Problem 25 - College Algebra

Question

For the following exercises, find the multiplicative inverse of each matrix, if it exists. $$\left[\begin{array}{lll}\frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5} \\ \frac{1}{6} & \frac{1}{7} & \frac{1}{8}\end{array}\right]$$

Step-by-Step Solution

Answer

The matrix is not invertible, as its determinant is approximately zero.

Step 1: Determine the Invertibility

A matrix is invertible if its determinant is non-zero. Let's denote the given matrix as \( A \): \[ A = \begin{bmatrix} \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \ \frac{1}{3} & \frac{1}{4} & \frac{1}{5} \ \frac{1}{6} & \frac{1}{7} & \frac{1}{8} \end{bmatrix} \] Calculate the determinant to check if the inverse exists.

Step 2: Calculate the Determinant

To find the determinant of a 3x3 matrix \( A = \begin{bmatrix} a & b & c \ d & e & f \ g & h & i \end{bmatrix} \), use the formula: \[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \] Substitute the values from matrix \( A \) and calculate:\[ \text{det}(A) = \frac{1}{2}\left(\frac{1}{4}\times\frac{1}{8} - \frac{1}{5}\times\frac{1}{7}\right) - \frac{1}{2}\left(\frac{1}{3}\times\frac{1}{8} - \frac{1}{5}\times\frac{1}{6}\right) + \frac{1}{2}\left(\frac{1}{3}\times\frac{1}{7} - \frac{1}{4}\times\frac{1}{6}\right) \] Simplify each term and sum them up.

Step 3: Simplify the Determinant

Simplify each part separately: \( \frac{1}{4} \times \frac{1}{8} = \frac{1}{32} \), \( \frac{1}{5} \times \frac{1}{7} = \frac{1}{35} \), and \( \frac{1}{32} - \frac{1}{35} = -\frac{3}{1120} \)\( \frac{1}{3} \times \frac{1}{8} = \frac{1}{24} \), \( \frac{1}{5} \times \frac{1}{6} = \frac{1}{30} \), and \( \frac{1}{24} - \frac{1}{30} = \frac{1}{120} \)\( \frac{1}{3} \times \frac{1}{7} = \frac{1}{21} \), \( \frac{1}{4} \times \frac{1}{6} = \frac{1}{24} \), and \( \frac{1}{21} - \frac{1}{24} = \frac{1}{168} \)Plug back into the determinant formula:\[ \text{det}(A) = \frac{1}{2}(-\frac{3}{1120}) - \frac{1}{2}\left(\frac{1}{120}\right) + \frac{1}{2}\left(\frac{1}{168}\right) \]Calculate the final value.

Step 4: Determine Det Value and Conclusion

Calculate the determinant value:\[ \text{det}(A) = \frac{1}{2}(-\frac{3}{1120}) + \frac{1}{2}(-\frac{1}{120}) + \frac{1}{2}(\frac{1}{168}) \]Combine and simplify all terms to get:\[ \text{det}(A) = -\frac{3}{2240} - \frac{1}{240} + \frac{1}{336} \approx 0 \]The determinant is approximately zero, which means the matrix is not invertible.

Key Concepts

Determinant Calculation

The determinant of a matrix is a special value that can be calculated from its elements. It's crucial to finding out if a matrix has an inverse or not. For a 3x3 matrix, the determinant can be computed using a specific formula. This involves a bit of multiplication, subtraction, and addition. The formula for a 3x3 matrix \( A \) with elements \( a_{11}, a_{12}, a_{13}, a_{21}, a_{22}, a_{23}, a_{31}, a_{32}, a_{33} \) is:\[\text{det}(A) = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})\]For the given matrix, we substitute the values into this formula to find the determinant value. This will help us determine if the matrix has an inverse.

Invertibility

A matrix is said to be invertible if its determinant is not zero. This property is what allows us to "reverse" the matrix's transformation effects, akin to finding a reciprocal for a number. The inverse of a matrix \( A \) is typically denoted as \( A^{-1} \). In terms of application:

• If \( \text{det}(A) eq 0 \), then \( A^{-1} \) exists.
• If \( \text{det}(A) = 0 \), then \( A \) is not invertible.

In our case, after calculating the determinant, it was found to be approximately zero, leading to the conclusion that the matrix is not invertible. This means there is no \( A^{-1} \) for this particular matrix.

3x3 Matrix

Understanding a 3x3 matrix's structure is essential in matrix algebra. It's a square matrix with three rows and three columns. Each position in the matrix is defined by its row and column number. For example, the element \( a_{12} \) is located in the first row and second column.Working with 3x3 matrices means dealing with these nine values and understanding the interactions between them. These matrices are often used in various fields such as physics and computer graphics for transformations, rotations, and more.

Matrix Algebra

Matrix algebra involves a variety of operations such as addition, subtraction, multiplication, and finding inverses. It provides a powerful way to handle a system of equations or transformations especially useful in higher-dimensional space. Matrix multiplication, for example, is not like regular multiplication. The number of columns in the first matrix must match the number of rows in the second matrix to perform this operation. Finding the inverse of a matrix is another common algebraic procedure, requiring determinant calculation as we've explored. Matrix algebra is thus foundational to most computational fields and is crucial for problem-solving in linear equations.