Problem 1 - Applied Discrete Structures

Question

For the given matrices \(A\) find \(A^{-1}\) if it exists and verify that \(A A^{-1}=\) \(A^{-1} A=I .\) If \(A^{-1}\) does not exist explain why. (a) \(A=\left(\begin{array}{ll}1 & 3 \\ 2 & 1\end{array}\right)\) (b) \(A=\left(\begin{array}{ll}6 & -3 \\ 8 & -4\end{array}\right)\) (c) \(A=\left(\begin{array}{cc}1 & -3 \\ 0 & 1\end{array}\right)\) (d) \(A=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)\) (e) Use the definition of the inverse of a matrix to find \(A^{-1}: A=\) $$ \left(\begin{array}{ccc} 3 & 0 & 0 \\ 0 & \frac{1}{2} & 0 \\ 0 & 0 & -5 \end{array}\right) $$

Step-by-Step Solution

Answer

Solutions: (a) Invertible, inverse exists; (b) Not invertible; (c) Invertible, inverse found; (d) Identity matrix, self-inverse; (e) Invertible, inverse calculated.

Step 1: Calculate Determinant of Matrix A (a)

Given matrix \( A = \begin{pmatrix} 1 & 3 \ 2 & 1 \end{pmatrix} \). Calculate its determinant using the formula for a 2x2 matrix, \( \text{det}(A) = ad - bc \). Here, \( a = 1 \), \( b = 3 \), \( c = 2 \), and \( d = 1 \), thus \( \text{det}(A) = 1 \times 1 - 3 \times 2 = 1 - 6 = -5 \).

Step 2: Find Inverse of Matrix A (a)

Since the determinant is non-zero \( (-5) \), an inverse exists. For a 2x2 matrix \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \), the inverse is given by \( A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} d & -b \ -c & a \end{pmatrix} \). Thus, \( A^{-1} = \frac{1}{-5} \begin{pmatrix} 1 & -3 \ -2 & 1 \end{pmatrix} = \begin{pmatrix} -0.2 & 0.6 \ 0.4 & -0.2 \end{pmatrix} \).

Step 3: Verify AA^{-1} = A^{-1}A = I for (a)

Multiply \( A \) and \( A^{-1} \): \( AA^{-1} = \begin{pmatrix} 1 & 3 \ 2 & 1 \end{pmatrix} \begin{pmatrix} -0.2 & 0.6 \ 0.4 & -0.2 \end{pmatrix} = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} \). Do the same for \( A^{-1}A \) and it should result in the identity matrix \( I \). Both yield \( I \), verifying correctness.

Step 4: Calculate Determinant of Matrix A (b)

For \( A = \begin{pmatrix} 6 & -3 \ 8 & -4 \end{pmatrix} \), find \( \text{det}(A) = 6(-4) - (-3)(8) = -24 + 24 = 0 \).

Step 5: Determine Invertibility of Matrix A (b)

Since \( \text{det}(A) = 0 \), matrix \( A \) is not invertible.

Step 6: Calculate Determinant of Matrix A (c)

For \( A = \begin{pmatrix} 1 & -3 \ 0 & 1 \end{pmatrix} \), calculate \( \text{det}(A) = 1 \times 1 - (-3) \times 0 = 1 \).

Step 7: Find Inverse of Matrix A (c)

With a non-zero determinant \( (1) \), the inverse exists: \( A^{-1} = \begin{pmatrix} 1 & 3 \ 0 & 1 \end{pmatrix} \).

Step 8: Verify AA^{-1}=A^{-1}A=I for (c)

Multiply \( AA^{-1} = \begin{pmatrix} 1 & -3 \ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 3 \ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \) confirms the inverse.

Step 9: Identity Matrix Inverse (d)

The matrix \( A = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \) is already the identity matrix \( I \) so its inverse \( A^{-1} \) is itself.

Step 10: Verify AA^{-1}=A^{-1}A=I for (d)

Since the identity matrix is involved, multiplying yields \( I \) in both cases, confirming the inverse.

Step 11: Inverse of Diagonal Matrix (e)

Given the diagonal matrix \( A = \begin{pmatrix} 3 & 0 & 0 \ 0 & \frac{1}{2} & 0 \ 0 & 0 & -5 \end{pmatrix} \), the inverse is the reciprocal of each diagonal element: \( A^{-1} = \begin{pmatrix} \frac{1}{3} & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & -\frac{1}{5} \end{pmatrix} \).

Step 12: Verify AA^{-1}=A^{-1}A=I for (e)

Multiplying \( AA^{-1} \) and \( A^{-1}A \) yields the identity, confirming the inverse is correct.

Key Concepts

Determinant of a Matrix

The determinant is a special number that can be calculated from a square matrix. It's very important because it tells us a lot about the matrix itself. Simply put, the determinant can help in understanding whether the matrix has an inverse, and therefore if the matrix is invertible. For a 2x2 matrix, the formula for calculating the determinant is quite straightforward. Given a matrix \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \), the determinant is calculated as \( \text{det}(A) = ad - bc \). This simple arithmetic operation reveals an important property: if the determinant is zero, the matrix is singular, meaning it does not have an inverse.

Identity Matrix

The identity matrix acts as the 'do-nothing' element in matrix multiplication, much like the number 1 does in regular arithmetic. It's a square matrix with ones along the diagonal and zeros everywhere else. For example, in the 2x2 case, it looks like this: \( I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \). When you multiply any matrix by the identity matrix, the result is the original matrix. This is a crucial concept because when we find an inverse of a matrix \( A \), we are essentially looking for a matrix \( A^{-1} \) that satisfies \( AA^{-1} = A^{-1}A = I \), where \( I \) is the identity matrix.

Diagonal Matrix

A diagonal matrix is a type of matrix where all the entries outside the main diagonal are zero. These matrices are very handy in linear algebra because they are easy to work with, especially when finding the inverse. If the diagonal entries are non-zero, then the inverse of a diagonal matrix is just the reciprocals of these diagonal elements. For instance, if you have a diagonal matrix \( D = \begin{pmatrix} a & 0 & 0 \ 0 & b & 0 \ 0 & 0 & c \end{pmatrix} \), its inverse \( D^{-1} \) is \( \begin{pmatrix} \frac{1}{a} & 0 & 0 \ 0 & \frac{1}{b} & 0 \ 0 & 0 & \frac{1}{c} \end{pmatrix} \). This property greatly simplifies many calculations when dealing with diagonal matrices.

2x2 Matrix Inversion

The process of finding an inverse for a 2x2 matrix is quite mechanical when the determinant is non-zero. For a matrix \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \), the inverse is given by swapping the elements on the main diagonal, changing the signs of the off-diagonal elements, and dividing by the determinant. Thus, \( A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix} \). This formula works only if \( ad - bc eq 0 \), as a zero value would make the division undefined. Knowing this method for 2x2 matrices provides a foundation for understanding larger matrices.

Invertibility of Matrices

A matrix is described as invertible or non-singular if there exists another matrix, called the inverse, which results in the identity matrix when multiplied with the original matrix. Essentially, for a matrix \( A \), if there exists a matrix \( A^{-1} \) such that \( AA^{-1} = A^{-1}A = I \), then \( A \) is invertible. The invertibility of matrices hinges crucially on the determinant. If the determinant is zero, the matrix will not have an inverse. However, if the determinant is non-zero, the matrix has an inverse. This condition is a key criterion in linear algebra computations and applications, making it a vital aspect in diverse areas such as solving systems of linear equations.