Problem 22 - Algebra for College Students

Question

For Problems 21-36, use the technique discussed in this section to find the multiplicative inverse (if one exists) of each matrix. $$ \left[\begin{array}{rr} 1 & 2 \\ 2 & -3 \end{array}\right] $$

Step-by-Step Solution

Answer

The inverse is \( \left[\begin{array}{cc} \frac{3}{7} & \frac{2}{7} \\ \frac{2}{7} & -\frac{1}{7} \end{array}\right] \).

Step 1: Understand the Definition

The multiplicative inverse of a matrix \( A \) is another matrix \( A^{-1} \) such that \( AA^{-1} = I \), where \( I \) is the identity matrix of the same size. A 2x2 identity matrix is \( \left[ \begin{array}{cc} 1 & 0 \ 0 & 1 \end{array} \right] \).

Step 2: Check for Determinant

Calculate the determinant of the given matrix. The determinant of a 2x2 matrix \( \left[ \begin{array}{cc} a & b \ c & d \end{array} \right] \) is \( ad - bc \). For the matrix \( \left[ \begin{array}{rr} 1 & 2 \ 2 & -3 \end{array} \right] \), the determinant is: \( (1)(-3) - (2)(2) = -3 - 4 = -7 \).

Step 3: Determine Invertibility

A matrix has an inverse only if its determinant is non-zero. Since the determinant here is \(-7\), which is not zero, the matrix is invertible.

Step 4: Calculate the Inverse

For a 2x2 matrix \( \left[ \begin{array}{cc} a & b \ c & d \end{array} \right] \), the inverse is given by \( \frac{1}{ad-bc} \left[ \begin{array}{cc} d & -b \ -c & a \end{array} \right] \). Using the given matrix \( \left[ \begin{array}{rr} 1 & 2 \ 2 & -3 \end{array} \right] \), the inverse is: \[ \frac{1}{-7} \left[ \begin{array}{cc} -3 & -2 \ -2 & 1 \end{array} \right] = \left[ \begin{array}{cc} \frac{3}{7} & \frac{2}{7} \ \frac{2}{7} & -\frac{1}{7} \end{array} \right] \].

Key Concepts

Determinant Calculation

To find if a matrix has an inverse, the first step is to calculate its determinant. The determinant is a special number that is computed from the elements of a square matrix. For a 2x2 matrix \( \left[ \begin{array}{cc} a & b \ c & d \end{array} \right] \), the determinant is found using the formula: \( ad - bc \).
In this example, we have the matrix \( \left[ \begin{array}{rr} 1 & 2 \ 2 & -3 \end{array} \right] \). Calculating the determinant gives us:

• Multiply 1 (first element, top left) by -3 (second diagonal element, bottom right).
• Subtract the result of multiplying 2 (top right) by 2 (bottom left).

So, the determinant is \( (1)(-3) - (2)(2) = -3 - 4 = -7 \).
The determinant is non-zero, which means we can proceed to the next steps of finding the inverse.

Invertibility of Matrices

Not all matrices have an inverse. A key condition for a matrix to be invertible is that its determinant is not zero. If the determinant equals zero, the matrix is singular and cannot have an inverse.
In our case, the determinant of the matrix \( \left[ \begin{array}{rr} 1 & 2 \ 2 & -3 \end{array} \right] \) is \(-7\), which is clearly non-zero. This confirms that the matrix is invertible. Understanding invertibility is crucial for solving systems of linear equations, as inverse matrices often provide insight into finding solutions.

2x2 Matrices

When dealing with matrices, especially 2x2 ones, it's important to remember that they represent systems with two variables. The most common operations on these matrices include:

• Addition and subtraction of matrices.
• Multiplication of matrices.
• Determining the determinant.
• Finding the inverse if it exists.

The process for calculating the inverse of a 2x2 matrix is more straightforward compared to larger matrices. Once we know the matrix is invertible, we use the determinant calculated in the previous steps to find the inverse.

Identity Matrix

An identity matrix, often denoted as \( I \), is a square matrix with ones on the diagonal and zeros elsewhere. For a 2x2 matrix, the identity matrix looks like this: \[ \left[ \begin{array}{cc} 1 & 0 \ 0 & 1 \end{array} \right]\]
The identity matrix plays a crucial role in the concept of matrix inverses. If \( A \) is a matrix and \( A^{-1} \) is its inverse, then the product \( AA^{-1} \) must equal the identity matrix of the same dimensions. This property is essential because it essentially "undoes" the matrix \( A \) and yields a neutral result, similar to multiplying a number by one in arithmetic.