Problem 1 - Algebra for College Students

Question

For Problems \(1-18\), find the multiplicative inverse (if one exists) of each matrix. $$ \left[\begin{array}{ll} 5 & 7 \\ 2 & 3 \end{array}\right] $$

Step-by-Step Solution

Answer

The inverse of the matrix is \( \begin{bmatrix} 3 & -7 \\ -2 & 5 \end{bmatrix} \).

Step 1: Define the Multiplicative Inverse

The multiplicative inverse (also known as the inverse matrix) of a given matrix \( A \) is another matrix \( A^{-1} \) such that when it is multiplied with \( A \), it results in the identity matrix \( I \). For a 2x2 matrix \( \begin{bmatrix}a & b \ c & d \end{bmatrix} \), the inverse exists if the determinant is non-zero.

Step 2: Calculate the Determinant

For the matrix \( A = \begin{bmatrix} 5 & 7 \ 2 & 3 \end{bmatrix} \), the determinant \( \text{det}(A) \) is calculated as \( ad - bc \). Substituting the given values, we have \( \text{det}(A) = (5 \cdot 3) - (7 \cdot 2) = 15 - 14 = 1 \). Since the determinant is not zero, the inverse exists.

Step 3: Find the Matrix of Minors

For our 2x2 matrix, this step involves swapping the elements on the main diagonal and changing signs of the off-diagonal elements. Hence, the matrix of minors becomes:\[ \begin{bmatrix} 3 & -7 \ -2 & 5 \end{bmatrix} \]

Step 4: Transpose the Matrix of Minors

For a 2x2 matrix, the transpose involves exchanging rows and columns. However, the matrix from Step 3 \( \begin{bmatrix} 3 & -7 \ -2 & 5 \end{bmatrix} \) already matches its transpose.

Step 5: Multiply by the Reciprocal of Determinant

The inverse matrix is obtained by multiplying the matrix from Step 4 with \( \frac{1}{\text{det}(A)} \). Since \( \text{det}(A) = 1 \), the inverse matrix is:\[ \frac{1}{1} \begin{bmatrix} 3 & -7 \ -2 & 5 \end{bmatrix} = \begin{bmatrix} 3 & -7 \ -2 & 5 \end{bmatrix} \].

Step 6: Verification: Confirming the Result

To ensure correctness, multiply the computed inverse with the original matrix and check if the product is the identity matrix \( \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \). Compute:\[ \begin{bmatrix} 5 & 7 \ 2 & 3 \end{bmatrix} \begin{bmatrix} 3 & -7 \ -2 & 5 \end{bmatrix} = \begin{bmatrix} (5 \times 3) + (7 \times -2) & (5 \times -7) + (7 \times 5) \ (2 \times 3) + (3 \times -2) & (2 \times -7) + (3 \times 5) \end{bmatrix} \]Calculating gives:\[ \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \]Therefore, the inverse is correct.

Key Concepts

Determinants

Determinants play a crucial role in determining the invertibility of a matrix. For a 2x2 matrix, the determinant is calculated using the formula \( ext{det}(A) = ad - bc \) for a matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \). In simple terms, the determinant is a single number that provides a lot of information about the matrix.

To decide if a matrix has an inverse, we first need to find its determinant. If the result is zero, it means the matrix does not have an inverse. If the determinant is non-zero, an inverse does exist, making the calculation worth proceeding.

• In our exercise, the matrix is \( \begin{bmatrix} 5 & 7 \ 2 & 3 \end{bmatrix} \).
• The determinant is calculated as \( 5 \times 3 - 7 \times 2 = 1 \).
• Since the determinant is 1, which is not zero, this matrix has an inverse.

Understanding determinants is essential as they determine the step forward for different matrix operations, especially inversions.

Inverse Matrix Calculation

Once we know a matrix is invertible, we can calculate its inverse. For a 2x2 matrix, finding the inverse involves several steps.

We use the matrix of minors, transpose it, and finally multiply the result by the reciprocal of the determinant.

• First, swap the elements on the main diagonal while changing the signs of the off-diagonal elements. For our matrix \( \begin{bmatrix} 5 & 7 \ 2 & 3 \end{bmatrix} \), the resulting matrix becomes \( \begin{bmatrix} 3 & -7 \ -2 & 5 \end{bmatrix} \).
• For 2x2 matrices, the transposed matrix looks the same, so no change happens here.
• Finally, multiply by the inverse of the determinant. Since our determinant is 1, the inverse matrix does not change: \( \begin{bmatrix} 3 & -7 \ -2 & 5 \end{bmatrix} \).

This sequence, though straightforward for a 2x2 matrix, requires more complex operations for larger matrices. However, the concept remains fundamental and manageable, especially when starting with matrix transformations.

Verification of Matrix Inversion

Verifying the inverse of a matrix ensures the computed result is accurate. The easiest way to verify is by multiplying the original matrix with its calculated inverse. The product should yield the identity matrix if everything is correct.

• Starting with the original matrix \( \begin{bmatrix} 5 & 7 \ 2 & 3 \end{bmatrix} \), we multiply it by its inverse \( \begin{bmatrix} 3 & -7 \ -2 & 5 \end{bmatrix} \).
• Perform the multiplication:
• First row: \( (5 \times 3) + (7 \times -2) = 15 - 14 = 1 \) and \( (5 \times -7) + (7 \times 5) = -35 + 35 = 0 \).
• Second row: \( (2 \times 3) + (3 \times -2) = 6 - 6 = 0 \) and \( (2 \times -7) + (3 \times 5) = -14 + 15 = 1 \).

These calculations confirm that the product is \( \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \), the identity matrix. Thus, the inversion is verified successfully, reaffirming the correctness of our earlier calculations.