Problem 6 - College Algebra with Corequisite Support

Question

In the following exercises, show that matrix \(A\) is the inverse of matrix \(B\). $$ A=\left[\begin{array}{cc} 1 & 0 \\ -1 & 1 \end{array}\right], \quad B=\left[\begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array}\right] $$

Step-by-Step Solution

Answer

Matrix \(A\) is the inverse of matrix \(B\) because \(AB = I\) and \(BA = I\).

Step 1: Understand the Problem

To prove that matrix \(A\) is the inverse of matrix \(B\), we need to show that the product \(AB = I\) and \(BA = I\), where \(I\) is the identity matrix of the same size as \(A\) and \(B\).

Step 2: Calculate AB

Multiply matrices \(A\) and \(B\): \[ A \times B = \left[ \begin{array}{cc} 1 & 0 \ -1 & 1 \end{array} \right] \times \left[ \begin{array}{cc} 1 & 0 \ 1 & 1 \end{array} \right] = \left[ \begin{array}{cc} (1 \times 1 + 0 \times 1) & (1 \times 0 + 0 \times 1) \ (-1 \times 1 + 1 \times 1) & (-1 \times 0 + 1 \times 1) \end{array} \right] = \left[ \begin{array}{cc} 1 & 0 \ 0 & 1 \end{array} \right] \]This is the identity matrix \(I\).

Step 3: Calculate BA

Multiply matrices \(B\) and \(A\): \[ B \times A = \left[ \begin{array}{cc} 1 & 0 \ 1 & 1 \end{array} \right] \times \left[ \begin{array}{cc} 1 & 0 \ -1 & 1 \end{array} \right] = \left[ \begin{array}{cc} (1 \times 1 + 0 \times (-1)) & (1 \times 0 + 0 \times 1) \ (1 \times 1 + 1 \times (-1)) & (1 \times 0 + 1 \times 1) \end{array} \right] = \left[ \begin{array}{cc} 1 & 0 \ 0 & 1 \end{array} \right] \]This is again the identity matrix \(I\).

Step 4: Conclusion

Since we have shown both \(AB = I\) and \(BA = I\), it confirms that \(A\) is indeed the inverse of \(B\). Therefore, matrix \(A\) is the inverse of matrix \(B\).

Key Concepts

Matrix Multiplication

Matrix multiplication is a fundamental operation in linear algebra. It involves multiplying two matrices to produce another matrix. The process of matrix multiplication is not as straightforward as regular multiplication. For two matrices to be multiplied, the number of columns in the first matrix must equal the number of rows in the second. This rule ensures compatibility between the matrices.

When multiplying matrices, each element of the resulting matrix is determined by a specific dot product. If you have a matrix \(A\) with dimensions \(m \times n\) and a matrix \(B\) with dimensions \(n \times p\), the resulting matrix \(C = A \times B\) will have dimensions \(m \times p\).

- To find each entry of the resulting matrix \(C\):
- Consider the row from matrix \(A\) and the column from matrix \(B\).
- Multiply corresponding elements from the row and column.
- Sum these products to obtain a single entry in matrix \(C\).

This process is repeated for each row of \(A\) and each column of \(B\), resulting in a complete matrix \(C\). In our original exercise, matrix \(A\) and matrix \(B\) were multiplied to check if their product is the identity matrix, hence proving the inverse relationship.

Identity Matrix

The identity matrix is a special type of matrix in linear algebra that acts like the number 1 in regular multiplication. For any square matrix \(A\), multiplying with the identity matrix will yield matrix \(A\) itself. The identity matrix is usually denoted by \(I\).

An identity matrix is essentially a square matrix having ones on the diagonal (from top left to bottom right) and zeros in all other positions. For example, a 2x2 identity matrix looks like this:
\[ I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \]

Identity matrices are crucial when discussing matrix inverses. Inverse matrices are defined such that when a matrix \(A\) is multiplied by its inverse \(B\), the result is the identity matrix \(I\). This property was a key part of the original exercise: showing \(AB = I\) and \(BA = I\) proved that \(A\) and \(B\) are inverses of each other.

Remember, identity matrices are pivotal in various applications, including solving linear equations, transforming data, and more.

Algebra Problems

Matrix algebra is a powerful tool in mathematics that extends basic algebraic concepts to matrices. Solving algebra problems involving matrices helps in understanding transformations, system of equations, and more.

When dealing with matrix algebra, some typical operations include addition, multiplication, and finding inverses. These are used to solve systems of linear equations, which can be represented in a matrix form as \(AX = B\), where \(A\) and \(B\) are known matrices, and \(X\) is a matrix of variables. The goal is to find \(X\).

The inverse of a matrix, if it exists, provides a straightforward way to solve these equations. By multiplying both sides of the equation \(AX = B\) by the inverse of \(A\) (notated as \(A^{-1}\)), you can isolate \(X\):
\[ A^{-1}AX = A^{-1}B \]
\[ IX = A^{-1}B \]
\[ X = A^{-1}B \]

The solution \(X\) can now be calculated, assuming the inverse \(A^{-1}\) exists. Understanding these operations can simplify complex algebra problems significantly.