Math Problem Solver

The system of equations has one solution.\[a = 0\] \[b = 0\] \[c = 0\]

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Solve by substitution:

Solve \(a + 2 b + 3 c = 0\) for \(a\). \[a = - 2 b - 3 c\]

Substitute \(- 2 b - 3 c\) for \(a\) in each of the remaining equations and simplify. $$\begin{aligned}2 a + 3 b + c &= 0 \\ 2 \left(- 2 b - 3 c\right) + 3 b + c &= 0 \\ b &= - 5 c \end{aligned}$$$$\begin{aligned}4 a + 5 b + 4 c &= 0 \\ 4 \left(- 2 b - 3 c\right) + 5 b + 4 c &= 0 \\ 3 b &= - 8 c \end{aligned}$$$$\begin{aligned}a + 2 b - 2 c &= 0 \\ \left(- 2 b - 3 c\right) + 2 b - 2 c &= 0 \\ c &= 0 \end{aligned}$$

Solve \(b = - 5 c\) for \(b\). \[b = - 5 c\]

Substitute \(- 5 c\) for \(b\) in \(3 b = - 8 c\) and simplify. $$\begin{aligned}3 b &= - 8 c \\ 3 \left(- 5 c\right) &= - 8 c \\ c &= 0 \end{aligned}$$

Use substitution of the numerical value of \(c\) to get the values of \(a\) and \( b\). $$\begin{aligned}b &= - 5 c \\ b &= - 5 \0dot 0 \\ b &= 0\end{aligned}$$$$\begin{aligned}a &= - 2 b - 3 c \\ a &= - 2 \0dot 0 - 3 \0dot 0 \\ a &= 0\end{aligned}$$

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Solve by Gauss-Jordan Elimination:

Begin by writing the augmented matrix of the system of equations. $$\begin{bmatrix}1 & 2 & 3 &\bigm |& 0\\2 & 3 & 1 &\bigm |& 0\\4 & 5 & 4 &\bigm |& 0\\1 & 2 & -2 &\bigm |& 0\end{bmatrix}$$

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.

$$\begin{bmatrix}4 & 5 & 4 &\bigm |& 0\\2 & 3 & 1 &\bigm |& 0\\1 & 2 & 3 &\bigm |& 0\\1 & 2 & -2 &\bigm |& 0\end{bmatrix}$$

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

$$\begin{bmatrix}1 & \frac{5}{4} & 1 &\bigm |& 0\\2 & 3 & 1 &\bigm |& 0\\1 & 2 & 3 &\bigm |& 0\\1 & 2 & -2 &\bigm |& 0\end{bmatrix}$$

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

$$\begin{bmatrix}1 & \frac{5}{4} & 1 &\bigm |& 0\\0 & \frac{1}{2} & -1 &\bigm |& 0\\1 & 2 & 3 &\bigm |& 0\\1 & 2 & -2 &\bigm |& 0\end{bmatrix}$$

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

$$\begin{bmatrix}1 & \frac{5}{4} & 1 &\bigm |& 0\\0 & \frac{1}{2} & -1 &\bigm |& 0\\0 & \frac{3}{4} & 2 &\bigm |& 0\\1 & 2 & -2 &\bigm |& 0\end{bmatrix}$$

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

$$\begin{bmatrix}1 & \frac{5}{4} & 1 &\bigm |& 0\\0 & \frac{1}{2} & -1 &\bigm |& 0\\0 & \frac{3}{4} & 2 &\bigm |& 0\\0 & \frac{3}{4} & -3 &\bigm |& 0\end{bmatrix}$$

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.

$$\begin{bmatrix}1 & \frac{5}{4} & 1 &\bigm |& 0\\0 & \frac{3}{4} & 2 &\bigm |& 0\\0 & \frac{1}{2} & -1 &\bigm |& 0\\0 & \frac{3}{4} & -3 &\bigm |& 0\end{bmatrix}$$

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

$$\begin{bmatrix}1 & \frac{5}{4} & 1 &\bigm |& 0\\0 & 1 & \frac{8}{3} &\bigm |& 0\\0 & \frac{1}{2} & -1 &\bigm |& 0\\0 & \frac{3}{4} & -3 &\bigm |& 0\end{bmatrix}$$

Multiply row \(2\) by scalar \(- \frac{5}{4}\) and add it to row \(1\).

$$\begin{bmatrix}1 & 0 & - \frac{7}{3} &\bigm |& 0\\0 & 1 & \frac{8}{3} &\bigm |& 0\\0 & \frac{1}{2} & -1 &\bigm |& 0\\0 & \frac{3}{4} & -3 &\bigm |& 0\end{bmatrix}$$

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

$$\begin{bmatrix}1 & 0 & - \frac{7}{3} &\bigm |& 0\\0 & 1 & \frac{8}{3} &\bigm |& 0\\0 & 0 & - \frac{7}{3} &\bigm |& 0\\0 & \frac{3}{4} & -3 &\bigm |& 0\end{bmatrix}$$

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

$$\begin{bmatrix}1 & 0 & - \frac{7}{3} &\bigm |& 0\\0 & 1 & \frac{8}{3} &\bigm |& 0\\0 & 0 & - \frac{7}{3} &\bigm |& 0\\0 & 0 & -5 &\bigm |& 0\end{bmatrix}$$

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.

$$\begin{bmatrix}1 & 0 & - \frac{7}{3} &\bigm |& 0\\0 & 1 & \frac{8}{3} &\bigm |& 0\\0 & 0 & -5 &\bigm |& 0\\0 & 0 & - \frac{7}{3} &\bigm |& 0\end{bmatrix}$$

Multiply row \(3\) by scalar \(- \frac{1}{5}\) to make the leading term \(1\).

$$\begin{bmatrix}1 & 0 & - \frac{7}{3} &\bigm |& 0\\0 & 1 & \frac{8}{3} &\bigm |& 0\\0 & 0 & 1 &\bigm |& 0\\0 & 0 & - \frac{7}{3} &\bigm |& 0\end{bmatrix}$$

Multiply row \(3\) by scalar \(\frac{7}{3}\) and add it to row \(1\).

$$\begin{bmatrix}1 & 0 & 0 &\bigm |& 0\\0 & 1 & \frac{8}{3} &\bigm |& 0\\0 & 0 & 1 &\bigm |& 0\\0 & 0 & - \frac{7}{3} &\bigm |& 0\end{bmatrix}$$

Multiply row \(3\) by scalar \(- \frac{8}{3}\) and add it to row \(2\).

$$\begin{bmatrix}1 & 0 & 0 &\bigm |& 0\\0 & 1 & 0 &\bigm |& 0\\0 & 0 & 1 &\bigm |& 0\\0 & 0 & - \frac{7}{3} &\bigm |& 0\end{bmatrix}$$

Multiply row \(3\) by scalar \(\frac{7}{3}\) and add it to row \(4\).

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

Once the matrix is in reduced-row echelon form, convert the matrix back into linear equations to find the solution. $$\begin{aligned}1 \cdot a+ 0 \cdot b+ 0 \cdot c = 0 \\ a = 0\end{aligned}$$$$\begin{aligned}0 \cdot a+ 1 \cdot b+ 0 \cdot c = 0 \\ b = 0\end{aligned}$$$$\begin{aligned}0 \cdot a+ 0 \cdot b+ 1 \cdot c = 0 \\ c = 0\end{aligned}$$$$\begin{aligned}0 \cdot a+ 0 \cdot b+ 0 \cdot c = 0 \\ None = 0\end{aligned}$$