Problem 4 - Precalculus Mathematics for Calculus

Question

Calculate the products \(A B\) and \(B A\) to verify that \(B\) is the inverse of \(A\). $$A=\left[\begin{array}{ll} 2 & -3 \\ 4 & -7 \end{array}\right] \quad B=\left[\begin{array}{ll} \frac{7}{2} & -\frac{3}{2} \\ 2 & -1 \end{array}\right]$$

Step-by-Step Solution

Answer

Both products AB and BA equal the identity matrix, so B is the inverse of A.

Step 1: Understanding the Problem

To verify that matrix \(B\) is the inverse of matrix \(A\), we need to calculate the products \(AB\) and \(BA\). If both results yield the identity matrix \(I\), then \(B\) is indeed the inverse of \(A\).

Step 2: Calculate the Product AB

Compute \(AB\) using matrix multiplication, where each element of the resulting matrix is the dot product of the corresponding row from matrix \(A\) and column from matrix \(B\). \[AB = \begin{bmatrix}2 & -3 4 & -7\end{bmatrix}\begin{bmatrix}\frac{7}{2} & -\frac{3}{2} 2 & -1\end{bmatrix}\]Calculate each element:- First row, first column: \(2 \times \frac{7}{2} + (-3) \times 2 = 7 - 6 = 1\)- First row, second column: \(2 \times -\frac{3}{2} + (-3) \times (-1) = -3 + 3 = 0\)- Second row, first column: \(4 \times \frac{7}{2} + (-7) \times 2 = 14 - 14 = 0\)- Second row, second column: \(4 \times -\frac{3}{2} + (-7) \times (-1) = -6 + 7 = 1\)Thus, \(AB = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\).

Step 3: Calculate the Product BA

Compute \(BA\) using matrix multiplication:\[BA = \begin{bmatrix}\frac{7}{2} & -\frac{3}{2} 2 & -1\end{bmatrix}\begin{bmatrix}2 & -3 4 & -7\end{bmatrix}\]Calculate each element:- First row, first column: \(\frac{7}{2} \times 2 + (-\frac{3}{2}) \times 4 = 7 - 6 = 1\)- First row, second column: \(\frac{7}{2} \times -3 + (-\frac{3}{2}) \times (-7) = -10.5 + 10.5 = 0\)- Second row, first column: \(2 \times 2 + (-1) \times 4 = 4 - 4 = 0\)- Second row, second column: \(2 \times -3 + (-1) \times (-7) = -6 + 7 = 1\)Thus, \(BA = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\).

Step 4: Conclusion

Since both \(AB\) and \(BA\) result in the identity matrix \(\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\), we conclude that \(B\) is indeed the inverse of \(A\).

Key Concepts

Matrix Inverse

A matrix inverse is a concept from linear algebra that provides a way to "reverse" the effects of a matrix operation. Suppose you have a square matrix \(A\). If there exists another matrix \(B\) such that the product of \(A\) and \(B\) yields the identity matrix, then \(B\) is the inverse of \(A\). This is symbolically represented as:\[ AB = BA = I \]where \(I\) is the identity matrix.

• Finding an inverse is not always possible. Only non-singular (invertible) matrices have inverses.
• An invertible matrix has a nonzero determinant.

Matrix inversion is useful in solving systems of linear equations, among other applications. Instead of performing element-by-element arithmetic to find \(B\), various methods such as row reduction or matrix algebraic formulas can be utilized.

Identity Matrix

An identity matrix acts like the number 1 in matrix algebra. When used in multiplication, it leaves the other matrix it multiplies unchanged. The identity matrix is a square matrix, symbolized by \(I\), with 1s along its diagonal from top left to bottom right and 0s elsewhere. For a 2x2 identity matrix, it looks like:\[I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\]

• Matrix multiplication involving an identity matrix does not alter the original matrix: \(AI = IA = A\).
• It serves as the multiplicative identity in matrix algebra, similar to how multiplying any number by 1 leaves it unchanged.

The identity matrix is crucial in verifying the inverse. When you multiply a matrix by its inverse, you should get the identity matrix. In this exercise, confirming that both products \(AB\) and \(BA\) equal the identity matrix signals that \(B\) is the inverse of \(A\).

Matrix Algebra

Matrix algebra simplifies the manipulation and transformation of matrices, following defined operations and rules similar to those for numbers. Some fundamental operations in matrix algebra include addition, subtraction, scalar multiplication, and matrix multiplication.

• Matrix Multiplication: Involves multiplying rows by columns across two matrices to produce a new matrix.
• Associative Property: Matrix multiplication is associative, meaning \((AB)C = A(BC)\).
• Non-Commutative Property: Unlike numbers, matrices generally do not commute under multiplication, i.e., \(AB eq BA\).

A specialized operation in matrix algebra is finding a matrix's inverse. This operation is not always possible and requires that the matrix be square and have a non-zero determinant. These concepts are essential for understanding systems of equations, transformations, and other advanced topics in linear algebra.