Question

Solve each system of equations using matrices (row operations). If the system has no solution, say that it is inconsistent. $$ \left\\{\begin{array}{l} 2 x-y=-1 \\ x+\frac{1}{2} y=\frac{3}{2} \end{array}\right. $$

Short answer

The solution is \( (x, y) = \left( \frac{1}{2}, 2 \right) \).

Step-by-step solution

Write the Augmented Matrix

First, write the system of equations as an augmented matrix. The system is: \[\begin{array}{l} 2x - y = -1 \ x + \frac{1}{2} y = \frac{3}{2} \end{array}\] The augmented matrix is: \[ \begin{bmatrix} 2 & -1 & | & -1 \ 1 & \frac{1}{2} & | & \frac{3}{2} \end{bmatrix} \]

Row Operation to Make Leading 1 in First Row

Divide the first row by 2 to make the leading coefficient of the first row 1: \[ \frac{1}{2} \begin{bmatrix} 2 & -1 & | & -1 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & -\frac{1}{2} & | & -\frac{1}{2} \end{bmatrix} \] Resulting matrix: \[ \begin{bmatrix} 1 & -\frac{1}{2} & | & -\frac{1}{2} \ 1 & \frac{1}{2} & | & \frac{3}{2} \end{bmatrix} \]

Eliminate First Column Entry in Second Row

Subtract the first row from the second row to eliminate the first column entry in the second row: \[ \begin{bmatrix} 1 & -\frac{1}{2} & | & -\frac{1}{2} \end{bmatrix} - \begin{bmatrix} 1 & -\frac{1}{2} & | & -\frac{1}{2} \end{bmatrix} \rightarrow \begin{bmatrix} 0 & 1 & | & 2 \end{bmatrix} \] Resulting matrix: \[ \begin{bmatrix} 1 & -\frac{1}{2} & | & -\frac{1}{2} \ 0 & 1 & | & 2 \end{bmatrix} \]

Back Substitution

Interpret the resulting matrix back into system of equations: \[ \begin{array}{l} x - \frac{1}{2} y = -\frac{1}{2} \ y = 2 \end{array} \] Solve the second equation for y to get: \[ y = 2 \] Substitute \(y = 2\) into the first equation to find x: \[ x - \frac{1}{2}(2) = -\frac{1}{2} \rightarrow x - 1 = -\frac{1}{2} \rightarrow x = \frac{1}{2} \]

Key concepts

Augmented Matrix

When solving systems of linear equations, an augmented matrix is a very useful tool. It combines the coefficients of the variables and the constants from the system into one matrix. For example, consider the system: \[\begin{array}{l} 2x - y = -1 \ x + \frac{1}{2} y = \frac{3}{2} \end{array}\] This system can be written as an augmented matrix: \[\begin{bmatrix} 2 & -1 & | & -1 \ 1 & \frac{1}{2} & | & \frac{3}{2} \end{bmatrix}\] The vertical bar separates the coefficients from the constants, making it easier to manipulate the equations for solving.

Row Operations

Row operations are essential techniques used to simplify an augmented matrix. These operations include:

• Row swapping – exchanging two rows
• Row multiplication – multiplying all entries in a row by a non-zero scalar
• Row addition – adding a multiple of one row to another row

The goal is to transform the augmented matrix into a simpler form, usually row-echelon form or reduced row-echelon form. In our example, we start by making the leading coefficient of the first row 1: \[ \frac{1}{2} \begin{bmatrix} 2 & -1 & | & -1 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & -\frac{1}{2} & | & -\frac{1}{2} \end{bmatrix} \] Then, we use row subtraction to eliminate the first column entry in the second row.

Back Substitution

After transforming the augmented matrix into an easier form, we interpret it back to a system of equations. In our case, the matrix: \[ \begin{bmatrix} 1 & -\frac{1}{2} & | & -\frac{1}{2} \ 0 & 1 & | & 2 \end{bmatrix} \] translates to: \[ \begin{array}{l} x - \frac{1}{2} y = -\frac{1}{2} \ y = 2 \end{array} \] We first solve for \(y\): \ y = 2 \ Substituting \(y = 2\) into the first equation allows us to find \(x\): \[ x - \frac{1}{2}(2) = -\frac{1}{2} \rightarrow x - 1 = -\frac{1}{2} \rightarrow x = \frac{1}{2} \]

Linear Algebra

Linear Algebra is a branch of mathematics focusing on vector spaces and linear mappings between these spaces. It is the foundation for solving systems of linear equations using matrices. Understanding Linear Algebra concepts like vector spaces, matrices, transformations, and row operations is crucial for solving complex problems efficiently. The process of transforming an augmented matrix and applying back substitution are fundamental skills in Linear Algebra. Applying these skills allows us to solve systems of equations quickly and with less error compared to traditional methods.