Problem 32 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Let $$A=\left[\begin{array}{rr}2 & 4 \\\5 & -6\end{array}\right] \text { and } B=\left[\begin{array}{rr}4 & 8 \\ -7 & 3\end{array}\right]$$ a. Find \(A^{T}\) and show that \(\left(A^{T}\right)^{T}=A\). b. Show that \((A+B)^{T}=A^{T}+B^{T}\). c. Show that \((A B)^{T}=B^{T} A^{T}\).

Step-by-Step Solution

Answer

In this exercise, we found the transpose of matrices A and B, and demonstrated the following properties of matrix transpositions: a. \(A^T = \left[\begin{array}{rr} 2 & 5 \\ 4 & -6 \end{array}\right]\), and \((A^T)^T = A\). b. \((A+B)^T = A^T + B^T\), with \((A+B)^T = \left[\begin{array}{rr} 6 & -2 \\ 12 & -3 \end{array}\right]\) and \(A^T + B^T = \left[\begin{array}{rr} 6 & -2 \\ 12 & -3 \end{array}\right]\). c. \((AB)^T = B^T A^T\), with \((AB)^T = \left[\begin{array}{rr} (-14) & 47 \\ 20 & 28 \end{array}\right]\) and \(B^T A^T = \left[\begin{array}{rr} (-14) & 47 \\ 20 & 28 \end{array}\right]\).

Step 1: Find A^T

Let's first find the transpose of matrix A. To find the transpose of a matrix, we swap rows with columns (i.e., the element in row i, column j of the original matrix becomes the element in row j, column i of the transposed matrix). So, we have: $$A^T = \left[\begin{array}{rr} 2 & 5 \\ 4 & -6 \end{array}\right]$$

Step 2: Show (A^T)^T = A

Now, let's find the transpose of A^T in the same way we found the transpose of A: $$\left(A^T\right)^T = \left[\begin{array}{rr} 2 & 4 \\ 5 & -6 \end{array}\right]$$ As we can see, \((A^T)^T = A\).

Step 3: Find B^T

Let's also find the transpose of matrix B: $$B^T = \left[\begin{array}{rr} 4 & -7 \\ 8 & 3 \end{array}\right]$$

Step 4: Show (A+B)^T = A^T + B^T

First, let's find the sum of A and B: $$A + B = \left[\begin{array}{rr} 2+4 & 4+8 \\ 5-7 & -6+3 \end{array}\right] = \left[\begin{array}{rr} 6 & 12 \\ -2 & -3 \end{array}\right]$$ Now, let's find the transpose of (A+B): $$(A+B)^T = \left[\begin{array}{rr} 6 & -2 \\ 12 & -3 \end{array}\right]$$ We can confirm that \((A+B)^T = A^T + B^T\) as: $$\left[\begin{array}{rr} 6 & -2 \\ 12 & -3 \end{array}\right] = \left[\begin{array}{rr} 2 & 5 \\ 4 & -6 \end{array}\right] + \left[\begin{array}{rr} 4 & -7 \\ 8 & 3 \end{array}\right]$$

Step 5: Show (AB)^T = B^T A^T

First, let's find the product of A and B: $$AB = \left[\begin{array}{rr} 2 & 4 \\ 5 & -6 \end{array}\right] \left[\begin{array}{rr} 4 & 8 \\ -7 & 3 \end{array}\right] = \left[\begin{array}{rr} (-14) & 20 \\ 47 & 28 \end{array}\right]$$ Now, let's find the transpose of (AB): $$(AB)^T = \left[\begin{array}{rr} (-14) & 47 \\ 20 & 28 \end{array}\right]$$ Now let's find the product of B^T and A^T: $$B^T A^T = \left[\begin{array}{rr} 4 & -7 \\ 8 & 3 \end{array}\right] \left[\begin{array}{rr} 2 & 5 \\ 4 & -6 \end{array}\right] = \left[\begin{array}{rr} (-14) & 47 \\ 20 & 28 \end{array}\right]$$ We can confirm that \((AB)^T = B^T A^T\): $$\left[\begin{array}{rr} (-14) & 47 \\ 20 & 28 \end{array}\right] = \left[\begin{array}{rr} (-14) & 47 \\ 20 & 28 \end{array}\right]$$