Question

Each matrix is nonsingular. Find the inverse of each matrix. $$ \left[\begin{array}{ll} b & 3 \\ b & 2 \end{array}\right] \quad b \neq 0 $$

Short answer

The inverse is \(\begin{pmatrix} \frac{-2}{b} & \frac{3}{b} \ 1 & -1 \end{pmatrix}\).

Step-by-step solution

Recall the formula for the inverse of a 2x2 matrix

For a 2x2 matrix the invertible matrix \[A = \begin{pmatrix} a & b \ c & d \end{pmatrix}\] the inverse is given by \[A^{-1} = \frac{1}{{ad - bc}} \begin{pmatrix} d & -b \ -c & a \end{pmatrix}\]

Identify the elements of the matrix

In the given matrix \[\begin{pmatrix} b & 3 \ b & 2 \end{pmatrix}\] the elements are: a = b, b = 3, c = b, and d = 2.

Calculate the determinant

The determinant of the matrix is calculated as follows: \[\text{det}(A) = ad - bc = b \times 2 - 3 \times b = 2b - 3b = -b\]

Substitute into the inverse formula

Replace a, b, c, d, and the determinant into the inverse formula: \[A^{-1} = \frac{1}{{-b}} \begin{pmatrix} 2 & -3 \ -b & b \end{pmatrix} = \begin{pmatrix} \frac{2}{-b} & \frac{-3}{-b} \ \frac{-b}{-b} & \frac{b}{-b} \end{pmatrix} = \begin{pmatrix} \frac{-2}{b} & \frac{3}{b} \ 1 & -1 \end{pmatrix}\]

Key concepts

nonsingular matrix

A nonsingular matrix is a square matrix that has an inverse. This means that when you multiply the matrix by its inverse, the result is the identity matrix. To check if a matrix is nonsingular, you need to find its determinant. If the determinant is not zero, then the matrix is nonsingular. For example, in our exercise, the matrix \(\begin{pmatrix} b & 3 \ b & 2 \end{pmatrix}\), with the condition \( b eq 0 \), is nonsingular because its determinant is not zero. Let's understand this concept further in the following sections.

2x2 matrix inverse formula

The formula for finding the inverse of a 2x2 matrix is straightforward but essential. For a 2x2 matrix \(A = \begin{pmatrix} a & b \ c & d \end{pmatrix}\), the inverse \(A^{-1} \) is given by: \[ A^{-1} = \frac{1}{{ad - bc}} \begin{pmatrix} d & -b \ -c & a \end{pmatrix} \] This formula only works if the determinant (\(ad - bc\)) is not zero. In other words, the matrix must be nonsingular. The formula highlights the crucial role of the determinant and how the elements of the matrix are rearranged and negated to find the inverse. Understanding this formula is essential for working with linear equations and other advanced matrix applications.

determinant calculation

Calculating the determinant of a 2x2 matrix is simple yet crucial. For a matrix \(A = \begin{pmatrix} a & b \ c & d \end{pmatrix}\), the determinant is calculated as: \( \text{det}(A) = ad - bc \).
In our example, the matrix \(M = \begin{pmatrix} b & 3 \ b & 2 \end{pmatrix}\) has elements \(a = b\), \(b = 3\), \(c = b\), and \(d = 2\).
Substituting these values into the determinant formula: \( \text{det}(A) = b \times 2 - 3 \times b = 2b - 3b = -b \).
Remember, since \( b eq 0 \), the determinant is non-zero and thus confirms that the matrix is nonsingular.

matrix elements substitution

For matrix operations, identifying and substituting elements accurately is crucial. Let's apply this to our matrix \(M = \begin{pmatrix} b & 3 \ b & 2 \end{pmatrix}\). The elements of the matrix are:

• \(a = b\)
• \(b = 3\)
• \(c = b\)
• \(d = 2\)

Using the inverse formula \(A^{-1} = \frac{1}{{ad - bc}} \begin{pmatrix} d & -b \ -c & a \end{pmatrix}\), substitute the matrix elements and determinant:
\(A^{-1} = \frac{1}{{-b}} \begin{pmatrix} 2 & -3 \ -b & b \end{pmatrix} = \begin{pmatrix} \frac{2}{-b} & \frac{-3}{-b} \ \frac{-b}{-b} & \frac{b}{-b} \end{pmatrix} = \begin{pmatrix} \frac{-2}{b} & \frac{3}{b} \ 1 & -1 \end{pmatrix}\).
Accurate substitution ensures the correct inverse matrix, highlighting the importance of each matrix element in the calculations.