Answer to Question #133221 in Linear Algebra for Galata

Question

Show necessary step for the following questions
("For the given matrix"
A=the first row [1 -1] second row [3 2]
B=the first row [3 -2] second row [0 1]

1) show that (A^t)^t=A
2) show that (A+B)^t=A^t+B^t
3) show that (4A)^t=4(A^t)
4) show that (AB)^t=A^tB^t)

Answer 1

$$A=\begin{bmatrix} 1 & -1 \\ 3 & 2 \end{bmatrix}, B=\begin{bmatrix} 3 & -2 \\ 0 & 1 \end{bmatrix}$$

1)

$$A^T=\begin{bmatrix} 1 & 3 \\ -1 & 2 \end{bmatrix}$$

$$(A^T)^T=\begin{bmatrix} 1 & -1 \\ 3 & 2 \end{bmatrix}=A$$

2)

$$A+B=\begin{bmatrix} 1 & -1 \\ 3 & 2 \end{bmatrix}+\begin{bmatrix} 3 & -2 \\ 0 & 1 \end{bmatrix}=$$

$$=\begin{bmatrix} 1+3 & -1-2 \\ 3+0 & 2+1 \end{bmatrix}=\begin{bmatrix} 4 & -3 \\ 3 & 3 \end{bmatrix}$$

$$(A+B)^T=\begin{bmatrix} 4 & 3 \\ -3 & 3 \end{bmatrix}$$

$$A^T=\begin{bmatrix} 1 & 3 \\ -1 & 2 \end{bmatrix}, B^T=\begin{bmatrix} 3 & 0 \\ -2 & 1 \end{bmatrix}$$

$$A^T+B^T=\begin{bmatrix} 1 &3 \\ -1 & 2 \end{bmatrix}+\begin{bmatrix} 3 & 0 \\ -2 & 1 \end{bmatrix}=$$

$$=\begin{bmatrix} 1+3 & 3+0 \\ -1-2 & 2+1 \end{bmatrix}=\begin{bmatrix} 4 & 3\\ -3 & 3 \end{bmatrix}=(A+B)^T$$

3)

$$4A=4\begin{bmatrix} 1 & -1 \\ 3 & 2 \end{bmatrix}=\begin{bmatrix} 4(1) & 4(-1) \\ 4(3) & 4(2) \end{bmatrix}=\begin{bmatrix} 4 & -4 \\ 12 & 8 \end{bmatrix}$$

$$(4A)^T=\begin{bmatrix} 4 & 12 \\ -4 & 8 \end{bmatrix}$$

$$A^T=\begin{bmatrix} 1 & 3 \\ -1 & 2 \end{bmatrix}$$

$$4A^T=4\begin{bmatrix} 1 & 3 \\ -1 & 2 \end{bmatrix}=\begin{bmatrix} 4(1) & 4(3) \\ 4(-1) & 4(2) \end{bmatrix}=\begin{bmatrix} 4 & 12\\ -4 & 8 \end{bmatrix}$$

$$(4A)^T=\begin{bmatrix} 4 & 12 \\ -4 & 8 \end{bmatrix}=4A^T$$

4)

$$AB=\begin{bmatrix} 1 & -1 \\ 3 & 2 \end{bmatrix}\cdot\begin{bmatrix} 3 & -2 \\ 0 & 1 \end{bmatrix}=$$

$$=\begin{bmatrix} 1(3)+(-1)(0) & 1(-2)+(-1)(1) \\ 3(3)+2(0) & 3(-2)+2(1) \end{bmatrix}=\begin{bmatrix} 3 & -3 \\ 9 & -4 \end{bmatrix}$$

$$(AB)^T=\begin{bmatrix} 3 & 9 \\ -3 & -4 \end{bmatrix}$$

$$B^TA^T=\begin{bmatrix} 3 & 0 \\ -2 & 1 \end{bmatrix}\cdot\begin{bmatrix} 1 & 3 \\ -1 & 2 \end{bmatrix}=$$

$$=\begin{bmatrix} 3(1)+0(-1) & 3(3)+0(2) \\ -2(1)+1(-1) & -2(3)+1(2) \end{bmatrix}=\begin{bmatrix} 3 & 9 \\ -3 & -4 \end{bmatrix}$$

$$(AB)^T=B^TA^T$$