Problem 50 - Precalculus

Question

determine whether \(B\) is the multiplicative inverse of \(A\) using \(A A^{-1}=I\) $$A=\left[\begin{array}{lll}-1 & 0 & -1 \\\\-1 & 1 & -2 \\\\-1 & 1 & -1\end{array}\right] \quad B=\left[\begin{array}{rrr}-1 & 1 & -1 \\\\-1 & 0 & 1 \\\0 & -1 & 1\end{array}\right]$$

Step-by-Step Solution

Answer

Matrix B is not the multiplicative inverse of A since AB ≠ I.

Step 1: Understanding the Multiplicative Inverse Concept

The multiplicative inverse of a matrix \( A \) is a matrix \( A^{-1} \) such that when \( A \) is multiplied by \( A^{-1} \), it results in the identity matrix \( I \). For a matrix to be the inverse, \( AB = I \) where \( I \) is the identity matrix of the same size as \( A \).

Step 2: Perform Matrix Multiplication

Calculate the product of matrices \( A \) and \( B \). Arrange the two given matrices as follows and utilize matrix multiplication rules:\[ A B = \begin{bmatrix} -1 & 0 & -1 \ -1 & 1 & -2 \ -1 & 1 & -1 \end{bmatrix} \begin{bmatrix} -1 & 1 & -1 \ -1 & 0 & 1 \ 0 & -1 & 1 \end{bmatrix} \]Each entry in the resulting product matrix \( C \) is given by \[ C_{ij} = \sum_{k=1}^{3} A_{ik} \times B_{kj} \].

Step 3: Calculate Each Element of the Product Matrix

Compute the elements of the product matrix \( C \):- For \( C_{11} \): \( (-1)(-1) + (0)(-1) + (-1)(0) = 1 \)- For \( C_{12} \): \( (-1)(1) + (0)(0) + (-1)(-1) = 0 \)- For \( C_{13} \): \( (-1)(-1) + (0)(1) + (-1)(1) = 0 \)- For \( C_{21} \): \( (-1)(-1) + (1)(-1) + (-2)(0) = 0 \)- For \( C_{22} \): \( (-1)(1) + (1)(0) + (-2)(-1) = 3 \)- For \( C_{23} \): \( (-1)(-1) + (1)(1) + (-2)(1) = 0 \)- For \( C_{31} \): \( (-1)(-1) + (1)(-1) + (-1)(0) = 0 \)- For \( C_{32} \): \( (-1)(1) + (1)(0) + (-1)(-1) = 0 \)- For \( C_{33} \): \( (-1)(-1) + (1)(1) + (-1)(1) = 1 \)

Step 4: Form Resulting Product Matrix

Combine the calculated elements from Step 3 into the resulting matrix:\[ C = \begin{bmatrix} 1 & 0 & 0 \ 0 & 3 & 0 \ 0 & 0 & 1 \end{bmatrix} \]

Step 5: Check Against the Identity Matrix

The identity matrix \( I \) for a 3x3 matrix is:\[ I = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} \]Compare this with the product matrix \( C \) from Step 4. Since they do not match (\( C eq I \)), \( B \) is not the inverse of \( A \).

Key Concepts

Identity Matrix

An identity matrix, often denoted as \( I \), is a special type of square matrix. In an identity matrix, all the elements of the principal diagonal are ones while all other elements are zeros.
This matrix serves as the multiplicative identity in matrix algebra, just like the number 1 does for real numbers.

Consider a 3x3 identity matrix, which looks like this:

• The main diagonal consists of 1's: \[ \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} \]
• All off-diagonal elements are 0.

When any 3x3 matrix \( A \) is multiplied by a 3x3 identity matrix, the result is matrix \( A \) itself:
\( AI = IA = A \).
This property makes the identity matrix vital when determining the multiplicative inverse of a matrix.

Multiplicative Inverse

In the context of matrices, the multiplicative inverse of a matrix \( A \) is another matrix, \( A^{-1} \), such that when multiplied together, they yield the identity matrix \( I \).
This relationship can be written mathematically as:

• \( AA^{-1} = A^{-1}A = I \)

Not all matrices have inverses. Only square matrices (e.g., 2x2, 3x3) can have an inverse, and even then, not all square matrices do.
A matrix must be non-singular, meaning it has a non-zero determinant, to have an inverse.

For the inverse to exist in the example exercise, the matrices \( A \) and \( B \) would need to multiply together to produce the identity matrix \( I \). Since they do not, \( B \) is not the inverse of \( A \). This illustrates the importance of verifying the inverse through multiplication results.

3x3 Matrices

A 3x3 matrix consists of three rows and three columns, giving it nine elements.
Each element is designated by two indices: the row number and the column number. For instance, matrix element \( a_{ij} \) is located in the \( i \)-th row and \( j \)-th column.

When you perform matrix multiplication, like in the step-by-step solution for matrices \( A \) and \( B \), calculation involves taking each element from the row of the first matrix and multiplying it with the corresponding element from the column of the second matrix. Then, sum up these products to obtain the element in the result matrix.
For instance, for the upper left corner (\( C_{11} \)) of the resulting matrix \( C \), the calculation followed is:

• Multiply corresponding elements of the first row of \( A \) by the first column of \( B \): \((-1) \cdot (-1) + 0 \cdot (-1) + (-1) \cdot 0 = 1\).

This method is repeated for each element of the resulting matrix. Understanding this process is crucial to grasp complex concepts like multiplicative inverses and the role of identity matrices in linear algebra.