Problem 40 - Algebra 2

Question

Determine whether each matrix has an inverse. If an inverse matrix exists, find it. If it does not exist, explain why not. $$ \left[\begin{array}{ll}{1} & {2} \\ {2} & {1}\end{array}\right] $$

Step-by-Step Solution

Answer

The matrix \( \left[\begin{array}{ll}{1} & {2} \ {2} & {1}\end{array}\right] \) has an inverse since its determinant is -3. The inverse matrix is \( \left[\begin{array}{cc} -\frac{1}{3} & \frac{2}{3} \ \frac{2}{3} & -\frac{1}{3} \end{array}\right] \).

Step 1: Check Determinant

Calculate the determinant of the matrix. If the determinant is 0, the matrix does not have an inverse. The determinant is found by subtracting the product of the elements of the leading diagonal and the product of the elements of the other diagonal. So for matrix \( A = \left[\begin{array}{cc} 1 & 2 \ 2 & 1 \end{array}\right] \), the determinant \( \text{det}(A) \) is \( (1)(1) - (2)(2) = 1 - 4 = -3 \).

Step 2: Determine Inversibility

Since the determinant of the matrix is not 0 (\( \text{det}(A) = -3 \)), the matrix has an inverse.

Step 3: Find the Inverse

The formula for finding the inverse of a 2x2 matrix \( A = \left[\begin{array}{cc} a & b \ c & d \end{array}\right] \) is \( A^{-1} = \frac{1}{\text{det}(A)} \left[\begin{array}{cc} d & -b \ -c & a \end{array}\right] \). Applying this to matrix A: \( A^{-1} = \frac{1}{-3} \left[\begin{array}{cc} 1 & -2 \ -2 & 1 \end{array}\right] = \left[\begin{array}{cc} -\frac{1}{3} & \frac{2}{3} \ \frac{2}{3} & -\frac{1}{3} \end{array}\right] \).

Key Concepts

Matrix Inversibility

Understanding whether a matrix is invertible is a cornerstone in linear algebra. A matrix that is invertible, also known as a nonsingular or nondegenerate matrix, effectively has a counterpart that, when multiplied by the original, yields the identity matrix.

For a square matrix to be invertible, it must meet specific requirements. One of the primary conditions is that its determinant must not be zero. Determinants serve as an indicator for many properties of a matrix, including the existence of an inverse. If the determinant is zero, this implies that the matrix cannot be inverted, which is also equivalent to saying that the matrix is singular.

In the context of the provided exercise, the determinant of the matrix was calculated and found to be -3, not zero. This non-zero determinant is the first confirmation that the matrix in question is indeed invertible, leading us to the next steps, where we can find the actual inverse of the matrix.

Determinant of a Matrix

The determinant of a matrix provides powerful insights into the matrix's characteristics and its behavior in various mathematical settings. In a tangible sense, the determinant can be seen as a scalar value that reflects the volume scaling factor that a matrix, when considered as a transformation, applies to a spanned space. For a 2x2 matrix, like the one we are focusing on, the determinant is calculated by taking the product of the elements on the leading diagonal and subtracting the product of the off-diagonal elements.

Mathematically, for a matrix with elements \[ A = \left[\begin{array}{cc} a & b \ c & d \end{array}\right] \], the determinant is given by \(\text{det}(A) = ad - bc\).

This step is essential not only to assess inversibility but also when solving systems of linear equations, finding eigenvalues, and in various applications across different fields such as physics and engineering.

2x2 Matrix Inverse

Calculating the inverse of a 2x2 matrix is a straightforward process once its determinant is known and confirmed to be non-zero. The formula for the inverse of a 2x2 matrix is quite elegant and utilizes the elements of the original matrix in a specific rearrangement, combined with the multiplicative inverse (or reciprocal) of the determinant.

For matrix \[ A = \left[\begin{array}{cc} a & b \ c & d \end{array}\right] \], the inverse \( A^{-1} \) is given by \[ A^{-1} = \frac{1}{\text{det}(A)} \left[\begin{array}{cc} d & -b \-c & a \end{array}\right] \].

The elements a and d swap places, and elements b and c are negated, followed by scalar multiplication of each element by the reciprocal of the determinant. As shown in the solution for the exercise, applying these steps using the determinant earlier calculated, yields the inverse of the given matrix. This process is fundamental when solving linear systems and for understanding concepts such as eigenvectors and linear transformations in higher dimensions.