Math Problem Solver

The system of linear equations has exactly one solution.

$$r = 3$$

Begin by writing the augmented matrix of the system of equations. $$\begin{bmatrix}0 & 1 & 1 &\bigm |& 20\\1 & 0 & 1 &\bigm |& 10\\1 & 1 & 0 &\bigm |& 24\end{bmatrix}$$

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.

$$\begin{bmatrix}1 & 0 & 1 &\bigm |& 10\\0 & 1 & 1 &\bigm |& 20\\1 & 1 & 0 &\bigm |& 24\end{bmatrix}$$

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

$$\begin{bmatrix}1 & 0 & 1 &\bigm |& 10\\0 & 1 & 1 &\bigm |& 20\\1 & 1 & 0 &\bigm |& 24\end{bmatrix}$$

Multiply row $1$ by scalar $-1$ and add it to row $3$.

$$\begin{bmatrix}1 & 0 & 1 &\bigm |& 10\\0 & 1 & 1 &\bigm |& 20\\0 & 1 & -1 &\bigm |& 14\end{bmatrix}$$

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.

$$\begin{bmatrix}1 & 0 & 1 &\bigm |& 10\\0 & 1 & 1 &\bigm |& 20\\0 & 1 & -1 &\bigm |& 14\end{bmatrix}$$

The leading term of row $2$ is already $1$ so this row does not need to be multiplied by a scalar.

$$\begin{bmatrix}1 & 0 & 1 &\bigm |& 10\\0 & 1 & 1 &\bigm |& 20\\0 & 1 & -1 &\bigm |& 14\end{bmatrix}$$

Multiply row $2$ by scalar $-1$ and add it to row $3$.

$$\begin{bmatrix}1 & 0 & 1 &\bigm |& 10\\0 & 1 & 1 &\bigm |& 20\\0 & 0 & -2 &\bigm |& -6\end{bmatrix}$$

Multiply row $3$ by scalar $- \frac{1}{2}$ to make the leading term $1$.

$$\begin{bmatrix}1 & 0 & 1 &\bigm |& 10\\0 & 1 & 1 &\bigm |& 20\\0 & 0 & 1 &\bigm |& 3\end{bmatrix}$$

Multiply row $3$ by scalar $-1$ and add it to row $1$.

$$\begin{bmatrix}1 & 0 & 0 &\bigm |& 7\\0 & 1 & 1 &\bigm |& 20\\0 & 0 & 1 &\bigm |& 3\end{bmatrix}$$

Multiply row $3$ by scalar $-1$ and add it to row $2$.

$$\begin{bmatrix}1 & 0 & 0 &\bigm |& 7\\0 & 1 & 0 &\bigm |& 17\\0 & 0 & 1 &\bigm |& 3\end{bmatrix}$$

Once the matrix is in reduced-row echelon form, convert the matrix back into linear equations to find the solution. $$\begin{aligned}1 \cdot c+ 0 \cdot d+ 0 \cdot r = 7 \\ c = 7\end{aligned}$$$$\begin{aligned}0 \cdot c+ 1 \cdot d+ 0 \cdot r = 17 \\ d = 17\end{aligned}$$$$\begin{aligned}0 \cdot c+ 0 \cdot d+ 1 \cdot r = 3 \\ r = 3\end{aligned}$$