& \frac{1}{3}\\0 & 0 & 0 & - \frac{3}{2} &\bigm |& - \frac{1}{2}\end{bmatrix}$$

Multiply row \(4\) by scalar \(- \frac{1}{2}\) and add it to row \(1\).

$$\begin{bmatrix}1 & 0 & 0 & 0 &\bigm |& \frac{1}{3}\\0 & 1 & 0 & 1 &\bigm |& \frac{1}{2}\\0 & 0 & 1 & - \frac{1}{2} &\bigm |& 0\\0 & 0 & 0 & 1 &\bigm |& \frac{1}{3}\\0 & 0 & 0 & - \frac{3}{2} &\bigm |& - \frac{1}{2}\end{bmatrix}$$

Multiply row \(4\) by scalar \(-1\) and add it to row \(2\).

$$\begin{bmatrix}1 & 0 & 0 & 0 &\bigm |& \frac{1}{3}\\0 & 1 & 0 & 0 &\bigm |& \frac{1}{6}\\0 & 0 & 1 & - \frac{1}{2} &\bigm |& 0\\0 & 0 & 0 & 1 &\bigm |& \frac{1}{3}\\0 & 0 & 0 & - \frac{3}{2} &\bigm |& - \frac{1}{2}\end{bmatrix}$$

Multiply row \(4\) by scalar \(\frac{1}{2}\) and add it to row \(3\).

$$\begin{bmatrix}1 & 0 & 0 & 0 &\bigm |& \frac{1}{3}\\0 & 1 & 0 & 0 &\bigm |& \frac{1}{6}\\0 & 0 & 1 & 0 &\bigm |& \frac{1}{6}\\0 & 0 & 0 & 1 &\bigm |& \frac{1}{3}\\0 & 0 & 0 & - \frac{3}{2} &\bigm |& - \frac{1}{2}\end{bmatrix}$$

Multiply row \(4\) by scalar \(\frac{3}{2}\) and add it to row \(5\).

$$\begin{bmatrix}1 & 0 & 0 & 0 &\bigm |& \frac{1}{3}\\0 & 1 & 0 & 0 &\bigm |& \frac{1}{6}\\0 & 0 & 1 & 0 &\bigm |& \frac{1}{6}\\0 & 0 & 0 & 1 &\bigm |& \frac{1}{3}\\0 & 0 & 0 & 0 &\bigm |& 0\end{bmatrix}$$

Once the matrix is in reduced-row echelon form, convert the matrix back into linear equations to find the solution. $$\begin{aligned}1 \cdot t+ 0 \cdot x+ 0 \cdot y+ 0 \cdot z = \frac{1}{3} \\ t = \frac{1}{3}\end{aligned}$$$$\begin{aligned}0 \cdot t+ 1 \cdot x+ 0 \cdot y+ 0 \cdot z = \frac{1}{6} \\ x = \frac{1}{6}\end{aligned}$$$$\begin{aligned}0 \cdot t+ 0 \cdot x+ 1 \cdot y+ 0 \cdot z = \frac{1}{6} \\ y = \frac{1}{6}\end{aligned}$$$$\begin{aligned}0 \cdot t+ 0 \cdot x+ 0 \cdot y+ 1 \cdot z = \frac{1}{3} \\ z = \frac{1}{3}\end{aligned}$$$$\begin{aligned}0 \cdot t+ 0 \cdot x+ 0 \cdot y+ 0 \cdot z = 0 \\ None = 0\end{aligned}$$