Question

The following problem illustrates a danger that occurs because of round-off error when nearly equal numbers are subtracted, and the difference then multiplied by a large number. Evaluate the quantity $$ 1000 \cdot\left|\begin{array}{cc}{6.010} & {18.04} \\ {2.004} & {6.000}\end{array}\right| $$ as follows. (a) First round each entry in the determinant to two digits. (b) First round each entry in the determinant to three digits. (c) Retain all four digits. Compare this value with the results in parts (a) and (b).

Short answer

Answer: The round-off error can cause a significant difference in the calculated determinant value, as seen when comparing the results from rounding to two digits versus retaining more digits. This highlights the potential danger of round-off errors when nearly equal numbers are subtracted and the difference is then multiplied by a large number.

Step-by-step solution

Round the determinant entries as per the given instructions.

For each case, follow the instructions and round the entries of the determinant accordingly. For example, in case (a), round each entry to two digits.

Calculate the determinant value.

Use the formula for a 2x2 determinant: If the determinant has entries $$ \left|\begin{array}{cc}{a} & {b} \\\ {c} & {d}\end{array}\right| $$ its value is given by (a * d) - (b * c).

Multiply by 1000.

Multiply the calculated determinant value by 1000 to find the final answer for each case. Now let's perform these steps for each case.

Part (a): Round each entry to two digits.

We have $$ \left|\begin{array}{cc}{6.01} & {18.04} \\\ {2.00} & {6.00}\end{array}\right| $$ Calculating the determinant value, we get: $$ (6.01 * 6.00) - (18.04 * 2.00) = 36.06 - 36.08 = -0.02 $$ Multiplying by 1000, we obtain the final answer for part (a): $$ -0.02 * 1000 = -20 $$

Part (b): Round each entry to three digits.

We have $$ \left|\begin{array}{cc}{6.010} & {18.040} \\\ {2.004} & {6.000}\end{array}\right| $$ Calculating the determinant value, we get: $$ (6.010 * 6.000) - (18.040 * 2.004) = 36.06 - 36.14408 = -0.08408 $$ Multiplying by 1000, we obtain the final answer for part (b): $$ -0.08408 * 1000 = -84.08 $$

Part (c): Retain all four digits.

We have $$ \left|\begin{array}{cc}{6.0100} & {18.0400} \\\ {2.0040} & {6.0000}\end{array}\right| $$ Calculating the determinant value, we get: $$ (6.0100 * 6.0000) - (18.0400 * 2.0040) = 36.0600 - 36.14416 = -0.08416 $$ Multiplying by 1000, we obtain the final answer for part (c): $$ -0.08416 * 1000 = -84.16 $$

Comparing Results:

Comparing the values obtained in parts (a), (b), and (c): - Part (a): -20 - Part (b): -84.08 - Part (c): -84.16 We see a significant difference in values between parts (a) and (b)/(c) due to the round-off error caused by rounding to two digits in part (a). This illustrates the danger of round-off error when nearly equal numbers are subtracted, and the difference is then multiplied by a large number.

Key concepts

Determinant Calculation

Calculating a determinant is a fundamental operation in linear algebra, particularly important when dealing with matrices. For a 2x2 matrix, the determinant helps determine the scale factor for the area transformation represented by the matrix. To compute the determinant for a 2x2 matrix \[\begin{bmatrix} a & b \ c & d \end{bmatrix},\]use the formula: \[(a \cdot d) - (b \cdot c).\] This calculation gives us a single numerical value. It's crucial to follow the precise mathematical operations as this value is often used in more complex calculations, such as solving systems of linear equations by Cramer's Rule or determining if a matrix is invertible.

Numerical Methods

Numerical methods are techniques used to approximate mathematical procedures, which is crucial when dealing with complex calculations that cannot be solved analytically. They provide tools for problems where exact calculation is either impossible or impractical.

One common numerical method is the use of rounding during computation, which simplifies complex data but at the cost of precision. This simplification can lead, as seen in the original exercise, to a significant distortion of results in operations like determinant calculations when small values are subtracted and then magnified (by multiplying with a large number).

• When numbers are rounded too much, it can introduce round-off errors, adversely affecting the accuracy of the final result.
• Various numerical methods are developed to minimize errors and provide better accuracy in computations by using finely-tuned algorithms.

Understanding these principles is essential for students to grasp how slight changes in computation methods can lead to varied results.

Precision and Accuracy

Precision and accuracy are two fundamental aspects of numerical computation. While precision refers to the number of significant digits maintained in calculations, accuracy refers to how close a computed value is to the actual true value.

In the provided exercise, rounding the entries of a matrix significantly affects the final determinant value. By comparing parts (a), (b), and (c), we observe:

• Part (a) calculated with lesser precision (two digits) yields the most significant deviation from parts (b) and (c).
• Part (c) with maximum precision (all four digits) gives a determinant closest to the most accurate result.

This difference illustrates the interplay between precision and accuracy, showing how lower precision diminishes the accuracy of results.

Thus, maintaining a higher level of precision is crucial, especially in situations where results are sensitive to changes in the input, as even small errors can escalate in processes involving subtraction of nearly equal numbers followed by multiplication.