Problem 72 - Algebra 2

Question

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

Step-by-Step Solution

Answer

The inverse of the matrix is \( \begin{bmatrix} 1 & -3 \\ -1 & 4 \end{bmatrix} \).

Step 1: Check Matrix Invertibility

To find the inverse of a matrix, we first need to check if the matrix is invertible. A matrix is invertible if its determinant is non-zero. For the matrix \( A = \begin{bmatrix} 4 & 3 \ 1 & 1 \end{bmatrix} \), the determinant is calculated as:\[\text{det}(A) = (4)(1) - (3)(1) = 4 - 3 = 1\]Since the determinant is not zero, the matrix is invertible.

Step 2: Use the Inverse Formula for a 2x2 Matrix

For a 2x2 matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the inverse is given by:\[A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]Substitute \( a = 4, b = 3, c = 1, d = 1 \) into the formula:\[A^{-1} = \frac{1}{1} \begin{bmatrix} 1 & -3 \ -1 & 4 \end{bmatrix}\]Simplifying, we have:\[A^{-1} = \begin{bmatrix} 1 & -3 \ -1 & 4 \end{bmatrix}\]

Step 3: Verify by Multiplication

To ensure the calculated inverse is correct, multiply the original matrix \( A \) by its inverse \( A^{-1} \):\[A \times A^{-1} = \begin{bmatrix} 4 & 3 \ 1 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & -3 \ -1 & 4 \end{bmatrix} = \begin{bmatrix} (4\cdot1 + 3\cdot(-1)) & (4\cdot(-3) + 3\cdot4) \ (1\cdot1 + 1\cdot(-1)) & (1\cdot(-3) + 1\cdot4) \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\]Since the product is the identity matrix, the inverse is correctly found.

Key Concepts

Determinant

To find the inverse of a matrix, we begin by calculating its determinant. The determinant helps us know if a matrix is invertible. For a 2x2 matrix like the one given:
\[ A = \begin{bmatrix} 4 & 3 \ 1 & 1 \end{bmatrix}\]
We calculate the determinant using the formula:
\[ \text{det}(A) = a \cdot d - b \cdot c\]
where

• \( a = 4 \),
• \( b = 3 \),
• \( c = 1 \),
• \( d = 1 \).

The determinant is:
\[ (4)(1) - (3)(1) = 4 - 3 = 1\]
Since this result is not zero, our matrix can have an inverse.

2x2 Matrix

A 2x2 matrix is a simple form of a matrix that contains two rows and two columns. It is usually shown like this:
\[ \begin{bmatrix} a & b \ c & d \end{bmatrix}\]
In a 2x2 matrix, there are four elements of numbers. The positions of these numbers are very important because they dictate how operations like matrix multiplication and finding the inverse are performed.
For example, our matrix:
\[ \begin{bmatrix} 4 & 3 \ 1 & 1 \end{bmatrix}\]
The first row is \(4, 3\) and the second row is \(1, 1\). Understanding this layout is key to performing further calculations.

Inverse Matrix Calculation

Once we have confirmed that the determinant is not zero, we can proceed to find the inverse of the matrix. For the general 2x2 matrix:
\[ A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\]
The formula for finding the inverse is:
\[ A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]
Substituting our specific values:

• \( a = 4 \)
• \( b = 3 \)
• \( c = 1 \)
• \( d = 1 \)

We get:
\[ A^{-1} = \frac{1}{1} \begin{bmatrix} 1 & -3 \ -1 & 4 \end{bmatrix} = \begin{bmatrix} 1 & -3 \ -1 & 4 \end{bmatrix}\]
This result is the inverse of our 2x2 matrix.

Matrix Multiplication Verification

To ensure that the calculated inverse is accurate, we perform matrix multiplication with the original matrix and its inverse. If the multiplication yields the identity matrix, our inverse is correct.
Using our original matrix:
\[ A = \begin{bmatrix} 4 & 3 \ 1 & 1 \end{bmatrix}\]
and its inverse:
\[ A^{-1} = \begin{bmatrix} 1 & -3 \ -1 & 4 \end{bmatrix}\]
The multiplication is carried out as follows:
\[ A \times A^{-1} = \begin{bmatrix} 4 & 3 \ 1 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & -3 \ -1 & 4 \end{bmatrix}\]\[ = \begin{bmatrix} (4 \times 1 + 3 \times (-1)) & (4 \times (-3) + 3 \times 4) \ (1 \times 1 + 1 \times (-1)) & (1 \times (-3) + 1 \times 4) \end{bmatrix}\]
This results in:
\[ \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\]
The product is indeed the identity matrix, confirming our inverse matrix calculation is correct.