Problem 4 - Algebra and Trigonometry with Analytic Geometry

Question

Find the inverse of the matrix if it exists. $$ \left[\begin{array}{ll} 3 & 2 \\ 4 & 5 \end{array}\right] $$

Step-by-Step Solution

Answer

The inverse is \( \begin{bmatrix} \frac{5}{7} & -\frac{2}{7} \\ -\frac{4}{7} & \frac{3}{7} \end{bmatrix} \).

Step 1: Determine the Matrix Format

The given matrix is a 2x2 matrix. For a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the formula for the inverse is \( \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \), provided that \( ad-bc eq 0 \). This is called the determinant.

Step 2: Calculate the Determinant

Calculate the determinant for the matrix \( \begin{bmatrix} 3 & 2 \ 4 & 5 \end{bmatrix} \) using the formula \( ad-bc \). In our case, \( a=3, b=2, c=4, \) and \( d=5 \). Thus, the determinant is \( 3(5) - 4(2) = 15 - 8 = 7 \).

Step 3: Check Determinant's Condition

The determinant is 7, which is not zero, hence the inverse exists. Proceed with finding the inverse.

Step 4: Apply the Inverse Formula

Substitute the values into the inverse formula: \( \text{If the determinant is } 7, \text{ then the inverse of } \begin{bmatrix} 3 & 2 \ 4 & 5 \end{bmatrix}\) is \( \frac{1}{7} \begin{bmatrix} 5 & -2 \ -4 & 3 \end{bmatrix} \).

Step 5: Simplify the Inverse

Multiply each element in the matrix \( \begin{bmatrix} 5 & -2 \ -4 & 3 \end{bmatrix} \) by \( \frac{1}{7} \) to obtain the inverse: \( \begin{bmatrix} \frac{5}{7} & -\frac{2}{7} \ -\frac{4}{7} & \frac{3}{7} \end{bmatrix} \).

Key Concepts

Determinant Calculation

When you want to find the inverse of a matrix, calculating the determinant is the first crucial step. For a 2x2 matrix, the determinant tells us if the inverse exists. You use the formula for the determinant of a 2x2 matrix: \[ ad - bc \] where \( a \), \( b \), \( c \), and \( d \) are the elements of the matrix:\[ \begin{bmatrix} a & b \ c & d \end{bmatrix} \]
In our case, we have the matrix: \[ \begin{bmatrix} 3 & 2 \ 4 & 5 \end{bmatrix} \] Plug in the numbers to get \( 3 \times 5 - 4 \times 2 = 15 - 8 = 7 \). The determinant is 7, which is not zero, indicating that the matrix does have an inverse. If this had been zero, the matrix would be singular, and we could not find an inverse.

2x2 Matrix

A 2x2 matrix is the simplest type of square matrix, made up of two rows and two columns. It is commonly used in linear algebra because its operations are straightforward.
For the matrix \[ \begin{bmatrix} a & b \ c & d \end{bmatrix} \], each entry represents a component of the matrix, organized in rows and columns. 'a' and 'b' are the elements of the first row, while 'c' and 'd' are in the second row.

• 'a' is the first row, first column element.
• 'b' is in the first row, second column.
• 'c' is in the second row, first column.
• 'd' is in the second row, second column.

A basic calculation involves finding the inverse, which is essential in solving systems of linear equations and other algebraic tasks.

Matrix Algebra

Matrix algebra is a fundamental concept in linear algebra involving matrices and their operations. Matrices are essential in various applications ranging from computer graphics to engineering.
To manipulate a matrix or solve problems, you often need to perform operations such as addition, subtraction, multiplication, and finding inverses. When dealing with matrix inverses, especially for a 2x2 matrix, the formula \[ \text{Inverse} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \] is applied, given that the determinant (ad-bc) is not zero. The inverse essentially 'undoes' a matrix operation. If you multiply a matrix by its inverse, you get an identity matrix, which serves a similar purpose to the number 1 in regular arithmetic.

• Identity matrix for 2x2 is \[ \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \]

Understanding these concepts is key to mastering matrix algebra and its applications.