Math Problem Solver

The system of equations has one solution.\[a = \frac{2195200}{3}\] \[b = \frac{219520}{3}\]

Solve by substitution:

Solve \(\frac{a}{10} = b\) for \(a\). \[a = 10 b\]

Substitute \(10 b\) for \(a\) in each of the remaining equations and simplify. $$\begin{aligned}a &= b + 658560 \\ \left(10 b\right) &= b + 658560 \\ b &= \frac{219520}{3} \end{aligned}$$$$\begin{aligned}658560 &= a - b \\ 658560 &= \left(10 b\right) - b \\ b &= \frac{219520}{3} \end{aligned}$$

Use substitution of the numerical value of \(b\) to get the values of \(a\). $$\begin{aligned}a &= 10 b \\ a &= 10 \cdot \frac{219520}{3} \\ a &= \frac{2195200}{3}\end{aligned}$$

Solve by Gauss-Jordan Elimination:

Begin by writing the augmented matrix of the system of equations. $$\begin{bmatrix}\frac{1}{10} & -1 &\bigm |& 0\\1 & -1 &\bigm |& 658560\\-1 & 1 &\bigm |& -658560\end{bmatrix}$$

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.

$$\begin{bmatrix}1 & -1 &\bigm |& 658560\\\frac{1}{10} & -1 &\bigm |& 0\\-1 & 1 &\bigm |& -658560\end{bmatrix}$$

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

$$\begin{bmatrix}1 & -1 &\bigm |& 658560\\\frac{1}{10} & -1 &\bigm |& 0\\-1 & 1 &\bigm |& -658560\end{bmatrix}$$

Multiply row \(1\) by scalar \(- \frac{1}{10}\) and add it to row \(2\).

$$\begin{bmatrix}1 & -1 &\bigm |& 658560\\0 & - \frac{9}{10} &\bigm |& -65856\\-1 & 1 &\bigm |& -658560\end{bmatrix}$$

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

$$\begin{bmatrix}1 & -1 &\bigm |& 658560\\0 & - \frac{9}{10} &\bigm |& -65856\\0 & 0 &\bigm |& 0\end{bmatrix}$$

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.

$$\begin{bmatrix}1 & -1 &\bigm |& 658560\\0 & - \frac{9}{10} &\bigm |& -65856\\0 & 0 &\bigm |& 0\end{bmatrix}$$

Multiply row \(2\) by scalar \(- \frac{10}{9}\) to make the leading term \(1\).

$$\begin{bmatrix}1 & -1 &\bigm |& 658560\\0 & 1 &\bigm |& \frac{219520}{3}\\0 & 0 &\bigm |& 0\end{bmatrix}$$

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

$$\begin{bmatrix}1 & 0 &\bigm |& \frac{2195200}{3}\\0 & 1 &\bigm |& \frac{219520}{3}\\0 & 0 &\bigm |& 0\end{bmatrix}$$

Once the matrix is in reduced-row echelon form, convert the matrix back into linear equations to find the solution. $$\begin{aligned}1 \cdot a+ 0 \cdot b = \frac{2195200}{3} \\ a = \frac{2195200}{3}\end{aligned}$$$$\begin{aligned}0 \cdot a+ 1 \cdot b = \frac{219520}{3} \\ b = \frac{219520}{3}\end{aligned}$$$$\begin{aligned}0 \cdot a+ 0 \cdot b = 0 \\ None = 0\end{aligned}$$