Question

Solve each system of equations using matrices (row operations). If the system has no solution, say that it is inconsistent. $$ \left\\{\begin{array}{r} x+2 y=4 \\ 2 x+4 y=8 \end{array}\right. $$

Short answer

The system has infinitely many solutions.

Step-by-step solution

- Write the augmented matrix

Convert the system of equations into an augmented matrix form. The system o \[ x + 2y = 4 \] \[ 2x + 4y = 8 \]This converts to the augmented matrix: \[ \begin{bmatrix} 1 & 2 & | & 4 \ 2 & 4 & | & 8 \end{bmatrix} \]

- Row operations to achieve an upper triangular form

Use row operations to create zeros below the leading 1 in the first column. Subtract 2 times the first row from the second row: \( R_2 \rightarrow R_2 - 2R_1 \).\[ \begin{bmatrix} 1 & 2 & | & 4 \ 0 & 0 & | & 0 \end{bmatrix} \]

- Interpret the reduced row echelon form

The second row consists of all zeros, which means it does not provide any new information. The augmented matrix translates back to the system of equations: \[ x + 2y = 4 \].

- Determine the solution or inconsistency

The single remaining equation, \( x + 2y = 4 \), represents a line.Any point \((x, y)\) on this line is a solution to the system, implying infinitely many solutions. The system is consistent.

Key concepts

Matrices

Matrices are rectangular grids of numbers arranged in rows and columns. They help represent and solve systems of equations in a more organized way. For example, the system\( x + 2y = 4 \) and \( 2x + 4y = 8 \) can be written as a matrix:

• First row: Coefficients of the first equation, \(1, 2\), and the constant term \(4\).
• Second row: Coefficients of the second equation, \(2, 4\), and the constant term \(8\).

This converts to the augmented matrix: \[ \begin{bmatrix} 1 & 2 & | & 4 \ 2 & 4 & | & 8 \ \ \ \end{bmatrix} \] The vertical bar \(|\) separates the coefficients from the constants. By using matrices, we can employ various techniques to solve systems of equations more efficiently.

Row Operations

Row operations are used to manipulate the rows of a matrix to simplify and eventually solve a system of equations. There are three main types of row operations:

• Swapping Rows: Changing the order of two rows.
• Multiplying a Row by a Constant: Scaling all elements of a row by a non-zero constant.
• Adding/Subtracting Rows: Adding or subtracting the elements of one row from another.

For the given system, we used the operation: \[ R_2 \rightarrow R_2 - 2R_1 \]Here, we subtracted 2 times the first row from the second row: \[ \begin{bmatrix} 1 & 2 & | & 4 \ 0 & 0 & | & 0 \ \end{bmatrix} \]This helped us achieve zero below the leading 1 in the first row. Row operations make it easier to simplify matrices and eventually find the solutions to the system of equations.

Reduced Row Echelon Form

The Reduced Row Echelon Form (RREF) of a matrix is a simplified version where each leading entry is 1, and all other elements in the column with the leading 1 are zeros. For our system, we transformed the matrix to look like this:\[ \begin{bmatrix} 1 & 2 & | & 4 \ 0 & 0 & | & 0 \ \end{bmatrix} \]This matrix form is almost in RREF. The second row does not provide new information. The first row translates back to the equation \( x + 2y = 4 \). This means any point \((x, y)\) on this line is a solution: implying an infinite number of solutions. Hence, our system is consistent.In general, RREF is helpful because:

• It clearly shows the solutions to the system.
• It lets us identify inconsistent systems, which have no solutions.
• It makes solving larger systems systematic and efficient.