Answer to Question #192278 in Linear Algebra for prince

Question

Consider the matrices

A = 3 0 2 , B = -5 1 1 , C = 1 1 1

4 -6 3 0 3 0 2 3 -1

-2 1 8 7 6 2 3 -5 -7

Verify the following expressions (where possible and give reasons)

(i) A + (B + C) = (A + B) + C and A(BC) = (AB)C.

(ii) (a − b)C = aC + bC and a(B − C) = aB − aC, where a = −2, b = 3 .

(iii) (A^T)^T = A and (A − B)^T = A ^T− B^T

Answer 1

(i)

$$A+(B+C)=\begin{pmatrix} 3 & 0 & 2 \\ 4 & -6 & 3 \\ -2 & 1 & 8 \end{pmatrix}$$

$$+\bigg(\begin{pmatrix} -5 & 1 & 1 \\ 0 & 3 & 0 \\ 7 & 6 & 2 \end{pmatrix}+\begin{pmatrix} 1 & 1 & 1\\ 2 & 3 & -1 \\ 3 & -5 & -7 \end{pmatrix}\bigg)$$

$$=\begin{pmatrix} 3 & 0 & 2 \\ 4 & -6 & 3 \\ -2 & 1 & 8 \end{pmatrix}+\begin{pmatrix} -4 & 2 & 2 \\ 2 & 6 & -1 \\ 10 & 1 & -5 \end{pmatrix}$$

$$=\begin{pmatrix} -1 & 2 & 4 \\ 6 & 0 & 2 \\ 8 & 2 & 3 \end{pmatrix}$$

$$(A+B)+C=\bigg(\begin{pmatrix} 3 & 0 & 2 \\ 4 & -6 & 3 \\ -2 & 1 & 8 \end{pmatrix}$$

$$+\begin{pmatrix} -5 & 1 & 1 \\ 0 & 3 & 0 \\ 7 & 6 & 2 \end{pmatrix}\bigg)+\begin{pmatrix} 1 & 1 & 1\\ 2 & 3 & -1 \\ 3 & -5 & -7 \end{pmatrix}$$

$$=\begin{pmatrix} -2 & 1 & 3 \\ 4 & -3 & 3 \\ 5 & 7 & 10 \end{pmatrix}+\begin{pmatrix} 1 & 1 & 1\\ 2 & 3 & -1 \\ 3 & -5 & -7 \end{pmatrix}$$

$$=\begin{pmatrix} -1 & 2 & 4 \\ 6 & 0 & 2 \\ 8 & 2 & 3 \end{pmatrix}$$

$$A+(B+C)=\begin{pmatrix} -1 & 2 & 4 \\ 6 & 0 & 2 \\ 8 & 2 & 3 \end{pmatrix}=(A+B)+C$$

$$BC=\begin{pmatrix} -5 & 1 & 1 \\ 0 & 3 & 0 \\ 7 & 6 & 2 \end{pmatrix}\begin{pmatrix} 1 & 1 & 1\\ 2 & 3 & -1 \\ 3 & -5 & -7 \end{pmatrix}$$

$$=\begin{pmatrix} 0 & -7 & -13 \\ 6 & 9 & -3 \\ 25 & 15 & -13 \end{pmatrix}$$

$$A\big(BC\big)=\begin{pmatrix} 3 & 0 & 2 \\ 4 & -6 & 3 \\ -2 & 1 & 8 \end{pmatrix}\begin{pmatrix} 0 & -7 & -13 \\ 6 & 9 & -3 \\ 25 & 15 & -13 \end{pmatrix}$$

$$=\begin{pmatrix} 50 & 9 & -65 \\ 39 & -37 & -73 \\ 206 & 143 &-81 \end{pmatrix}$$

$$AB=\begin{pmatrix} 3 & 0 & 2 \\ 4 & -6 & 3 \\ -2 & 1 & 8 \end{pmatrix}\begin{pmatrix} -5 & 1 & 1 \\ 0 & 3 & 0 \\ 7 & 6& 2 \end{pmatrix}$$

$$=\begin{pmatrix} -1 & 15 & 7 \\ 1 & 4 &10 \\ 66 & 49 & 14 \end{pmatrix}$$

$$\big(AB\big)C=\begin{pmatrix} -1 & 15& 7 \\ 1 & 4 & 10 \\ 66 & 49 & 14 \end{pmatrix}\begin{pmatrix} 1 & 1 & 1\\ 2 & 3 & -1 \\ 3 & -5 & -7 \end{pmatrix}$$

$$=\begin{pmatrix} 50 & 9 & -65\\ 39 & -37 & -73 \\ 206 & 143 & -81 \end{pmatrix}$$

$$A\big(BC\big)=\begin{pmatrix} 50 & 9 & -65\\ 39 & -37 & -73 \\ 206 & 143 & -81 \end{pmatrix}=A\big(BC\big)$$

