Math Problem Solver

The system of linear equations has exactly one solution.

The substitutions are \( C = 7 \), \( D = 17 \), \( R = 3 \).

\[C + R + D = 27\]

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

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. $$\begin{aligned}1 \cdot C+ 0 \cdot D+ 0 \cdot R = 7 \\ C = 7\end{aligned}$$$$\begin{aligned}0 \cdot C+ 1 \cdot D+ 0 \cdot R = 17 \\ D = 17\end{aligned}$$$$\begin{aligned}0 \cdot C+ 0 \cdot D+ 1 \cdot R = 3 \\ R = 3\end{aligned}$$

By substituting these values into the expression, we find that:$$\begin{aligned} C + R + D &= 3 + 7 + 17\\ &=27\end{aligned}$$