Problem 20 - Algebra 2

Question

Find the inverse of each matrix, if it exists. $$ \left[\begin{array}{rr}{-2} & {0} \\ {5} & {6}\end{array}\right] $$

Step-by-Step Solution

Answer

The inverse of the matrix is \( \begin{bmatrix} -\frac{1}{2} & 0 \\ \frac{5}{12} & \frac{1}{6} \end{bmatrix} \).

Step 1: Check Determinant

First, calculate the determinant of the matrix. The formula for the determinant \( (a, b, c, d) \) of a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is given by \( \text{det} = ad - bc \). For the matrix \( \begin{bmatrix} -2 & 0 \ 5 & 6 \end{bmatrix} \), substitute the respective values: \( \text{det} = (-2)(6) - (0)(5) = -12 \). Since the determinant is not zero, the inverse exists.

Step 2: Set Up Inverse Formula

The inverse of a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is given by \( \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \). Using our matrix \( \begin{bmatrix} -2 & 0 \ 5 & 6 \end{bmatrix} \) and determinant \( \text{det} = -12 \), the formula becomes \( -\frac{1}{12} \begin{bmatrix} 6 & 0 \ -5 & -2 \end{bmatrix} \).

Step 3: Compute the Inverse

Perform the scalar multiplication \( -\frac{1}{12} \) with each element of the matrix \( \begin{bmatrix} 6 & 0 \ -5 & -2 \end{bmatrix} \). The inverse matrix is \( \begin{bmatrix} -\frac{1}{2} & 0 \ \frac{5}{12} & \frac{1}{6} \end{bmatrix} \). Factor \( -\frac{1}{12} \) through each corresponding element: \(-\frac{1}{2} = -\frac{1}{12}(6), \frac{5}{12} = -\frac{1}{12}(-5), \frac{1}{6} = -\frac{1}{12}(-2) \).

Key Concepts

Determinant Calculation

The determinant is a special number that can tell us a lot about a matrix, including whether it's possible to find an inverse. For a 2x2 matrix like the one in the exercise, calculating the determinant is straightforward. You start with the formula \( \text{det} = ad - bc \), where \( a \), \( b \), \( c \), and \( d \) are the elements of the matrix. In this exercise, the matrix is \( \begin{bmatrix} -2 & 0 \ 5 & 6 \end{bmatrix} \).

Plugging in the values, you compute the determinant: \( (-2)(6) - (0)(5) = -12 \). The main thing to remember is that if the determinant is zero, the matrix doesn't have an inverse. But here, since \( -12 \) is not zero, we can proceed to find the inverse.

2x2 Matrix

A 2x2 matrix consists of four numbers arranged in two rows and two columns. Here's the general form:

• First Row: \( [a \ b] \)
• Second Row: \( [c \ d] \)

The matrix in your exercise is an example of a 2x2 matrix, \( \begin{bmatrix} -2 & 0 \ 5 & 6 \end{bmatrix} \).
These matrices are particularly nice because they're simpler to deal with than larger matrices, but they still allow us to explore some powerful concepts like finding inverses or calculating determinants.
2x2 matrices have an inverse only if their determinant isn't zero. Always calculate the determinant first before attempting to find an inverse.

Inverse Formula

The inverse of a matrix is a special matrix that, when multiplied by the original matrix, results in the identity matrix. For a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the formula for the inverse is \( \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \).
This formula hinges on the determinant \( ad-bc \); since we know it for this matrix is \( -12 \), the inverse becomes \( -\frac{1}{12} \begin{bmatrix} 6 & 0 \ -5 & -2 \end{bmatrix} \).
Basically, you switch the places of \( a \) and \( d \), make \( b \) and \( c \) negative, and multiply the entire matrix by \( \frac{1}{ad-bc} \). If the determinant were zero, this formula wouldn't work, because you can't divide by zero. This will prevent us from finding an inverse.

Scalar Multiplication in Matrices

Scalar multiplication involves multiplying every element of a matrix by a single number (the scalar). In the exercise, after determining the inverse matrix structure \( \begin{bmatrix} 6 & 0 \ -5 & -2 \end{bmatrix} \), each element is multiplied by \( -\frac{1}{12} \).

• \( 6 \times -\frac{1}{12} = -\frac{1}{2} \)
• \( 0 \times -\frac{1}{12} = 0 \)
• \( -5 \times -\frac{1}{12} = \frac{5}{12} \)
• \( -2 \times -\frac{1}{12} = \frac{1}{6} \)

The resulting matrix \( \begin{bmatrix} -\frac{1}{2} & 0 \ \frac{5}{12} & \frac{1}{6} \end{bmatrix} \) is the inverse of the original matrix. Scalar multiplication is like distribution; you apply the scalar to every part of the matrix. This process is crucial when finding the inverse, as the matrix needs to be adjusted by the reciprocal of its determinant.