Problem 31 - Algebra 2

Question

Determine whether each pair of matrices are inverses of each other. $$ J=\left[\begin{array}{lll}{1} & {2} & {3} \\ {2} & {3} & {1} \\ {1} & {1} & {2}\end{array}\right], K=\left[\begin{array}{rrr}{-\frac{5}{4}} & {\frac{1}{4}} & {\frac{7}{4}} \\ {\frac{3}{4}} & {\frac{1}{4}} & {-\frac{5}{4}} \\ {\frac{1}{4}} & {-\frac{1}{4}} & {\frac{1}{4}}\end{array}\right] $$

Step-by-Step Solution

Answer

Yes, matrices J and K are inverses of each other.

Step 1: Understanding Inverses

Two matrices, \( A \) and \( B \), are inverses of each other if their product results in the identity matrix \( I \). For a 3x3 matrix, \( I \) is \( \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} \).

Step 2: Multiply the Matrices

Multiply matrix \( J \) by matrix \( K \):\[J \times K = \begin{bmatrix} 1 & 2 & 3 \ 2 & 3 & 1 \ 1 & 1 & 2 \end{bmatrix} \times \begin{bmatrix} -\frac{5}{4} & \frac{1}{4} & \frac{7}{4} \ \frac{3}{4} & \frac{1}{4} & -\frac{5}{4} \ \frac{1}{4} & -\frac{1}{4} & \frac{1}{4} \end{bmatrix}\]

Step 3: Calculate Product Terms

Compute each element of the product matrix:- First row, first column: \((1)(-\frac{5}{4}) + (2)(\frac{3}{4}) + (3)(\frac{1}{4}) = -\frac{5}{4} + \frac{6}{4} + \frac{3}{4} = 1\)- First row, second column: \((1)(\frac{1}{4}) + (2)(\frac{1}{4}) + (3)(-\frac{1}{4}) = \frac{1}{4} + \frac{2}{4} - \frac{3}{4} = 0\)- First row, third column: \((1)(\frac{7}{4}) + (2)(-\frac{5}{4}) + (3)(\frac{1}{4}) = \frac{7}{4} - \frac{10}{4} + \frac{3}{4} = 0\)Repeat this process for the other rows and columns as well.

Step 4: Resulting Product Matrix

The resulting product matrix \( J \times K \) is:\[\begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}\]This is the identity matrix \( I \), indicating that \( J \) and \( K \) are inverses of each other.

Key Concepts

Matrix Multiplication

Matrix multiplication is a process that involves the dot product of rows and columns from two matrices to form a new matrix. It follows specific steps due to rules defined by linear algebra.
For matrix multiplication to be possible, the number of columns in the first matrix must equal the number of rows in the second matrix.

• When multiplying a 3x3 matrix, each element in the resulting matrix is the sum of products of corresponding elements from rows and columns.
• For instance, to find the element in the first row and first column of the product, calculate: \( (1) \times (-\frac{5}{4}) + (2) \times (\frac{3}{4}) + (3) \times (\frac{1}{4}) \), which simplifies to 1.

It’s essential to carefully perform these calculations to avoid errors, as small mistakes can lead to incorrect results.

Identity Matrix

An identity matrix is a special kind of matrix that acts as a multiplicative identity in matrix multiplication. This means when any matrix is multiplied by an identity matrix of the same size, it retains its original values.
The identity matrix for a 3x3 matrix is given by:

• It has ones on the diagonal from the top left to the bottom right and zeros elsewhere: e.g., \[ \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}. \]
• Its role is to verify if two matrices are inverses: if their product is an identity matrix, they are inverses.

In this example, after calculating the product of matrices \( J \) and \( K \), the resulting identity matrix confirmed they are inverses.

3x3 Matrix

A 3x3 matrix is a square matrix containing three rows and three columns. Each element is typically a real number and is positioned in a row and column format.
Working with 3x3 matrices involves understanding several matrix operations, such as addition, multiplication, and finding inverses.

• To determine if two 3x3 matrices are inverses, their product must yield the 3x3 identity matrix.
• Each position in a 3x3 matrix impacts the overall behavior and attributes of that matrix significantly.

Mastering operations with 3x3 matrices is fundamental in linear algebra, as it builds the grounds for handling more complex matrices used in advanced mathematical computations and applications.