Question

\(\mathrm{A}=\mid \begin{array}{lll}1 & 1 \mid \text { and } \mathrm{P}=\mid 1 & 1 \mid \\ \mid \begin{array}{ll}0 & 1 \mid\end{array} & \mid 1 & -1 \mid \text { . }\end{array}\) (a) Find \(\mathrm{P}^{-1}\). (b) Find \(\mathrm{P}^{-1} \mathrm{AP}\). (c) Verify that, if \(\mathrm{B}\) is similar to \(\mathrm{A}\) then \(\mathrm{A}\) is similar to \(\mathrm{B}\). (d) Show that \(\mathrm{B}^{\mathrm{k}}=\mathrm{P}^{-1} \mathrm{~A}^{\mathrm{k} \mathrm{P}}\) if \(\mathrm{B}=\mathrm{P}^{-1}\) AP where \(\mathrm{k}\) is any positive integer.

Short answer

(a) \(P^{-1} = \begin{pmatrix} 1 & 1 \\ 1 & -1\end{pmatrix}\) (b) \(B = \begin{pmatrix} 1 & 3 \\ 1 & 1 \end{pmatrix}\) (c) A is similar to B. (d) \(B^k = P^{-1} A^k P\) for any positive integer k.

Step-by-step solution

Find P^{-1} (matrix inverse of P)

To find the inverse of matrix P, we need to calculate its determinant and adjoint. Since it is a 2x2 matrix, we can find the inverse directly using the following formula: \[P^{-1} = \frac{1}{\text{det}(P)} \cdot \text{adj}(P)\] where the determinant (det(P)) is calculated as Ι1∙(-1) - 1∙0Ι = -1, and the adjoint (adj(P)) is obtained by swapping the diagonal elements and negating off-diagonal elements: \[\text{adj}(P) = \begin{pmatrix} -1 & -1 \\ -1 & 1 \end{pmatrix}\] Now we can find P^{-1}: \[P^{-1} = \frac{1}{-1} \cdot \begin{pmatrix} -1 & -1 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & -1\end{pmatrix}\]

Find P^{-1}AP (matrix B)

To find the matrix B, we need to compute the matrix product P^{-1}AP. The multiplication is done in the following order: First, calculate R = P^{-1}A, and then calculate B = RP. \[R = P^{-1}A = \begin{pmatrix} 1 & 1 \\ 1 & -1\end{pmatrix}\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 1 & 0 \end{pmatrix}\] Now, calculate B: \[B = RP = \begin{pmatrix} 1 & 2 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 3 \\ 1 & 1 \end{pmatrix}\]

Verify if A is similar to B

Two matrices are similar if there exists an invertible matrix P such that B = P^{-1}AP. From step 2, we found that B = P^{-1}AP, so we can conclude that matrix A is indeed similar to matrix B.

Show that B^{k} = P^{-1} A^{k} P for any positive integer k

We will use mathematical induction to prove this statement. Base case (k=1): \[B^1 = B = P^{-1}AP = P^{-1}A^1P\] Now consider the induction step. Assume the statement holds for k=n, that is: \[B^n = P^{-1}A^nP\] Now we need to show that it holds for k=n+1. We have: \[B^{n+1} = B^n \cdot B = (P^{-1}A^nP) \cdot B = (P^{-1}A^nP)(P^{-1}AP) = P^{-1}A^nPP^{-1}AP\] Since the product of a matrix and its inverse is the identity matrix (I), we can simplify the expression: \[B^{n+1} = P^{-1}A^n(AP) = P^{-1}A^nA(P) = P^{-1}(A^{n+1})P\] As the statement holds for k=n+1, we can conclude that B^k = P^{-1}A^kP for any positive integer k.

Key concepts

Matrix Inverse

Finding the inverse of a matrix, particularly a 2x2 matrix, is one of the fundamental operations in linear algebra. The inverse of a matrix \( P \), denoted \( P^{-1} \), is such that when it is multiplied with \( P \), it yields the identity matrix. For a 2x2 matrix, it's quite simple to find the inverse, provided the matrix is invertible, i.e., its determinant is not zero.

