Problem 7 - Algebra and Trigonometry

Question

\(1-8\) Find the determinant of the matrix, if it exists. $$ \left[\begin{array}{cc}{\frac{1}{2}} & {\frac{1}{8}} \\ {1} & {\frac{1}{2}}\end{array}\right] $$

Step-by-Step Solution

Answer

The determinant of the matrix is \( \frac{1}{8} \).

Step 1: Check Matrix Size

First, we note that the given matrix is a 2x2 square matrix. The matrix is: \( A = \begin{bmatrix} \frac{1}{2} & \frac{1}{8} \ 1 & \frac{1}{2} \end{bmatrix} \). For a 2x2 matrix, the determinant can be calculated, so let's proceed with finding the determinant.

Step 2: Apply Determinant Formula for 2x2 Matrix

For a matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the determinant is given by the formula \( \text{det}(A) = ad - bc \).

Step 3: Substitute Values into the Determinant Formula

In our matrix \( A = \begin{bmatrix} \frac{1}{2} & \frac{1}{8} \ 1 & \frac{1}{2} \end{bmatrix} \), substitute \( a = \frac{1}{2}, b = \frac{1}{8}, c = 1, d = \frac{1}{2} \) into the formula: \[ \text{det}(A) = \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) - \left(\frac{1}{8}\right)(1) \].

Step 4: Perform Calculations

Calculate each part of the formula: \[ \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) = \frac{1}{4} \] and \[ \left(\frac{1}{8}\right)(1) = \frac{1}{8} \]. Now, subtract the second result from the first: \[ \frac{1}{4} - \frac{1}{8} = \frac{2}{8} - \frac{1}{8} = \frac{1}{8} \].

Step 5: Confirm the Determinant Exists

The determinant of matrix \( A \) is \( \frac{1}{8} \). Since it is not zero, the matrix is invertible and the determinant exists.

Key Concepts

Matrix size

Before we dive into calculating anything for a matrix, it's important to determine the matrix's size. The size of a matrix is noted in the format of rows by columns, for instance, a 2x2 matrix. This simply means the matrix has 2 rows and 2 columns.

Recognizing a 2x2 matrix is straightforward. For example, the matrix in the exercise is:

• First row: \( \begin{bmatrix} \frac{1}{2} & \frac{1}{8} \end{bmatrix} \)
• Second row: \( \begin{bmatrix} 1 & \frac{1}{2} \end{bmatrix} \)

A matrix must be square to have a determinant, which is why noting the size is essential. In this case, being 2x2 means we can proceed to find its determinant.

2x2 matrix

The 2x2 matrix is among the simplest types of matrices you'll encounter. This type of matrix makes it easier to perform various calculations, such as finding a determinant.

Here's the layout of a general 2x2 matrix:

• The first element \( a \) is at the top left.
• The second element \( b \) is at the top right.
• The third element \( c \) is at the bottom left.
• The fourth element \( d \) is at the bottom right.

Thus forming a matrix: \[\begin{bmatrix}a & b\c & d\end{bmatrix}\] Recognizing this layout is crucial to apply formulas or transformations that relate to such matrices.

Determinant formula

The determinant is a special number that can be calculated from a square matrix. For a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the determinant is calculated using the formula:
\[\text{det}(A) = ad - bc\] This formula is simple, yet it tells you whether the matrix is invertible or not. If the determinant equals zero, the matrix does not have an inverse.

Applying this to our example:

• Substitute \( a = \frac{1}{2} \), \( b = \frac{1}{8} \), \( c = 1 \), and \( d = \frac{1}{2} \) into the formula.
• Perform the multiplication: \( \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \).
• Then calculate: \( \frac{1}{8} \times 1 = \frac{1}{8} \).
• Subtract the second result from the first: \( \frac{1}{4} - \frac{1}{8} = \frac{1}{8} \).

Thus, the determinant is \( \frac{1}{8} \), confirming that this matrix is useful for further operations like finding its inverse.