Problem 16 - College Algebra

Question

For the following exercises, find the multiplicative inverse of each matrix, if it exists. $$\left[\begin{array}{rr}-4 & -3 \\ -5 & 8\end{array}\right]$$

Step-by-Step Solution

Answer

The inverse is \( \begin{bmatrix} -\frac{8}{47} & -\frac{3}{47} \\ -\frac{5}{47} & \frac{4}{47} \end{bmatrix} \).

Step 1: Understand the Multiplicative Inverse Condition

A matrix has a multiplicative inverse if it is a square matrix and its determinant is non-zero. The inverse of a matrix \( A \) is denoted \( A^{-1} \) and satisfies \( A \cdot A^{-1} = I \), where \( I \) is the identity matrix.

Step 2: Calculate the Determinant of the Matrix

The matrix given is \[ \begin{bmatrix} -4 & -3 \ -5 & 8 \end{bmatrix} \]. The determinant \( \text{det}(A) \) of a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is calculated as \( ad - bc \). For this matrix, \[ \text{det}(A) = (-4)(8) - (-3)(-5) = -32 - 15 = -47 \]. Since \( \text{det}(A) eq 0 \), the matrix is invertible.

Step 3: Use the Formula for the Inverse of a 2x2 Matrix

For a matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the inverse is given by \( \frac{1}{ad - bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \). Using the calculated determinant, the inverse is \[ A^{-1} = \frac{1}{-47} \begin{bmatrix} 8 & 3 \ 5 & -4 \end{bmatrix} \].

Step 4: Multiply the Elements by the Reciprocal of the Determinant

Now multiply each element in the matrix \( \begin{bmatrix} 8 & 3 \ 5 & -4 \end{bmatrix} \) by \( \frac{1}{-47} \) to find the inverse:\[ A^{-1} = \begin{bmatrix} \frac{8}{-47} & \frac{3}{-47} \ \frac{5}{-47} & \frac{-4}{-47} \end{bmatrix} = \begin{bmatrix} -\frac{8}{47} & -\frac{3}{47} \ -\frac{5}{47} & \frac{4}{47} \end{bmatrix} \].

Step 5: Conclude the Inverse Matrix

The multiplicative inverse of the given matrix is \( \begin{bmatrix} -\frac{8}{47} & -\frac{3}{47} \ -\frac{5}{47} & \frac{4}{47} \end{bmatrix} \). This is indeed the matrix that, when multiplied by the original matrix, yields the identity matrix.

Key Concepts

Understanding the Determinant

The determinant is a special number that can be calculated from a square matrix. It's a value that helps us to determine whether a matrix has an inverse. For a 2x2 matrix, the determinant is found using a simple formula. Given a 2x2 matrix\[ \begin{bmatrix} a & b \ c & d \end{bmatrix} \], the determinant is calculated as \( ad - bc \). In our exercise, the matrix is\[ \begin{bmatrix} -4 & -3 \ -5 & 8 \end{bmatrix} \]. By plugging the values into the determinant formula, we get:

• \( a = -4 \), \( b = -3 \), \( c = -5 \), \( d = 8 \)
• Determinant: \((-4)(8) - (-3)(-5) = -32 - 15 = -47 \)

The determinant is \(-47\), which is not zero. Therefore, this matrix is invertible, meaning that it has an inverse matrix.

What is a 2x2 Matrix?

A 2x2 matrix is a type of square matrix that has two rows and two columns:\[ \begin{bmatrix} a & b \ c & d \end{bmatrix} \]. Each matrix holds specific values that we call "elements". For the given exercise, the matrix \[ \begin{bmatrix} -4 & -3 \ -5 & 8 \end{bmatrix} \] is a 2x2 matrix. Working with 2x2 matrices is a foundational skill in linear algebra. It's often where students first learn about matrix operations like addition, subtraction, and multiplication.

In addition, 2x2 matrices are often used to represent transformations in 2D space, like rotations and reflections. A key concept is that 2x2 matrices can have inverses, provided they have a non-zero determinant. The inverse, when multiplied by the original matrix, results in the identity matrix.

The Role of the Identity Matrix

An identity matrix acts like the number 1 in matrix mathematics. It is a special square matrix with 1's on the diagonal and 0's elsewhere. For a 2x2 identity matrix, it looks like:\[ \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \]. This matrix has the property that any matrix multiplied by the identity matrix results in the original matrix itself.

When finding the inverse of a matrix, we want the product of the original matrix and its inverse to equal the identity matrix. If you multiply the matrix from our exercise\[ \begin{bmatrix} -4 & -3 \ -5 & 8 \end{bmatrix} \] by its inverse\[ \begin{bmatrix} -\frac{8}{47} & -\frac{3}{47} \ -\frac{5}{47} & \frac{4}{47} \end{bmatrix} \], the result should be the identity matrix\[ \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \]. This property helps verify the correctness of a calculated inverse.