Problem 42 - Algebra and Trigonometry

Question

Find the inverse of the matrix. For what value(s) of \(x\), if any, does the matrix have no inverse? \(\left[\begin{array}{cc}{e^{x}} & {-e^{2 x}} \\ {e^{2 x}} & {e^{3 x}}\end{array}\right]\)

Step-by-Step Solution

Answer

The matrix is invertible for all values of \(x\); the inverse is \(\frac{1}{2} \begin{pmatrix} e^{-x} & e^{-2x} \\ -e^{-2x} & e^{-3x} \end{pmatrix}\).

Step 1: Determine if the inverse exists

A square matrix has an inverse if and only if its determinant is non-zero. Let's compute the determinant of the given matrix:\[\text{det} = e^x \cdot e^{3x} - (-e^{2x}) \cdot e^{2x} = e^{4x} + e^{4x} = 2e^{4x}\]The matrix will not have an inverse if the determinant is 0. This occurs when \(2e^{4x} = 0\), but since \(e^{4x}\) is never zero, the determinant will never be zero.

Step 2: Calculate the inverse of the matrix

Since the determinant is non-zero for all \(x\), we can find the inverse of the matrix. The inverse of a 2x2 matrix \[A = \begin{pmatrix} a & b \ c & d \end{pmatrix}\] is given by:\[A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix}\]Substituting from our matrix, we have:- \(a = e^x\), \(b = -e^{2x}\), \(c = e^{2x}\), \(d = e^{3x}\)Thus, the inverse is:\[A^{-1} = \frac{1}{2e^{4x}} \begin{pmatrix} e^{3x} & e^{2x} \ -e^{2x} & e^x \end{pmatrix} = \frac{1}{2} \begin{pmatrix} e^{-x} & e^{-2x} \ -e^{-2x} & e^{-3x} \end{pmatrix}\]

Step 3: Conclusion: Analyze the result

The matrix has an inverse for all values of \(x\) because its determinant, \(2e^{4x}\), is never zero. The inverse matrix exists for every real number \(x\).

Key Concepts

Determinant of a Matrix

The determinant of a matrix is a scalar value that provides important insights about the matrix itself. In the context of a 2x2 matrix, and specifically for our exercise, it determines whether or not the matrix has an inverse.
The determinant of a matrix \(A = \begin{pmatrix} a & b \ c & d \end{pmatrix}\) is calculated with the formula:

• Determinant, \(\text{det}(A) = ad - bc\).

In our example, the matrix has elements \(a = e^x, b = -e^{2x}, c = e^{2x},\) and \(d = e^{3x}\). Thus the determinant is:

• \(\text{det}(A) = e^{4x} + e^{4x} = 2e^{4x}\).

Determining the sign and zero status of this determinant helps us know if the matrix is invertible.

Conditions for Matrix Invertibility

For a matrix to have an inverse, it is crucial for its determinant to be non-zero. This is a fundamental condition of matrix invertibility.
In our problem, we determine that the determinant, \(2e^{4x}\), is never zero for any real value of \(x\). This is because the exponential function \(e^{4x}\) can never equal zero.
Therefore, under these conditions, the matrix is invertible for all real numbers.

• A zero determinant means the rows of the matrix are linearly dependent.
• Non-zero determinant implies linearly independent rows and thus, invertibility.

Inverse of 2x2 Matrix

Finding the inverse of a 2x2 matrix is a structured process. When a matrix is invertible, the inverse of the matrix \(A = \begin{pmatrix} a & b \ c & d \end{pmatrix}\) is given by:

• \(A^{-1} = \frac{1}{ad-bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix}\).

For our specific example, substituting in the calculated determinant and the matrix components gives:

• \(A^{-1} = \frac{1}{2e^{4x}} \begin{pmatrix} e^{3x} & e^{2x} \ -e^{2x} & e^x \end{pmatrix}\).
• This simplifies as the inverse: \(\frac{1}{2} \begin{pmatrix} e^{-x} & e^{-2x} \ -e^{-2x} & e^{-3x} \end{pmatrix}\).

The inverse process flips and negates specific elements while also dividing by the determinant, ensuring all necessary calculations respect the equations for matrix operations.