Problem 7 - Advanced Engineering Mathematics

Question

\(S^{-1}(T(z))=\frac{a z+b}{c z+d}\) where $$\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]=\operatorname{adj}\left(\left[\begin{array}{ll} 1 & -2 \\ 1 & -1 \end{array}\right]\right)\left[\begin{array}{ll} 2 & -3 \\ 1 & -3 \end{array}\right]=\left[\begin{array}{rr} 0 & -3 \\ -1 & 0 \end{array}\right].$$ Therefore, $$S^{-1}(T(z))=\frac{-3}{-z}=\frac{3}{z} \quad \text { and } \quad S^{-1}(w)=\frac{-w+2}{-w+1}=\frac{w-2}{w-1}.$$

Step-by-Step Solution

Answer

\(S^{-1}(T(z))=\frac{3}{z}\) and \(S^{-1}(w)=\frac{w-2}{w-1}\).

Step 1: Understand the Problem

We are given the equation \(S^{-1}(T(z))=\frac{a z+b}{c z+d}\) along with an expression for the matrix on the right hand side of the equality involving the adjugate matrix and another matrix. We need to find the expression for \(S^{-1}(T(z))\) using these matrices.

Step 2: Calculate the Adjugate of the Given Matrix

Given matrix for adjugate calculation:\[\begin{bmatrix}1 & -2 \1 & -1\end{bmatrix}\]The adjugate (or adjoint) of a 2x2 matrix \(\begin{bmatrix}p & q \ r & s\end{bmatrix}\) is \(\begin{bmatrix}s & -q \ -r & p\end{bmatrix}\). Applying this, we have:\[\operatorname{adj}\left(\begin{bmatrix} 1 & -2 \ 1 & -1 \end{bmatrix}\right) = \begin{bmatrix} -1 & 2 \ -1 & 1 \end{bmatrix}\]

Step 3: Multiply the Adjugate Matrix by the Second Matrix

We must multiply the adjugate matrix by the given matrix:\[\begin{bmatrix} -1 & 2 \ -1 & 1 \end{bmatrix} \begin{bmatrix} 2 & -3 \ 1 & -3 \end{bmatrix} = \begin{bmatrix} (-1)(2) + (2)(1) & (-1)(-3) + (2)(-3) \ (-1)(2) + (1)(1) & (-1)(-3) + (1)(-3) \end{bmatrix} = \begin{bmatrix} 0 & -3 \ -1 & 0 \end{bmatrix}\]

Step 4: Determine Values of a, b, c, d

From the multiplication in the previous step, the resulting matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\) where \(a = 0\), \(b = -3\), \(c = -1\), and \(d = 0\).

Step 5: Substitute Values into the Fraction Expression

Substitute \(a, b, c, d\) into the equation \(S^{-1}(T(z))=\frac{a z+b}{c z+d}\), resulting in:\[S^{-1}(T(z)) = \frac{0 \cdot z + (-3)}{-1 \cdot z + 0} = \frac{-3}{-z} = \frac{3}{z}\]

Step 6: Analyze the Second Part of the Exercise

We are given: \(S^{-1}(w)=\frac{-w+2}{-w+1}\). Simplifying this, we have:\[S^{-1}(w) = \frac{w-2}{w-1}\] where the coefficients are already in place and do not need further simplification.

Key Concepts

Adjugate Matrix

The adjugate matrix is a special type of matrix that is derived from any given square matrix. Specifically, for a 2x2 matrix, producing its adjugate is relatively straightforward. If your original matrix is \[\begin{bmatrix}p & q \r & s\end{bmatrix}\]then the adjugate matrix is formulated by switching the positions of the diagonal elements and changing the signs of the off-diagonal elements. Thus, the adjugate of our matrix becomes:\[\operatorname{adj} \begin{bmatrix}1 & -2 \1 & -1\end{bmatrix}= \begin{bmatrix}-1 & 2 \-1 & 1\end{bmatrix}\]This technique is frequently used for finding the inverse of matrices and in transformation problems. The adjugate is pivotal as it provides a way to construct matrices that are easier to manipulate in inverse calculations.

Matrix Multiplication

Matrix multiplication is a process of combining two matrices to produce a third matrix. The importance of being able to multiply matrices is evident in many mathematical and practical applications.When multiplying two 2x2 matrices, you use the rule: \[\begin{bmatrix}a_{11} & a_{12} \a_{21} & a_{22}\end{bmatrix}\begin{bmatrix}b_{11} & b_{12} \b_{21} & b_{22}\end{bmatrix}\]to get:\[\begin{bmatrix}a_{11}b_{11} + a_{12}b_{21} & a_{11}b_{12} + a_{12}b_{22} \a_{21}b_{11} + a_{22}b_{21} & a_{21}b_{12} + a_{22}b_{22}\end{bmatrix}\]Applying this to our example, multiplying:\[\begin{bmatrix}-1 & 2 \-1 & 1\end{bmatrix}\begin{bmatrix}2 & -3 \1 & -3\end{bmatrix}= \begin{bmatrix}0 & -3 \-1 & 0\end{bmatrix}\]this computation helps us retrieve the coefficients for our subsequent functions. Each entry of the resulting matrix is a dot product of the rows and columns from the multiplying matrices. Understanding matrix multiplication is essential for any advanced applications in linear algebra.

Inverse Transformation

Inverse transformations are critical when solving equations that involve functions or mappings, particularly in linear algebra. An inverse function essentially "undoes" the effect of another function. In matrix terms, finding an inverse means locating a matrix that, when multiplied with your given matrix, yields the identity matrix. For the matrix equation\[S^{-1}(T(z))=\frac{a z+b}{c z+d}\]we determine the inverse transformation using a rearranged matrix expression derived from the multiplication of the adjugate with another matrix.The concept of an inverse transformation also extends to function composition where the inverse function essentially reverses the transformation applied by the original function, bringing us back to the original input value.

Fraction Simplification

Fraction simplification is an integral skill in solving algebraic expressions. It involves reducing fractions to their simplest form where the numerator and the denominator have no common factors other than one.In our scenario, the expression\[S^{-1}(T(z)) = \frac{-3}{-z} = \frac{3}{z}\]is simplified by recognizing that multiplying both the numerator and the denominator by -1 does not change the fraction's value, leading to its simplified form \(\frac{3}{z}\).Similarly, the transformation\[S^{-1}(w) = \frac{-w+2}{-w+1}\]simplifies to \(\frac{w-2}{w-1}\),which often requires careful manipulation to ensure proper simplification. Mastery of fraction simplification helps in making complex algebraic problems more approachable and solvable, especially in equations involving transformations.