(ii)

$$a=-2, b=3$$

$$(a-b)C=(-2-3)\begin{pmatrix} 1 & 1 & 1\\ 2 & 3 & -1 \\ 3 & -5 & -7 \end{pmatrix}$$

$$=-5\begin{pmatrix} 1 & 1 & 1\\ 2 & 3 & -1 \\ 3 & -5 & -7 \end{pmatrix}=\begin{pmatrix} -5 & -5 & -5\\ -10 & -15 & 5 \\ -15 & 25 & 35\end{pmatrix}$$

$$aC=-2\begin{pmatrix} 1 & 1 & 1\\ 2 & 3 & -1 \\ 3 & -5 & -7 \end{pmatrix}=\begin{pmatrix} -2 & -2 & -2\\ -4 & -6 & 2 \\ -6 & 10 & 14 \end{pmatrix}$$

$$bC=3\begin{pmatrix} 1 & 1 & 1\\ 2 & 3 & -1 \\ 3 & -5 & -7 \end{pmatrix}=\begin{pmatrix} 3 & 3 & 3\\ 6 & 9 & -3 \\ 9 & -15 & -21 \end{pmatrix}$$

$$aC-bC=\begin{pmatrix} -2 & -2 & -2\\ -4 & -6 & 2 \\ -6 & 10 & 14 \end{pmatrix}-\begin{pmatrix} 3 & 3 & 3\\ 6 & 9 & -3 \\ 9 & -15 & -21 \end{pmatrix}$$

$$=\begin{pmatrix} -5 & -5 & -5\\ -10 & -15 & 5 \\ -15 & 25 & 35\end{pmatrix}$$

$$(a-b)C=\begin{pmatrix} -5 & -5 & -5\\ -10 & -15 & 5 \\ -15 & 25 & 35\end{pmatrix}=aC-bC$$

$$B-C=\begin{pmatrix} -5 & 1 & 1 \\ 0 & 3 & 0 \\ 7 & 6 & 2 \end{pmatrix}-\begin{pmatrix} 1 & 1 & 1\\ 2 & 3 & -1 \\ 3 & -5 & -7 \end{pmatrix}$$

$$=\begin{pmatrix} -6 & 0 & 0\\ -2 & 0 & 1 \\ 4 & 11& 9 \end{pmatrix}$$

$$a(B-C)=-2\begin{pmatrix} -6 & 0 & 0\\ -2 & 0 & 1 \\ 4 & 11& 9 \end{pmatrix}=\begin{pmatrix} 12 & 0 & 0\\ 4 & 0 & -2 \\ -8 & -22 & -18 \end{pmatrix}$$

$$aB=-2\begin{pmatrix} -5 & 1 & 1 \\ 0 & 3 & 0 \\ 7 & 6 & 2 \end{pmatrix}=\begin{pmatrix} 10 & -2 & -2 \\ 0 & -6 & 0 \\ -14 & -12 &-4 \end{pmatrix}$$

$$aB-aC=\begin{pmatrix} 10 & -2 & -2 \\ 0 & -6 & 0 \\ -14 & -12 &-4 \end{pmatrix}$$

$$-\begin{pmatrix} -2 & -2 & -2\\ -4 & -6 & 2 \\ -6 & 10 & 14 \end{pmatrix}=\begin{pmatrix} 12 & 0 & 0\\ 4 & 0 & -2 \\ -8 & -22& -18 \end{pmatrix}$$

$$a(B-C)=\begin{pmatrix} 12 & 0 & 0\\ 4 & 0 & -2 \\ -8 & -22& -18 \end{pmatrix}=aB-aC$$

(iii)

$$A^T=\begin{pmatrix} 3 & 0 & 2 \\ 4 & -6 & 3 \\ -2 & 1 & 8 \end{pmatrix}^T=\begin{pmatrix} 3 & 4 & -2 \\ 0 & -6 & 1 \\ 2 & 3 & 8 \end{pmatrix}$$

$$(A^T)^T=\begin{pmatrix} 3 & 4 & -2 \\ 0 & -6 & 1 \\ 2 & 3 & 8 \end{pmatrix}^T=\begin{pmatrix} 3 & 0 & 2 \\ 4 & -6 & 3 \\ -2 & 1 & 8 \end{pmatrix}=A$$

$$(A^T)^T=\begin{pmatrix} 3 & 0 & 2 \\ 4 & -6 & 3 \\ -2 & 1 & 8 \end{pmatrix}=A$$

$$A-B=\begin{pmatrix} 3 & 0 & 2 \\ 4 & -6 & 3 \\ -2 & 1 & 8 \end{pmatrix}-\begin{pmatrix} -5 & 1 & 1 \\ 0 & 3 & 0 \\ 7 & 6 & 2 \end{pmatrix}$$

$$=\begin{pmatrix} 8 & -1 & 1 \\ 4 & -9 & 3 \\ -9 & -5 & 6 \end{pmatrix}$$

$$(A-B)^T=\begin{pmatrix} 8 & -1 & 1 \\ 4 & -9 & 3 \\ -9 & -5 & 6 \end{pmatrix}^T=\begin{pmatrix} 8 & 4 & -9 \\ -1 & -9 & -5 \\ 1 & 3 & 6 \end{pmatrix}$$

$$B^T=\begin{pmatrix} -5 & 1 & 1 \\ 0 & 3 & 0 \\ 7 & 6 & 2 \end{pmatrix}^T=\begin{pmatrix} -5 & 0 & 7 \\ 1 & 3 & 6 \\ 1 & 0& 2 \end{pmatrix}$$

$$A^T-B^T=\begin{pmatrix} 3 & 4 & -2 \\ 0 & -6 & 1 \\ 2 & 3 & 8 \end{pmatrix}-\begin{pmatrix} -5 & 0 & 7 \\ 1 & 3 & 6 \\ 1 & 0& 2 \end{pmatrix}$$

$$=\begin{pmatrix} 8 & 4 & -9 \\ -1 & -9 & -5 \\ 1 & 3 &6 \end{pmatrix}$$

$$(A-B)^T=\begin{pmatrix} 8 & 4 & -9 \\ -1 & -9 & -5 \\ 1 & 3 & 6 \end{pmatrix}=A^T-B^T$$