Problem 10 - Algebra and Trigonometry

Question

Find the inverse of the matrix if it exists. $$ \left[\begin{array}{ll}{3} & {4} \\ {7} & {9}\end{array}\right] $$

Step-by-Step Solution

Answer

The inverse of the matrix is \( \begin{bmatrix} -9 & 4 \\ 7 & -3 \end{bmatrix} \).

Step 1: Verify the Matrix is Square

A matrix must be square to have an inverse. This matrix \[A = \begin{bmatrix} 3 & 4 \ 7 & 9 \end{bmatrix}\] is a 2x2 matrix, which is square, so we can proceed to find its inverse.

Step 2: Calculate the Determinant

Find the determinant of matrix A. For a 2x2 matrix \[ \begin{bmatrix} a & b \ c & d \end{bmatrix}, \] the determinant is calculated as \[ det(A) = ad - bc. \] Thus, \[ det(A) = (3)(9) - (4)(7) = 27 - 28 = -1. \] Since the determinant is non-zero, the inverse exists.

Step 3: Apply the Formula for the Inverse of a 2x2 Matrix

The inverse of a 2x2 matrix \[ \begin{bmatrix} a & b \ c & d \end{bmatrix} \] can be found using the formula: \[ A^{-1} = \frac{1}{det(A)} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}. \] Inserting the values from our matrix and the determinant:\[ A^{-1} = \frac{1}{-1} \begin{bmatrix} 9 & -4 \ -7 & 3 \end{bmatrix} = \begin{bmatrix} -9 & 4 \ 7 & -3 \end{bmatrix}. \]

Step 4: Verify the Inverse is Correct

Verify the solution by checking that the product of the matrix and its inverse yields the identity matrix. Calculate:\[ \begin{bmatrix} 3 & 4 \ 7 & 9 \end{bmatrix} \begin{bmatrix} -9 & 4 \ 7 & -3 \end{bmatrix} = \begin{bmatrix} (3)(-9) + (4)(7) & (3)(4) + (4)(-3) \ (7)(-9) + (9)(7) & (7)(4) + (9)(-3) \end{bmatrix} = \begin{bmatrix} -27 + 28 & 12 - 12 \ -63 + 63 & 28 - 27 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}. \] The product equals the identity matrix, confirming the inverse is correct.

Key Concepts

2x2 Matrix

A 2x2 matrix is a collection of numbers arranged in two rows and two columns. It looks like this:

\[\begin{bmatrix} a & b \ c & d \end{bmatrix}\]

Each number in the matrix is known as an element or entry.
In matrix mathematics, only square matrices (those with the same number of rows and columns) can have an inverse.
This is why checking if a matrix is square is the first step in determining whether it can have an inverse. In the exercise, we work with a matrix:

• First row: 3, 4
• Second row: 7, 9

This matrix is indeed square as it has 2 rows and 2 columns, qualifying it as a 2x2 matrix.

Determinant Calculation

Before finding the inverse of a 2x2 matrix, one needs to calculate its determinant. The determinant provides crucial information, such as if the matrix has an inverse at all.
For a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\), the determinant is calculated as:

\[det(A) = ad - bc\]

Where:

• \(a\), \(b\), \(c\), \(d\) are elements of the matrix

In our example, the determinant is:

\[(3 \times 9) - (4 \times 7) = 27 - 28 = -1\]

Since the determinant is not zero, the matrix has an inverse.

Matrix Multiplication

Matrix multiplication involves calculating the dot products of the rows of the first matrix with the columns of the second matrix. This is a key skill in verifying the inverse of a matrix.
When two matrices are multiplied, their dimensions must align; that is, the number of columns in the first matrix must equal the number of rows in the second.
In our exercise, we multiplied the matrix by its calculated inverse:

• Original Matrix: \(\begin{bmatrix} 3 & 4 \ 7 & 9 \end{bmatrix}\)
• Inverse Matrix: \(\begin{bmatrix} -9 & 4 \ 7 & -3 \end{bmatrix}\)

Performing the multiplication confirms the result is the identity matrix, which is explained next.

Identity Matrix

The identity matrix is fundamental in linear algebra. It acts like the number 1 in regular arithmetic.
For a 2x2 identity matrix, it's represented as:

\[\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\]

Multiplying any matrix by its inverse will result in the identity matrix.
In our exercise, after finding the inverse of the original matrix, we confirmed its correctness by ensuring that:

\[\begin{bmatrix} 3 & 4 \ 7 & 9 \end{bmatrix} \times \begin{bmatrix} -9 & 4 \ 7 & -3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\]

This proves that the calculated inverse is correct.