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Chapter 12: Problem 3

Given $$ A=\left(\begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & -1 \end{array}\right), \quad B=\left(\begin{array}{cc} -2 & 3 \\ 0 & 4 \\ 7 & -1 \end{array}\right) $$ calculate (a) \(A+B^{\mathrm{T}}\) (b) \(B-2 A^{\mathrm{T}}\)

Short Answer

Expert verified
The resulting matrices are: (a) A + B^T: $$ \left(\begin{array}{ccc} -1 & 2 & 10 \\ 7 & 9 & -2 \end{array}\right) $$ (b) B - 2A^T: $$ \left(\begin{array}{cc} -4 & -5 \\ -4 & -6 \\ 1 & 1 \end{array}\right) $$

Step by step solution

01

Find the transpose of matrix B

To find the transpose of a matrix, simply switch its rows and columns, so the new matrix will have the same elements but with swapped indices. Therefore, the transpose of matrix B will be: $$ B^T = \left(\begin{array}{ccc} -2 & 0 & 7 \\ 3 & 4 & -1 \end{array}\right) $$
02

Add matrix A and the transpose of matrix B

To add two matrices, we add their corresponding elements. In this case, we are adding matrix A and the transpose of matrix B. Therefore, the operation looks like this: $$ A + B^T = \left(\begin{array}{ccc} 1 + (-2) & 2 + 0 & 3 + 7 \\ 4 + 3 & 5 + 4 & -1 + (-1) \end{array}\right) $$ Now, perform the addition to obtain the resulting matrix: $$ A + B^T = \left(\begin{array}{ccc} -1 & 2 & 10 \\ 7 & 9 & -2 \end{array}\right) $$ Now let's perform the second operation, which is subtracting twice the transpose of matrix A from matrix B.
03

Find the transpose of matrix A

Using the same process as in Step 1, we find the transpose of matrix A: $$ A^T = \left(\begin{array}{cc} 1 & 4 \\ 2 & 5 \\ 3 & -1 \end{array}\right) $$
04

Subtract twice the transpose of matrix A from matrix B

First, we need to calculate twice the transpose of matrix A: $$ 2A^T = \left(\begin{array}{cc} 2 & 8 \\ 4 & 10 \\ 6 & -2 \end{array}\right) $$ Now, to subtract this matrix from matrix B, we subtract the corresponding elements: $$ B - 2A^T = \left(\begin{array}{cc} (-2) - 2 & 3 - 8 \\ 0 - 4 & 4 - 10 \\ 7 - 6 & (-1) - (-2) \end{array}\right) $$ Perform the subtraction to obtain the resulting matrix: $$ B - 2A^T = \left(\begin{array}{cc} -4 & -5 \\ -4 & -6 \\ 1 & 1 \end{array}\right) $$ To summarize, the results are: (a) \(A + B^T\): $$ \left(\begin{array}{ccc} -1 & 2 & 10 \\ 7 & 9 & -2 \end{array}\right) $$ (b) \(B - 2A^T\): $$ \left(\begin{array}{cc} -4 & -5 \\ -4 & -6 \\ 1 & 1 \end{array}\right) $$

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Most popular questions from this chapter

State the number of elements in an \(n \times m\) matrix.

Given $$ A=\left(\begin{array}{ll} 6 & 1 \\ 3 & 7 \end{array}\right) $$ state (a) \(A-2 I_{2}\) (b) \(A-\lambda I_{2}\) where \(\lambda\) is a constant.

The matrix, \(H\), is defined by $$ H=\left(\begin{array}{cc} 2 & 1 \\ \alpha & 0 \\ -3 & \beta \end{array}\right) $$ (a) State the size of \(H\). (b) State \(h_{11}, h_{21}, h_{32}\)

State the transpose of \(C\) where $$ C=\left(\begin{array}{cc} 9 & 3 \\ 1 & -2 \\ \alpha & 4 \end{array}\right) $$

State the transpose of \(I_{3}\).
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