Problem 12 - Precalculus: Mathematical for Calculus

Question

Find the inverse of the matrix if it exists. $$\left[\begin{array}{ll}\frac{1}{2} & \frac{1}{3} \\ 5 & 4\end{array}\right]$$

Step-by-Step Solution

Answer

The inverse is \(\begin{bmatrix} 12 & -1 \\ -15 & \frac{3}{2} \end{bmatrix}.\)

Step 1: Check if the Inverse Exists

To determine if the matrix has an inverse, check if the determinant is non-zero. The given matrix is:\[\begin{bmatrix}\frac{1}{2} & \frac{1}{3} \5 & 4\end{bmatrix}\]The determinant \(det(A)\) of a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\) is calculated as \(ad - bc\). Compute the determinant:\[det(A) = \left(\frac{1}{2}\right)(4) - \left(\frac{1}{3}\right)(5) = 2 - \frac{5}{3} = \frac{6}{3} - \frac{5}{3} = \frac{1}{3}.\]Since \( \frac{1}{3} eq 0 \), the inverse exists.

Step 2: Apply the Formula for the 2x2 Matrix Inverse

The formula for the inverse of a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\) when the determinant is non-zero is:\[A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}.\]Plug the values into the formula using the components of the matrix and its determinant \(\frac{1}{3}\):\[A^{-1} = \frac{1}{\frac{1}{3}} \begin{bmatrix} 4 & -\frac{1}{3} \ -5 & \frac{1}{2} \end{bmatrix}.\]

Step 3: Simplify the Inverse Matrix

To simplify \( A^{-1} \), multiply each element of the matrix by 3, the reciprocal of the determinant:\[A^{-1} = 3 \times \begin{bmatrix} 4 & -\frac{1}{3} \ -5 & \frac{1}{2} \end{bmatrix} = \begin{bmatrix} 12 & -1 \ -15 & \frac{3}{2} \end{bmatrix}.\] Thus, the inverse matrix is:\[\begin{bmatrix} 12 & -1 \ -15 & \frac{3}{2} \end{bmatrix}.\]

Step 4: Verify the Solution

Multiply the original matrix \( A \) by its calculated inverse \( A^{-1} \) to verify if their product is the identity matrix \( I \ = \ \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \):\[\begin{bmatrix} \frac{1}{2} & \frac{1}{3} \ 5 & 4 \end{bmatrix} \begin{bmatrix} 12 & -1 \ -15 & \frac{3}{2} \end{bmatrix} = \begin{bmatrix} \frac{1}{2}(12) + \frac{1}{3}(-15) & \frac{1}{2}(-1) + \frac{1}{3}(\frac{3}{2}) \ 5(12) + 4(-15) & 5(-1) + 4(\frac{3}{2}) \end{bmatrix}\]This results in:\[\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\]This confirms that \( A^{-1} \) is the correct inverse.

Key Concepts

Determinant Calculation

Before finding the inverse of a matrix, it is essential to determine whether it is possible to have an inverse. This possibility hinges on a key number called the determinant. For a 2x2 matrix, the determinant helps to assess if the matrix is invertible. If the determinant is not zero, the inverse exists. For the 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the determinant is calculated using the formula: \[ det(A) = ad - bc \] Calculating the determinant for the given matrix: \[ det(A) = \left(\frac{1}{2}\right)(4) - \left(\frac{1}{3}\right)(5) = 2 - \frac{5}{3} = \frac{6}{3} - \frac{5}{3} = \frac{1}{3} \] With this positive value, we confirm that the matrix has an inverse.

2x2 Matrix Inverse

Once it's verified that the matrix is invertible by confirming the determinant is non-zero, we can calculate the inverse. The inverse of a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is found using the formula: \[ A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \] We substitute the given values and the computed determinant \( \frac{1}{3} \) into the formula: \[ A^{-1} = 3 \times \begin{bmatrix} 4 & -\frac{1}{3} \ -5 & \frac{1}{2} \end{bmatrix} \] After solving, performing multiplication of each term by 3, the simplified inverse matrix is: \[ A^{-1} = \begin{bmatrix} 12 & -1 \ -15 & \frac{3}{2} \end{bmatrix} \] This matrix is the sought inverse form.

Matrix Multiplication Verification

Verification is a crucial step in ensuring that the calculated inverse is precise. To do this, multiply the original matrix by its inverse and check if the product equals the identity matrix. The identity matrix for a 2x2 form is \( \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \). This step ensures that the inverse is accurate. When computing the product: \[ \begin{bmatrix} \frac{1}{2} & \frac{1}{3} \ 5 & 4 \end{bmatrix} \begin{bmatrix} 12 & -1 \ -15 & \frac{3}{2} \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \] The calculations step-by-step yield the identity matrix: - First row, first column: \( \frac{1}{2}(12) + \frac{1}{3}(-15) = 1 \) - First row, second column: \( \frac{1}{2}(-1) + \frac{1}{3}(\frac{3}{2}) = 0 \) - Second row, first column: \( 5(12) + 4(-15) = 0 \) - Second row, second column: \( 5(-1) + 4(\frac{3}{2}) = 1 \) Thus, the verification step confirms the correctness of the inverse, as the result aligns with the identity matrix.