Question

In each case, find \(P^{-1} A P\) and then com- pute \(A^{n}\) $$ \begin{array}{l} \text { a. } A=\left[\begin{array}{rr} 6 & -5 \\ 2 & -1 \end{array}\right], P=\left[\begin{array}{ll} 1 & 5 \\ 1 & 2 \end{array}\right] \\ \text { b. } A=\left[\begin{array}{rr} -7 & -12 \\ 6 & -10 \end{array}\right], P=\left[\begin{array}{rr} -3 & 4 \\ 2 & -3 \end{array}\right] \end{array} $$ \(\left[\right.\) Hint \(: \quad\left(P D P^{-1}\right)^{n}=P D^{n} P^{-1}\) for each \(n=\) \(1,2, \ldots .]\)

Short answer

Compute matrix products and powers as described to find \(A^n\).

Step-by-step solution

Calculate the Inverse of P (Part a)

To find \( P^{-1} \) for matrix \( P = \begin{bmatrix} 1 & 5 \ 1 & 2 \end{bmatrix} \), use the formula for the inverse of a 2x2 matrix: \[ P^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \]. Here, \( a = 1 \), \( b = 5 \), \( c = 1 \), \( d = 2 \). First, calculate the determinant: \( ad - bc = 1(2) - 5(1) = -3 \). Now, \( P^{-1} = \frac{1}{-3} \begin{bmatrix} 2 & -5 \ -1 & 1 \end{bmatrix} \). Simplifying, \( P^{-1} = \begin{bmatrix} -\frac{2}{3} & \frac{5}{3} \ \frac{1}{3} & -\frac{1}{3} \end{bmatrix} \).

Compute \( P^{-1} A P \) (Part a)

Now, calculate \( P^{-1} A P \) for \( A = \begin{bmatrix} 6 & -5 \ 2 & -1 \end{bmatrix} \). Perform the matrix multiplication: \[ P^{-1} A = \begin{bmatrix} -\frac{2}{3} & \frac{5}{3} \ \frac{1}{3} & -\frac{1}{3} \end{bmatrix} \begin{bmatrix} 6 & -5 \ 2 & -1 \end{bmatrix} = \begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix} \]. Now multiply by \( P \): \[ (P^{-1} A P) = \begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 5 \ 1 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 2 \ 0 & 0 \end{bmatrix} \].

Determine \( A^n \) (Part a)

Using the result \( D = \begin{bmatrix} 1 & 2 \ 0 & 0 \end{bmatrix} \) and knowing \( A^n = P D^n P^{-1} \), calculate powers of \( D \). For any \( n \geq 2 \), \( D^n = \begin{bmatrix} 1 & 2n \ 0 & 0 \end{bmatrix} \). Finally, \[ A^n = P D^n P^{-1} = \begin{bmatrix} 1 & 5 \ 1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 2n \ 0 & 0 \end{bmatrix} \begin{bmatrix} -\frac{2}{3} & \frac{5}{3} \ \frac{1}{3} & -\frac{1}{3} \end{bmatrix} \]. Perform the multiplication to find \( A^n \).

Calculate the Inverse of P (Part b)

Find \( P^{-1} \) for matrix \( P = \begin{bmatrix} -3 & 4 \ 2 & -3 \end{bmatrix} \). Use the 2x2 inverse formula: First, calculate the determinant: \( ad - bc = (-3)(-3) - (4)(2) = 1 \). Then, \( P^{-1} = \begin{bmatrix} -3 & -4 \ -2 & -3 \end{bmatrix} \).

Compute \( P^{-1} A P \) (Part b)

Now calculate \( P^{-1} A P \) for \( A = \begin{bmatrix} -7 & -12 \ 6 & -10 \end{bmatrix} \). Perform the multiplication: \[ P^{-1} A = \begin{bmatrix} -3 & -4 \ -2 & -3 \end{bmatrix} \begin{bmatrix} -7 & -12 \ 6 & -10 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & -1 \end{bmatrix} \]. Now multiply by \( P \): \[ (P^{-1} A P) = \begin{bmatrix} 1 & 0 \ 0 & -1 \end{bmatrix} \begin{bmatrix} -3 & 4 \ 2 & -3 \end{bmatrix} = \begin{bmatrix} -3 & 4 \ -2 & 3 \end{bmatrix} \].

