Problem 18 - Algebra 2

Question

Find the inverse of each matrix, if it exists. $$ \left[\begin{array}{rr}{-3} & {7} \\ {2} & {-6}\end{array}\right] $$

Step-by-Step Solution

Answer

The inverse is \( \begin{bmatrix} -\frac{3}{2} & -\frac{7}{4} \\ -\frac{1}{2} & -\frac{3}{4} \end{bmatrix} \).

Step 1: Check if the matrix is invertible

To find the inverse of a 2x2 matrix, we need to ensure it is invertible. A matrix is invertible if its determinant is non-zero. For the given matrix \( \begin{bmatrix} -3 & 7 \ 2 & -6 \end{bmatrix} \), calculate the determinant using the formula: \( \text{det} = ad - bc \), where \( a = -3, b = 7, c = 2, \text{ and } d = -6 \).

Step 2: Calculate the determinant

Compute the determinant: \( \text{det} = (-3)(-6) - (7)(2) = 18 - 14 = 4 \). Since the determinant is 4 (not zero), the matrix is invertible.

Step 3: Use the formula for the inverse of a 2x2 matrix

For an invertible matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the inverse is given by: \[ \frac{1}{ad - bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \]. Substitute \( a = -3, b = 7, c = 2, d = -6 \).

Step 4: Substitute values into the formula

Apply the values to the inverse formula: \[ \frac{1}{4} \begin{bmatrix} -6 & -7 \ -2 & -3 \end{bmatrix} = \begin{bmatrix} -\frac{6}{4} & -\frac{7}{4} \ -\frac{2}{4} & -\frac{3}{4} \end{bmatrix} \]. Simplify the fractions: \( \begin{bmatrix} -\frac{3}{2} & -\frac{7}{4} \ -\frac{1}{2} & -\frac{3}{4} \end{bmatrix} \).

Key Concepts

Determinant

The determinant is a key concept when dealing with matrices, particularly when we want to determine if a matrix is invertible or not. For a 2x2 matrix, the determinant is calculated using a simple formula:

• Look at the matrix: \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \).
• Apply the formula for the determinant: \( \text{det} = ad - bc \).

In simpler terms, multiply the diagonal elements (\(a\) and \(d\)), then subtract the product of the other diagonal elements (\(b\) and \(c\)). For example, in the matrix \( \begin{bmatrix} -3 & 7 \ 2 & -6 \end{bmatrix} \), you would compute the determinant as \((-3)(-6) - (7)(2) = 18 - 14 = 4\).
The resulting value is 4, which is crucial because a non-zero determinant means the matrix is invertible.
In contrast, if the determinant were zero, the matrix would not have an inverse.

2x2 Matrix

A 2x2 matrix is one of the most basic forms of a matrix in linear algebra. It has two rows and two columns, with four elements in total. A typical 2x2 matrix looks like this:
\[\begin{bmatrix}a & b \ c & d \\end{bmatrix}\]Each position in the matrix represents an element of data.
These elements are typically denoted by letters \(a\), \(b\), \(c\), and \(d\).

• Element \(a\) is in the first row and first column.
• Element \(b\) is in the first row and second column.
• Element \(c\) is in the second row and first column.
• Element \(d\) is in the second row and second column.

This simple structure makes 2x2 matrices an excellent starting point for learning about more complex matrix operations, like finding determinants and inverses.

Invertible Matrix

An invertible matrix, also known as a non-singular matrix, is one that has an inverse.
An inverse is like a reciprocal in algebra - when a matrix is multiplied by its inverse, the result is the identity matrix.
The identity matrix for a 2x2 matrix looks like this:\[\begin{bmatrix}1 & 0 \0 & 1 \\end{bmatrix}\]To determine if a matrix is invertible, you must check its determinant.
If the determinant is non-zero, the matrix can be inverted.
For example, the matrix \( \begin{bmatrix} -3 & 7 \ 2 & -6 \end{bmatrix} \) has a determinant of 4.
Since 4 is not zero, the matrix is invertible, and an inverse can be calculated using the formula for a 2x2 matrix inverse:\[\text{Inverse} = \frac{1}{\text{det}} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]By substituting the specific values from the matrix, you can solve for the inverse. This is crucial in various applications, including solving systems of linear equations and in transformations in science and engineering.