Problem 14 - Calculus for Biology and Medicine

Question

Reduce the system of linear equations to upper triangular form and solve. $$ \begin{aligned} 2 x+3 y &=5 \\ -\quad y &=-2+\frac{2}{3} x \end{aligned} $$

Step-by-Step Solution

Answer

Solutions: \(x = 3\) and \(y = -\frac{1}{3}\).

Step 1: Express the System of Equations

Start with rewriting the given system of equations. The equations are: \[ \begin{aligned}2x + 3y &= 5 \y &= -2 + \frac{2}{3}x \end{aligned} \]The second equation can be rewritten in standard form as: \[-\frac{2}{3}x + y = -2\].

Step 2: Arrange the System in Matrix Form

Express the system as an augmented matrix:\[\begin{bmatrix}2 & 3 & | & 5 \-\frac{2}{3} & 1 & | & -2\end{bmatrix}\]This matrix represents the coefficients of the variables with the constants on the right side.

Step 3: Eliminate \(x\) from the Second Equation

To achieve an upper triangular form, eliminate \(x\) from the second equation. Multiply the first row by \(\frac{1}{3}\) and add it to the second row to eliminate the \(x\) term in the second row. The calculation will look like this:\[\begin{aligned}\text{New Row 2} &= \Big(-\frac{2}{3} \cdot \frac{1}{3}\Big) \cdot (2) + \Big(1 \cdot \frac{1}{3}\Big) \end{aligned}\] This results in the new system:\[\begin{bmatrix}2 & 3 & | & 5 \0 & 1 & | & -2 + \frac{5}{3} \end{bmatrix}\] Simplifying the constant term:\[-2 + \frac{5}{3} = -\frac{6}{3} + \frac{5}{3} = -\frac{1}{3}\].

Step 4: Write the Triangular Form Equations

The matrix has now been transformed into an upper triangular form:\[\begin{aligned}2x + 3y &= 5 \y &= -\frac{1}{3} \end{aligned}\]From the second equation, we have the value of \(y\).

Step 5: Back-substitution to Find \(x\)

Substitute \(y = -\frac{1}{3}\) into the first equation to solve for \(x\):\[2x + 3\Big(-\frac{1}{3}\Big) = 5\]Simplifying, it becomes:\[2x - 1 = 5\]Adding 1 to both sides, we get:\[2x = 6\]Dividing both sides by 2 results in:\[x = 3\].

Key Concepts

Augmented Matrix

An augmented matrix is a simple way to represent a system of linear equations using a tabular form. This form encapsulates all the coefficients of the variables and the constants from the system of equations. The beauty of the augmented matrix is that it helps systematically transform and solve the equations using matrix operations.

Consider our system:\[ \begin{aligned} 2x + 3y &= 5 \ -\frac{2}{3}x + y &= -2 \end{aligned} \] To express this as an augmented matrix, we write:\[ \begin{bmatrix} 2 & 3 & | & 5 \ -\frac{2}{3} & 1 & | & -2 \end{bmatrix} \]

• The vertical bar divides the coefficients of the variables on the left from the constants on the right.
• This structure allows us to perform row operations efficiently, guiding us towards the solution.

Upper Triangular Form

The upper triangular form of a matrix simplifies solving systems of equations. In this structure, all elements below the main diagonal are zeros. This form allows us to use simpler methods to find the variable solutions when it's placed in a matrix equation.

To achieve the upper triangular form:

• Our goal is to eliminate elements below the diagonal.
• In our example, this involved manipulating the rows to zero out the \( x \) term from the second row.

For our matrix:\[ \begin{bmatrix} 2 & 3 & | & 5 \ 0 & 1 & | & -\frac{1}{3} \end{bmatrix} \]We performed specific operations to clear the elements under the leading coefficient in the first row (eliminating \( x \) from the second equation). This makes solving the matrix by substitution straightforward, starting from the bottom row and working upwards.

Back-substitution

Back-substitution is the final step in solving a system of linear equations once the matrix is in upper triangular form. This process involves solving the equations starting from the bottom row upwards, as the upper triangle format allows for immediate solutions from the bottom row upwards.

For our obtained system:\[ \begin{aligned} y &= -\frac{1}{3} \ 2x + 3y &= 5 \end{aligned} \]

• We directly solve the bottom row equation since it contains only one variable \( y \), giving us \( y = -\frac{1}{3} \).
• Using the known value of \( y \), substitute it back into the first equation to solve for \( x \).
• This step-by-step substitution ensures that we find all solutions systematically and accurately: substituting \( y \), we find \( 2x = 6 \), which simplifies to \( x = 3 \).