Problem 45 - Discrete Mathematics with Applications

Question

A square matrix \(A\) of order \(n\) is invertible if there is a matrix \(B\) such that \(A B=I_{n}=B A .\) Then \(B\) is the inverse of \(A,\) denoted by \(A^{-1} .\) In Exercises 44 and \(45,\) verify that \(B=A^{-1} .\) Assume that \(k=a d-b c \neq 0\). $$A=\left[\begin{array}{rrr} 1 & -2 & 0 \\ 3 & 1 & -1 \\ 1 & 2 & -3 \end{array}\right], B=\frac{1}{17}\left[\begin{array}{rrr} 1 & 6 & -2 \\ -8 & 3 & -1 \\ -5 & 4 & -7 \end{array}\right]$$

Step-by-Step Solution

Answer

We are given matrices A and B, and we need to verify if B is the inverse of A. We multiply A and B in both orders and obtain the following results: Matrix AB: \(AB = \left[\begin{array}{rrr}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1} \end{array}\right]\) Matrix BA: \(BA = \left[\begin{array}{rrr}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1} \end{array}\right]\) Since both products are equal to the identity matrix, we can conclude that B is indeed the inverse of A, or \(B=A^{-1}\).

Step 1: Write down the given matrices A and B

Matrix A: \( A=\left[\begin{array}{rrr}{1} & {-2} & {0} \\ {3} & {1} & {-1} \\ {1} & {2} & {-3} \end{array}\right] \) Matrix B: \( B=\frac{1}{17}\left[\begin{array}{rrr}{1} & {6} & {-2} \\ {-8} & {3} & {-1} \\ {-5} & {4} & {-7} \end{array}\right] \)

Step 2: Perform matrix multiplication AB and BA

Matrix AB: \( AB = \left[\begin{array}{rrr}{1} & {-2} & {0} \\ {3} & {1} & {-1} \\ {1} & {2} & {-3} \end{array}\right]\frac{1}{17}\left[\begin{array}{rrr}{1} & {6} & {-2} \\ {-8} & {3} & {-1} \\ {-5} & {4} & {-7} \end{array}\right] \) Matrix BA: \( BA = \frac{1}{17}\left[\begin{array}{rrr}{1} & {6} & {-2} \\ {-8} & {3} & {-1} \\ {-5} & {4} & {-7} \end{array}\right]\left[\begin{array}{rrr}{1} & {-2} & {0} \\ {3} & {1} & {-1} \\ {1} & {2} & {-3} \end{array}\right] \)

Step 3: Compute the products AB and BA

Matrix AB: \( AB = \frac{1}{17}\left[\begin{array}{rrr}{17} & {0} & {0} \\ {0} & {17} & {0} \\ {0} & {0} & {17} \end{array}\right] = \left[\begin{array}{rrr}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1} \end{array}\right] \) Matrix BA: \( BA = \frac{1}{17}\left[\begin{array}{rrr}{17} & {0} & {0} \\ {0} & {17} & {0} \\ {0} & {0} & {17} \end{array}\right] = \left[\begin{array}{rrr}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1} \end{array}\right] \)

Step 4: Compare AB and BA to the identity matrix

We can see that the matrices AB and BA are equal to the identity matrix I: \( I = \left[\begin{array}{rrr}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1} \end{array}\right] \) Since \(AB=BA=I_n\), we can conclude that matrix B is indeed the inverse of matrix A, or \(B=A^{-1}\).

Key Concepts

Matrix Multiplication

Matrix multiplication is a fundamental concept in linear algebra, allowing us to combine two matrices to form a new matrix. The result is determined by taking the dot product of the rows of the first matrix with the columns of the second matrix. For two matrices, say \( C \) (of size \( m \times n \)) and \( D \) (of size \( n \times p \)), the resulting product \( CD \) is a matrix of size \( m \times p \). Each element \( (i, j) \) of the product matrix is calculated as follows:

• Multiply each element of the ith row of matrix \( C \) with the corresponding element of the jth column of matrix \( D \).
• Sum all these products to get the element \( (i, j) \) for the product matrix.

Matrix multiplication is not commutative, meaning \( CD eq DC \) unless \( C \) and \( D \) are special matrices like identity matrices.

Identity Matrix

An identity matrix, denoted as \( I_n \), is a special kind of square matrix that acts like the number \( 1 \) in matrix multiplication. It is composed of all zeros except for the diagonals that contain ones. For any square matrix \( A \) of order \( n \), multiplying it by the identity matrix results in the matrix itself:

• \( AI_n = I_nA = A \)

This property makes the identity matrix a vital component in defining matrix inverses, which are crucial when solving systems of linear equations.
When you multiply a matrix by an appropriate order identity matrix, it remains unchanged, showing the essence of identities in calculations.

Invertible Matrices

Invertible matrices, also known as non-singular or non-degenerate matrices, have a special property that they can "reverse" the operation performed by themselves when multiplied by their inverse. A square matrix \( A \) is invertible if there exists another matrix \( B \) such that:

• \( AB = BA = I_n \), where \( I_n \) is the identity matrix of order \( n \).

The matrix \( B \) is referred to as the inverse of \( A \) and is denoted as \( A^{-1} \).
Not all matrices are invertible. For a matrix to be invertible, its determinant must be non-zero, and the inverse matrix is unique for any given invertible matrix. In practical applications, computing the inverse allows the solution of linear systems by assisting in isolating variables.

Determinant of a Matrix

The determinant is a crucial scalar value associated with square matrices, providing important information about the matrix properties, especially regarding invertibility. For a matrix \( A \), the determinant is denoted as \( \det(A) \) or \( |A| \). A non-zero determinant indicates the matrix is invertible, while a determinant of zero implies the matrix is singular and not invertible.
The calculation of the determinant involves specific operations over the matrix elements, which, for a 3x3 matrix \( \left[\begin{array}{ccc} a & b & c \ d & e & f \ g & h & i \end{array}\right] \), is computed as:

• \( \det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \)

This calculated value affects the solutions of linear systems and provides insights into matrix transformations, indicating, for example, if volume preservation occurs during transformations.