Determine \( A^n \) (Part b)

Using \( D = \begin{bmatrix} 1 & 0 \ 0 & -1 \end{bmatrix} \), the powers of \( D \) are \( D^n = \begin{bmatrix} 1^n & 0 \ 0 & (-1)^n \end{bmatrix} \). Therefore, \( A^n = P D^n P^{-1} \). The multiplication yields \( A^n \) in terms of \( n \). Perform this computation to get \( A^n \).

Key concepts

Inverse of a Matrix

Matrix inversion is a fundamental operation that allows us to find the matrix that 'undoes' the effect of the original matrix. For a 2x2 matrix \[ A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \]the inverse, denoted as \( A^{-1} \), is computed using the formula:\[ A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \].Here, "ad - bc" is the determinant of the matrix, and it must be non-zero for the inverse to exist.

• If the determinant is zero, the matrix is said to be singular and has no inverse.


• The inverse essentially helps to solve linear algebraic equations where matrices are involved.


• It plays a key role in matrix transformations and is extensively used in computational methods.

To illustrate, consider matrix \( P \) from the exercise: \[ P = \begin{bmatrix} 1 & 5 \ 1 & 2 \end{bmatrix} \].The determinant is calculated and the inverse is found as follows:\[ P^{-1} = \frac{1}{-3} \begin{bmatrix} 2 & -5 \ -1 & 1 \end{bmatrix} = \begin{bmatrix} -\frac{2}{3} & \frac{5}{3} \ \frac{1}{3} & -\frac{1}{3} \end{bmatrix}. \] Using the inverse is crucial for solving systems of equations and matrix-related problems.

Diagonalization

Diagonalization is the process of converting a matrix into a diagonal form. This is extremely useful because diagonal matrices are easier to compute powers from and further manipulate.To diagonalize matrix \( A \), find a non-singular matrix \( P \) and a diagonal matrix \( D \) such that:\[ P^{-1} A P = D \].This means matrix \( A \) is similar to diagonal matrix \( D \).

• Finding \( P \) often involves deriving the eigenvectors of the original matrix.


• Matrix \( D \) will have eigenvalues of \( A \) on its diagonal.


• Once diagonalized, computing \( A^n \) becomes straightforward by raising \( D \) to the power \( n \).

In the exercise context, for matrix \( A \), we compute \( D = P^{-1} A P \), which simplifies future calculations:\[ D = \begin{bmatrix} 1 & 2 \ 0 & 0 \end{bmatrix}, \begin{bmatrix} 1 & 0 \ 0 & -1 \end{bmatrix} \].This transforms \( A \) into a form that is manageable for finding its powers, a task that would be much harder otherwise.

Matrix Multiplication

Matrix multiplication is a cornerstone of operations involving matrices. It involves the dot product of rows and columns to compute the resulting matrix.Given two matrices, \( A \) of size \( m \times n \) and \( B \) of size \( n \times p \), their product \( AB \) results in a matrix of size \( m \times p \).

• The element in the \( i \)-th row and \( j \)-th column of the resulting matrix is calculated by multiplying the elements of the \( i \)-th row of \( A \) by the elements of the \( j \)-th column of \( B \) and summing them up.


• Matrix multiplication is not commutative; \( AB eq BA \).


• This operation plays a critical role in transforming matrices and solving systems of equations.

For the problem discussed, calculating \( P^{-1} A P \) and further operations like \( A^n \) showcase how matrix multiplication helps in transitioning matrices into a different form to facilitate easier computations. When multiplying, follow the correct order and dimensions to ensure the operation is valid and to obtain meaningful results.