Problem 11 - Algebra 2

Question

Determine whether each pair of matrices are inverses of each other. $$ A=\left[\begin{array}{ll}{6} & {2} \\ {5} & {2}\end{array}\right], B=\left[\begin{array}{rr}{1} & {1} \\ {-\frac{5}{2}} & {-3}\end{array}\right] $$

Step-by-Step Solution

Answer

Matrices A and B are not inverses.

Step 1: Understand Matrix Inversion

Two matrices \( A \) and \( B \) are inverses of each other if their product results in the identity matrix \( I \). For a 2x2 matrix, the identity matrix is \( I=\left[ \begin{array}{cc} 1 & 0 \ 0 & 1 \end{array} \right] \). We will calculate both \( A \cdot B \) and \( B \cdot A \) to determine if the products are equal to the identity matrix.

Step 2: Multiply A and B

Calculate the product \( AB = \left[ \begin{array}{cc} a_{11} & a_{12} \ a_{21} & a_{22} \end{array} \right] \times \left[ \begin{array}{cc} b_{11} & b_{12} \ b_{21} & b_{22} \end{array} \right]\):\(AB = \left[ \begin{array}{ll}{6} & {2} \ {5} & {2}\end{array}\right] \cdot \left[\begin{array}{rr}{1} & {1} \ {-rac{5}{2}} & {-3}\end{array}\right] = \left[ \begin{array}{cc} (6 \times 1 + 2 \times -\frac{5}{2}) & (6 \times 1 + 2 \times -3) \ (5 \times 1 + 2 \times -\frac{5}{2}) & (5 \times 1 + 2 \times -3) \end{array} \right] \)\.

Step 3: Simplify the Result

Simplify the computation:1. First element \( 6 \times 1 + 2 \times -\frac{5}{2} = 6 - 5 = 1 \).2. Second element \( 6 \times 1 + 2 \times -3 = 6 - 6 = 0 \).3. Third element \( 5 \times 1 + 2 \times -\frac{5}{2} = 5 - 5 = 0 \).4. Fourth element \( 5 \times 1 + 2 \times -3 = 5 - 6 = -1 \).Thus, \( AB = \left[ \begin{array}{cc} 1 & 0 \ 0 & -1 \end{array} \right] \), which is not the identity matrix.

Step 4: Multiply B and A

Now calculate the product \( BA = \left[\begin{array}{rr}{1} & {1} \ {-rac{5}{2}} & {-3}\end{array}\right] \cdot \left[ \begin{array}{ll}{6} & {2} \ {5} & {2}\end{array}\right] \):\( BA = \left[ \begin{array}{cc} (1 \times 6 + 1 \times 5) & (1 \times 2 + 1 \times 2) \ (-\frac{5}{2} \times 6 + (-3) \times 5) & (-\frac{5}{2} \times 2 + (-3) \times 2) \end{array} \right] \)\.

Step 5: Simplify BA Result

Simplify the computation:1. First element \( 1 \times 6 + 1 \times 5 = 6 + 5 = 11 \).2. Second element \( 1 \times 2 + 1 \times 2 = 2 + 2 = 4 \).3. Third element \( -\frac{5}{2} \times 6 + (-3) \times 5 = -15 - 15 = -30 \).4. Fourth element \( -\frac{5}{2} \times 2 + (-3) \times 2 = -5 - 6 = -11 \).Thus, \( BA = \left[ \begin{array}{cc} 11 & 4 \ -30 & -11 \end{array} \right] \), which is also not the identity matrix.

Step 6: Conclusion

Since neither \( AB \) nor \( BA \) equals the identity matrix \( \left[ \begin{array}{cc} 1 & 0 \ 0 & 1 \end{array} \right] \), matrices \( A \) and \( B \) are not inverses of each other.

Key Concepts

Identity Matrix

The identity matrix plays a crucial role in understanding matrix inverses. In a matrix context, the identity matrix acts like the number 1 in regular arithmetic. For any given matrix, when you multiply it by its inverse, you should end up with the identity matrix. For 2x2 matrices, the identity matrix is represented as follows: \[ I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \]
The identity matrix is special because:

• It has 1s on the diagonal from the top left to the bottom right.
• All other entries are 0.
• Multiplying any 2x2 matrix by the identity matrix will return the original matrix itself.

To determine if two matrices are inverses of each other, their product must be the identity matrix. If you perform the multiplication and get the identity matrix, then congratulations, you've found inverses! But remember, both orders of multiplication (i.e., \( A \times B \) and \( B \times A \)) must be checked when testing for inverses.

Matrix Multiplication

Matrix multiplication can seem a bit daunting at first, but breaking it down step-by-step can help. When you multiply two matrices, you're essentially performing several smaller calculations to fill in the elements of a resulting matrix. Each element of the resulting matrix is calculated through a combination of elements from the rows of the first matrix and the columns of the second matrix. For the matrices: \[ A = \begin{bmatrix} 6 & 2 \ 5 & 2 \end{bmatrix}, \ B = \begin{bmatrix} 1 & 1 \ -\frac{5}{2} & -3 \end{bmatrix} \]
Let's calculate \( AB \):

• First element: Multiply the first row of \( A \) \([6, 2]\) by the first column of \( B \) \([1, -\frac{5}{2}]\), resulting in \( 6 \times 1 + 2 \times -\frac{5}{2} = 1 \).
• Second element: Multiply the first row of \( A \) \([6, 2]\) by the second column of \( B \) \([1, -3]\), resulting in \( 6 \times 1 + 2 \times -3 = 0 \).
• Third element: Multiply the second row of \( A \) \([5, 2]\) by the first column of \( B \) \([1, -\frac{5}{2}]\), resulting in \( 5 \times 1 + 2 \times -\frac{5}{2} = 0 \).
• Fourth element: Multiply the second row of \( A \) \([5, 2]\) by the second column of \( B \) \([1, -3]\), resulting in \( 5 \times 1 + 2 \times -3 = -1 \).

So, \( AB = \begin{bmatrix} 1 & 0 \ 0 & -1 \end{bmatrix} \). Note how this detailed method is applied element by element to the matrices.

2x2 Matrices

2x2 matrices are among the simplest matrices you'll encounter in linear algebra. Each matrix is composed of 4 entries arranged in 2 rows and 2 columns. This simplicity makes them an excellent starting point for learning more complex matrix operations. Key things to remember when working with 2x2 matrices:

• They're represented as: \[ \begin{bmatrix} a & b \ c & d \end{bmatrix} \]
• Operations on 2x2 matrices, such as addition and multiplication, involve element-by-element computations.
• The determinant of a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is \( ad - bc \). This quantity is crucial when determining whether a matrix has an inverse.
• A 2x2 matrix only has an inverse if its determinant is not zero. The inverse is given by \[ \begin{bmatrix} \frac{d}{ad-bc} & \frac{-b}{ad-bc} \ \frac{-c}{ad-bc} & \frac{a}{ad-bc} \end{bmatrix} \].

Understanding these basic properties and operations not only helps in calculating matrix inverses but also builds your foundation for dealing with larger matrices and more complex systems.