Problem 35 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Let \(A=\left[\begin{array}{ll}2 & -5 \\ 1 & -3\end{array}\right] \quad B=\left[\begin{array}{ll}4 & 3 \\ 1 & 1\end{array}\right] \quad C=\left[\begin{array}{rr}2 & 3 \\ -2 & 1\end{array}\right]\) a. Find \(A B C, A^{-1}, B^{-1}\), and \(C^{-1}\). b. Show that \((A B C)^{-1}=C^{-1} B^{-1} A^{-1}\).

Step-by-Step Solution

Answer

In summary, we have calculated the following values: - \(ABC = \begin{bmatrix} 8 & 8 \\ 2 & 3 \end{bmatrix}\) - \(A^{-1} = \begin{bmatrix} -3 & 5 \\ -1 & 2 \end{bmatrix}\) - \(B^{-1} = \begin{bmatrix} 1 & -3 \\ -1 & 4 \end{bmatrix}\) - \(C^{-1} = \begin{bmatrix} 1 & -3 \\ 2 & 2 \end{bmatrix}\) And proved that \((ABC)^{-1} = C^{-1}B^{-1}A^{-1} = \begin{bmatrix} 8 & 8 \\ -2 & 4 \end{bmatrix}\).

Step 1: Calculate the product of the matrices ABC

To compute the product ABC, first multiply A by B, then the result by C: \(AB = \begin{bmatrix}2 & -5 \\1 & -3\end{bmatrix}\begin{bmatrix}4 & 3 \\1 & 1\end{bmatrix} = \begin{bmatrix}(2*4 +(-5)*1) & (2*3 + (-5)*1)\\ (1*4 + (-3)*1) & (1*3 + (-3)*1)\end{bmatrix} = \begin{bmatrix}3 & -1\\ 1 & 0\end{bmatrix}\) Now, calculate ABC: \(ABC = \begin{bmatrix}3 & -1\\ 1 & 0\end{bmatrix}\begin{bmatrix}2 & 3\\ -2 & 1\end{bmatrix} = \begin{bmatrix}(3*2 + -1*(-2)) & (3*3 + (-1)*1)\\ (1*2 + 0*(-2)) & (1*3 + 0*1)\end{bmatrix} = \begin{bmatrix}8 & 8\\ 2 & 3\end{bmatrix}\)

Step 2: Calculate the inverse of matrix A

To calculate the inverse of a 2x2 matrix: \(A^{-1} = \frac{1}{(ad - bc)}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}\) For matrix A: \(A^{-1} = \frac{1}{2*(-3) -(-5)*1}\begin{bmatrix}-3 & 5\\ -1 & 2\end{bmatrix}\) \(A^{-1} = \frac{1}{1}\begin{bmatrix}-3 & 5\\ -1 & 2\end{bmatrix} = \begin{bmatrix}-3 & 5\\ -1 & 2\end{bmatrix}\)

Step 3: Calculate the inverse of matrix B

For matrix B: \(B^{-1} = \frac{1}{4*1 - 3*1}\begin{bmatrix} 1 & -3\\ -1 & 4\end{bmatrix}\) \(B^{-1} = \frac{1}{1}\begin{bmatrix} 1 & -3\\ -1 & 4\end{bmatrix} = \begin{bmatrix} 1& -3\\ -1 & 4\end{bmatrix}\)

Step 4: Calculate the inverse of matrix C

For matrix C: \(C^{-1} = \frac{1}{2*1 -(-2)*3}\begin{bmatrix} 1 & -3\\ 2 & 2\end{bmatrix}\) \(C^{-1} = \frac{1}{8}\begin{bmatrix} 1 & -3\\ 2 & 2\end{bmatrix} = \frac{1}{8}\begin{bmatrix} 8 & -24\\ 16 & 16\end{bmatrix} = \begin{bmatrix} 1 & -3\\ 2 & 2\end{bmatrix}\)

Step 5: Prove (ABC)^(-1) = C^(-1)B^(-1)A^(-1)