To find \( P^{-1} \), we use the formula: \[ P^{-1} = \frac{1}{\text{det}(P)} \cdot \text{adj}(P) \] First, calculate the determinant \( \text{det}(P) \) and the adjoint \( \text{adj}(P) \). For a matrix: \[ \begin{pmatrix} a & b \ c & d \end{pmatrix}, \] the determinant is given by \( ad - bc \), and the adjoint can be found by swapping the diagonal elements and negating the off-diagonal ones:

• Determinant: \( \text{det}(P) = d \cdot a - b \cdot c \)
• Adjoint: \( \text{adj}(P) = \begin{pmatrix} d & -b \ -c & a \end{pmatrix} \)

In the given problem, for \( P = \begin{pmatrix} 1 & 1 \ 0 & 1 \end{pmatrix} \), the determinant is \(-1\), and the adjoint is \( \begin{pmatrix} -1 & -1 \ -1 & 1 \end{pmatrix} \). Thus, \( P^{-1} \) becomes \( \begin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix} \).

Make sure your matrices are invertible before attempting these calculations!

2x2 Matrix Determinant

The determinant is a special number that can be calculated from a square matrix. It provides important properties about the matrix, such as whether the matrix is invertible or not. For a 2x2 matrix, the determinant is simpler to compute than for larger matrices.

For a matrix \( \begin{pmatrix} a & b \ c & d \end{pmatrix} \), the determinant is calculated as \( ad - bc \). This operation is essentially taking the product of the diagonals, subtracting the product of the off-diagonals.

• If the determinant is zero, the matrix is singular, meaning it doesn't have an inverse.
• If the determinant is non-zero, then the matrix has an inverse.

In our case, for the matrix \( P = \begin{pmatrix} 1 & 1 \ 0 & 1 \end{pmatrix} \), the determinant is calculated as \( (1)(1) - (1)(0) = 1 \). Thus, \( P \) is invertible, and we can proceed to calculate its inverse, as mentioned in the previous section.

Always check the determinant before proceeding with inversion or other such operations.

Matrix Adjoint

The adjoint of a matrix, also known as the adjugate or classical adjoint, is a matrix used in finding the inverse of a matrix. When dealing with a 2x2 matrix, understanding how to obtain the adjoint allows for easier calculation of the matrix inverse.

Given a 2x2 matrix \( \begin{pmatrix} a & b \ c & d \end{pmatrix} \), the adjoint is generated by:

• Switching the principal diagonal elements (positions \( a \) and \( d \)).
• Negating the off-diagonal elements (positions \( b \) and \( c \)).

This results in the adjoint matrix: \[ \text{adj}(P) = \begin{pmatrix} d & -b \ -c & a \end{pmatrix} \]Using the given example, for the matrix \( P = \begin{pmatrix} 1 & 1 \ 0 & 1 \end{pmatrix} \), its adjoint is \( \begin{pmatrix} 1 & -1 \ 0 & 1 \end{pmatrix} \).

Having the correct adjoint is crucial in ensuring the proper computation of the matrix inverse, particularly in the formula \( P^{-1} = \frac{1}{\text{det}(P)} \cdot \text{adj}(P) \).

Mathematical Induction

Mathematical induction is a powerful proof technique used to establish the truth of an infinite number of cases. It is especially useful in proving statements about integers and matrix powers. The process has two main steps:

1. **Base Case**: Verify the statement for the initial value (often \( n = 1 \)). This case serves as the foundation. - In the context of matrix power, check if \( B^1 = P^{-1}A^1P \), as a starting point.2. **Inductive Step**: Assume the statement holds for some arbitrary positive integer \( n \), and then show it holds for \( n+1 \). This step ensures the property holds indefinitely. - Assuming \( B^n = P^{-1}A^nP \), prove that \( B^{n+1} = B^nB = P^{-1}A^{n+1}P \).The inductive step typically involves algebraic manipulations that simplify the problem using the assumption and aim to demonstrate the statement for the next case.

For the exercise, we used induction to show that \( B^k = P^{-1}A^kP \) for any positive integer \( k \). This demonstration confirms that our transformation via matrix similarity holds for powers of matrices. Understanding this technique aids in tackling more complex mathematical problems efficiently.