Math Problem Solver

The system of linear equations has exactly one solution.

The substitutions are \( a = \frac{1}{2} \), \( b = \frac{3}{2} \).

\[a + b \cdot 2 = \frac{7}{2}\]

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

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.

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

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

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

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

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

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

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

Once the matrix is in reduced-row echelon form, convert the matrix back into linear equations to find the solution. $$\begin{aligned}1 \cdot a+ 0 \cdot b = \frac{1}{2} \\ a = \frac{1}{2}\end{aligned}$$$$\begin{aligned}0 \cdot a+ 1 \cdot b = \frac{3}{2} \\ b = \frac{3}{2}\end{aligned}$$

By substituting these values into the expression, we find that:$$\begin{aligned} a + b \cdot 2 &= \frac{1}{2} + 2 \cdot \frac{3}{2}\\ &=\frac{7}{2}\end{aligned}$$