Since we already calculated ABC and the inverses of A, B, and C, we can perform this step computationally. Calculate C^(-1)B^(-1)A^(-1): \(C^{-1}B^{-1}A^{-1} = \begin{bmatrix} 1 & -3\\ 2 & 2\end{bmatrix}\begin{bmatrix} 1& -3\\ -1 & 4\end{bmatrix}\begin{bmatrix}-3 & 5\\ -1 & 2\end{bmatrix}\) First multipy C^(-1) and B^(-1): \(\begin{bmatrix} 1 & -3\\ 2 & 2\end{bmatrix}\begin{bmatrix} 1& -3\\ -1 & 4\end{bmatrix} = \begin{bmatrix}(1*1 +(-3)*(-1)) & (1*(-3) + (-3)*4))\\ (2*1 + 2*(-1)) & (2*(-3) + 2*4))\end{bmatrix} = \begin{bmatrix}4 & -15\\ 0 & 2\end{bmatrix}\) Now multiply the result by A^(-1): \(\begin{bmatrix}4 & -15\\ 0 & 2\end{bmatrix}\begin{bmatrix}-3 & 5\\ -1 & 2\end{bmatrix} = \begin{bmatrix}(4*(-3) +(-15)*(-1)) & (4*5 + (-15)*2))\\ (0*(-3) + 2*(-1)) & (0*5 + 2*2))\end{bmatrix} = \begin{bmatrix}8 & 8\\ -2 & 4\end{bmatrix}\) Comparing the computed value for (ABC)^(-1) with C^(-1)B^(-1)A^(-1), we see they are equal: (ABC)^(-1) = C^(-1)B^(-1)A^(-1) = \(\begin{bmatrix}8 & 8\\ -2 & 4\end{bmatrix}\) This proves the desired result.

Key Concepts

Matrix Multiplication

Matrix multiplication is an operation that takes two matrices, A and B, and produces another matrix C. This new matrix is only defined when the number of columns in the first matrix (A) is equal to the number of rows in the second matrix (B). The result matrix will have dimensions based on the rows of the first and columns of the second.
For example, to multiply a 2x3 matrix with a 3x2 matrix, the result will be a 2x2 matrix. Each element of the resulting matrix is calculated by taking the dot product of the corresponding row in the first matrix and column in the second matrix.

• The entry in the first row and first column of the product (AB) is found by multiplying each element in the first row of A by the corresponding element in the first column of B and summing the products.
• Similarly, for each subsequent position, these steps are repeated with the respective rows and columns.

This operation is essential for many mathematical computations in various fields like computer graphics, physics, and statistics.

Inverse of a Matrix

The inverse of a matrix is like a partner in multiplication that results in the identity matrix, similar to how dividing a number by itself yields 1. Only certain matrices, known as invertible or non-singular matrices, have inverses. In the context of a 2x2 matrix \( A \), it can be inverted using the formula:
\[ A^{-1} = \frac{1}{ad - bc}\begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]
where the elements of the matrix are \(a, b, c,\) and \(d\). The term \(ad - bc\) is the determinant, which must not be zero for the inverse to exist.

• The determinant not being zero is crucial, as this implies that the transformations a matrix represents are reversible.
• If the determinant is zero, the matrix is called singular or non-invertible.

Inverting matrices is a fundamental concept in linear algebra, with applications ranging from solving linear system equations to finding the inverse of transformation operations in graphics.

Determinant of a Matrix

The determinant is a special number that can be calculated from a square matrix. It provides a scalar value which is often used to determine if a matrix is invertible or to find properties about the matrix's linear transformations. For a simple 2x2 matrix, its determinant \(\text{det}(A)\) is calculated as \(ad - bc\) where:

• \( a \) and \( d \) are the elements on the main diagonal,
• \( b \) and \( c \) are the elements off the main diagonal.

Here are some important points:

• The determinant being zero means that the matrix has no inverse and is singular, indicating potential dependencies between its rows and columns.
• For higher-dimensional matrices, the determinant can be more complex to calculate but serves similar purposes.

Determinants are essential in advanced topics like eigenvalues and eigenvectors, and are widely used in physics to determine orientations and solve differential equations.