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Chapter 12: Problem 1

Given find \(R=\left(\begin{array}{ll}4 & 1 \\ 6 & 2\end{array}\right), \quad S=\left(\begin{array}{cc}-3 & 2 \\ 7 & -4\end{array}\right)\) (a) \(R+S\left(\right.\) b) \(S-R\) (c) \(R+R^{\mathrm{T}}\) (d) \((R-S)^{\mathrm{T}}\) (e) \(\left(S^{\mathrm{T}}\right)^{\mathrm{T}}\)

Short Answer

Expert verified
Answer: The result of the addition of matrix R and its transpose is a 2x2 matrix: $$\begin{pmatrix}8 & 7\\ 7& 4\end{pmatrix}$$

Step by step solution

01

(a) Addition of matrices R and S

To add two matrices, their dimensions must be the same. Both \(R\) and \(S\) are \(2 \times 2\) matrices, so we can add them. Add corresponding elements of matrices R and S: $$R + S = \begin{pmatrix}4 & 1\\6 & 2\end{pmatrix} + \begin{pmatrix}-3 & 2\\7& -4\end{pmatrix} = \begin{pmatrix} (4-3) & (1+2)\\ (6+7) & (2-4) \end{pmatrix} = \begin{pmatrix}1 & 3\\ 13& -2\end{pmatrix}$$
02

(b) Subtraction of matrices S and R

To subtract two matrices, their dimensions must be the same. Both \(R\) and \(S\) are \(2 \times 2\) matrices, so we can subtract them. Subtract corresponding elements of matrices R and S: $$S - R = \begin{pmatrix}-3 & 2\\7& -4\end{pmatrix} - \begin{pmatrix}4 & 1\\6 & 2\end{pmatrix} = \begin{pmatrix} (-3-4) & (2-1)\\ (7-6) & (-4-2) \end{pmatrix} = \begin{pmatrix}-7 & 1\\ 1& -6\end{pmatrix}$$
03

(c) Addition of matrix R and its transpose

To find the transpose of a matrix, simply swap its rows and columns. Therefore, the transpose of matrix \(R\) is: $$R^T = \begin{pmatrix}4 & 6\\1 & 2\end{pmatrix}$$ Now add the matrices \(R\) and \(R^T\): $$R + R^T = \begin{pmatrix}4 & 1\\6 & 2\end{pmatrix} +\begin{pmatrix}4 & 6\\1 & 2\end{pmatrix} = \begin{pmatrix}(4+4) & (1+6)\\ (6+1) & (2+2) \end{pmatrix} = \begin{pmatrix}8 & 7\\ 7& 4\end{pmatrix}$$
04

(d) Transpose of the difference of matrices R and S

We have already calculated the difference between matrices R and S in (b): $$S-R = \begin{pmatrix}-7 & 1\\1 & -6\end{pmatrix}$$ Now, find the transpose of this matrix: $$(S-R)^T = \begin{pmatrix}-7 & 1\\1 & -6\end{pmatrix}^T = \begin{pmatrix}-7 & 1\\1 & -6\end{pmatrix}$$
05

(e) Transpose of the transpose of matrix S

First, let's find the transpose of matrix S: $$S^T = \begin{pmatrix}-3 & 7\\2 & -4\end{pmatrix}$$ Now, find the transpose of the transpose of matrix S: $$(S^T)^T = \begin{pmatrix}-3 & 7\\2 & -4\end{pmatrix}^T = \begin{pmatrix}-3 & 2\\7 & -4\end{pmatrix}\quad \text{(This is equal to the original matrix S)}$$

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Most popular questions from this chapter

Refer to matrices \(P, Q\) and \(R\) where \(P\) is a \(3 \times 2\) matrix, \(Q\) is a \(3 \times 3\) matrix and \(R\) is a \(2 \times 3\) matrix. State the size of each of the following: (a) \(P^{\mathrm{T}^{7}}\) (b) \(Q^{\mathrm{T}}\) (c) \(R^{\mathrm{T}}\) (d) \(R^{T} P^{\mathrm{T}}\) (e) \(P^{\mathrm{T}} Q^{\mathrm{T}}\)

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) \(B-2 A^{\mathrm{T}}\)

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.

Refer to matrices \(P, Q\) and \(R\) where \(P\) is a \(3 \times 2\) matrix, \(Q\) is a \(3 \times 3\) matrix and \(R\) is a \(2 \times 3\) matrix. State the size of the following products. If a product cannot be found then state this. (a) \(P Q R\) (b) \(P R Q\) (c) \(Q P R\) (d) \(R Q P\)

Refer to matrices \(P, Q\) and \(R\) where \(P\) is a \(3 \times 2\) matrix, \(Q\) is a \(3 \times 3\) matrix and \(R\) is a \(2 \times 3\) matrix. State the size of the following products if they can be found. If they cannot be found then state this. (a) \(P Q\) (f) \(Q^{2}\) (g) \(R P\) (c) \(Q R\). (d) \(R Q\) (e) \(P^{2}\)
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