Problem 87 - College Algebra and Calculus: An Applied Approach

Question

If \(A\) is a \(2 \times 2\) matrix \(A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\), then \(A\) is invertible if and only if \(a d-b c \neq 0 .\) If \(a d-b c \neq 0\), verify that the inverse is \(A^{-1}=\frac{1}{a d-b c}\left[\begin{array}{rr}d & -b \\ -c & a\end{array}\right]\).

Step-by-Step Solution

Answer

Upon the multiplication of the matrix A=\[\begin{array}{ll}a & b \\ c & d\end{array}\] with the proposed inverse formula \[ A^{-1}=\frac{1}{a d-b c}\left[\begin{array}{rr}d & -b \\ -c & a\end{array}\right] \], we get the identity matrix. Therefore, the proposed formula is indeed the inverse of matrix A, provided the determinant \(ad−bc ≠ 0\).

Step 1: Calculate the determinant of matrix A

The determinant of a 2x2 matrix A=\[\begin{array}{ll}a & b \\ c & d\end{array}\] is calculated by the formula \(ad−bc\). If \(ad−bc ≠ 0\), then matrix A is invertible.

Step 2: Construct the inverse formula

According to the task, provided the determinant is not zero, the inverse of the matrix can be calculated using the formula \[ A^{-1}=\frac{1}{a d-b c}\left[\begin{array}{rr}d & -b \\ -c & a\end{array}\right] \]

Step 3: Verify the formula

The result of the multiplication of a matrix by its inverse must be the identity matrix. We have to verify the multiplication of A with the alleged \(A^{-1}\). We calculate \[ A \cdot A^{-1} = \left[\begin{array}{ll}a & b \\ c & d\end{array}\right] \cdot \frac{1}{a d-b c}\left[\begin{array}{rr}d & -b \\ -c & a\end{array}\right] \]. If we obtain the identity matrix, then the formula is correct.

Key Concepts

Determinant of a Matrix

Understanding the determinant of a matrix is critical when delving into linear algebra, especially when dealing with matrix invertibility. For a 2x2 matrix such as

\[\begin{equation} A = \begin{bmatrix} a & b \ c & d \text{ the determinant is calculated by the formula } ad - bc. \end{equation}\]
The determinant gives us important information. If this value is zero, the matrix cannot be inverted, which has implications for solving systems of linear equations and understanding linear transformations. When the determinant is non-zero, it indicates that the matrix has an inverse and occupies a specific area or volume in space, depending on the dimension of the matrix.

Inverse of a Matrix

The inverse of a matrix is akin to the reciprocal of a number. It is a crucial concept in linear algebra because it allows us to solve matrix equations akin to division in arithmetic. For a matrix to have an inverse, it must be square, and its determinant must not be zero.

For a 2x2 matrix

\[\begin{equation} A = \begin{bmatrix} a & b \ c & d \text{, its inverse } A^{-1} \text{ is calculated using the formula: } \frac{1}{ad - bc}\begin{bmatrix} d & -b \ -c & a \text{, provided that } ad - bc eq 0. \end{equation}\]
Multiplying the matrix by its inverse yields the identity matrix, a critical property that confirms the correctness of the inverse.

Identity Matrix

The identity matrix serves as the multiplicative identity in matrix operations. Imagine it as the number 1 for matrices. When any matrix is multiplied by the identity matrix, it remains unchanged.

For 2x2 matrices, the identity matrix is

\[\begin{equation} I = \begin{bmatrix} 1 & 0 \ 0 & 1 \text{. Multiplying any 2x2 matrix by this identity matrix—hence the name—results in the original matrix. The existence of an inverse is established when the product of a matrix and its proposed inverse equals the identity matrix. } \end{equation}\]
This property is essential when verifying the validity of an inverse matrix.

2x2 Matrix Operations

Matrix operations for a 2x2 setup are foundational exercises in linear algebra. Addition and subtraction involve element-wise operations, while matrix multiplication requires the dot product of rows and columns.

Let's consider two 2x2 matrices,

\[\begin{equation} A = \begin{bmatrix} a & b \ c & d \text{ and } B = \begin{bmatrix} e & f \ g & h \text{. For multiplication, the resulting matrix is obtained by: } \end{equation}\]

\[\begin{equation} A \times B = \begin{bmatrix} ae+bg & af+bh \ ce+dg & cf+dh \text{. These operations are essential when it comes to applying linear transformations and understanding system behaviors in linear algebra.} \end{equation}\]