Problem 3 - College Algebra and Calculus: An Applied Approach

Question

Show that \(B\) is the inverse of \(A\). \(A=\left[\begin{array}{ll}2 & -1 \\ 5 & -4\end{array}\right], B=\left[\begin{array}{ll}\frac{4}{3} & -\frac{1}{3} \\ \frac{5}{3} & -\frac{2}{3}\end{array}\right]\)

Step-by-Step Solution

Answer

Yes, the matrix \(B\) is the inverse of matrix \(A\), as both \( A \cdot B \) and \( B \cdot A \) produce the Identity Matrix.

Step 1: Multiply A and B

The first step is to multiply the matrices \(A\) and \(B\) together in both possible orders, i.e., \(A \cdot B\) and \(B \cdot A\). This will give us two resultant matrices.

Step 2: Check if resultant is Identity Matrix

Examine the resultant matrices from Step 1. If each resultant matrix is the Identity Matrix, then you have confirmed that \(B\) is in fact the inverse of \(A\). The Identity Matrix is a square matrix in which all the elements of the principal (main) diagonal are ones and all other elements are zeros.

Step 3: Calculation of A.B and B.A

The multiplication of matrix A and B: \(A \cdot B = \left[\begin{array}{ll}2 & -1 \ 5 & -4\end{array}\right]\cdot \left[\begin{array}{ll}4/3 & 1/3 \ 5/3 & 2/3\end{array}\right]\) will give us \(\left[\begin{array}{ll}1 & 0 \ 0 & 1\end{array}\right]\). Likewise, multiplying matrix B and A i.e, \(B \cdot A = \left[\begin{array}{ll}4/3 & 1/3 \ 5/3 & 2/3\end{array}\right]\cdot \left[\begin{array}{ll}2 & -1 \ 5 & -4\end{array}\right]\) will also give us \(\left[\begin{array}{ll}1 & 0 \ 0 & 1\end{array}\right]\). Therefore, both \( A \cdot B \) and \( B \cdot A \) result in the Identity Matrix, proving B to be the inverse of A.

Key Concepts

Identity Matrix

The identity matrix is a fundamental concept in linear algebra. It's like the number 1 in regular multiplication. When you multiply any matrix by an identity matrix, the result is the original matrix itself.

An identity matrix is always a square matrix (same number of rows and columns), and it looks like this for a 2x2 matrix:

• Main diagonal: All elements are 1s
• Off-diagonals: All elements are 0s

For example, the 2x2 identity matrix looks like this:\[\begin{bmatrix}1 & 0 \0 & 1\end{bmatrix}\]Understanding identity matrices helps in confirming inverse matrices since multiplying a matrix by its inverse results in the identity matrix.

Matrix Multiplication

Matrix multiplication is not as straightforward as multiplying regular numbers. Instead, it involves a dot product between rows and columns.

To multiply two matrices, you follow these steps:

• Take the row from the first matrix.
• Multiply it element-wise with the column in the second matrix.
• Summing the results will give you one element in the resulting matrix.

For example, if we have two matrices \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \) and \( B = \begin{bmatrix} e & f \ g & h \end{bmatrix} \), matrix multiplication gives us:\[AB = \begin{bmatrix} ae + bg & af + bh \ ce + dg & cf + dh \end{bmatrix}\]Matrix multiplication is essential for operations like finding inverse matrices, where the order of multiplication matters.

Inverse Matrices

Inverse matrices are like a key that unlocks a matrix's identity. If you multiply a matrix by its inverse, you will get the identity matrix.

Finding the inverse is a bit complex and typically involves:

• Checking if a matrix is invertible by ensuring its determinant is not zero.
• Using formulas or methods, such as the adjugate matrix and determinant, for calculation.

For a 2x2 matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the inverse is:\[A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]However, you must remember, not every matrix has an inverse. The inverse only exists when the determinant (\( ad-bc \)) is non-zero.