Answer to Question #190364 in Linear Algebra for Regomoditswe Dibob

Question #190364

Consider the matrices A =   3 0 2 4 −6 3 −2 1 8   ,B =   −5 1 1 0 3 0 7 6 2   , C =   1 1 1 2 3 −1 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 .


1
Expert's answer
2021-05-10T13:11:10-0400

Given matrices-


"A=\\begin{bmatrix}\n 3 & 0&2 \\\\\n 4 & -6&3\\\\-2&1&8\n\\end{bmatrix},B=\\begin{bmatrix}\n -5 & 1&1 \\\\\n 0 & 3&0\\\\7&6&2\n\\end{bmatrix},C=\\begin{bmatrix}\n 1 & 1&1 \\\\\n 2 & 3&-1\\\\3&-5&-7\n\\end{bmatrix}"


"(i) LHS- A+(B+C)"


"=\\begin{bmatrix}\n 3 & 0&2 \\\\\n 4 & -6&3\\\\-2&1&8\n\\end{bmatrix}+(\\begin{bmatrix}\n -5 & 1&1 \\\\\n 0 & 3&0\\\\7&6&2\n\\end{bmatrix}+\\begin{bmatrix}\n 1 & 1&1 \\\\\n 2 & 3&-1\\\\3&-5&-7\n\\end{bmatrix})\\\\[9pt]=\\begin{bmatrix}\n 3 & 0&2 \\\\\n 4 & -6&3\\\\-2&1&8\n\\end{bmatrix}+\\begin{bmatrix}\n -4& 2&2 \\\\\n 2 & 6&-1\\\\10&1&-5\n\\end{bmatrix}=\\begin{bmatrix}\n -1 & 2&4 \\\\\n 6 & 0&2\\\\8&2&3\n\\end{bmatrix}"


RHS-(A+B)+C

"=(\\begin{bmatrix}\n 3 & 0&2 \\\\\n 4 & -6&3\\\\-2&1&8\n\\end{bmatrix}+\\begin{bmatrix}\n -5 & 1&1 \\\\\n 0 & 3&0\\\\7&6&2\n\\end{bmatrix})+\\begin{bmatrix}\n 1 & 1&1 \\\\\n 2 & 3&-1\\\\3&-5&-7\n\\end{bmatrix}\\\\[9pt]=\\begin{bmatrix}\n -2 & 2&2 \\\\\n 4 & -3&3\\\\5&7&10\n\\end{bmatrix}+\\begin{bmatrix}\n 1 & 1&1 \\\\\n 2 & 3&-1\\\\3&-5&-7\n\\end{bmatrix}=\\begin{bmatrix}\n -1 & 2&4 \\\\\n 6 & 0&2\\\\8&2&3\n\\end{bmatrix}"


Hence LHS=RHS.


"(ii) Lhs= (a-b)C=(-2-3)C=-5C=-5\\begin{bmatrix}\n 1 & 1&1 \\\\\n 2 & 3&-1\\\\3&-5&-7\n\\end{bmatrix}" ="\\begin{bmatrix}\n -5 & -5&-5 \\\\\n -10 & -15&5\\\\-15&25&49\n\\end{bmatrix}"


"RHS=aC-bC=-2C-3C=-2\\begin{bmatrix}\n 1 & 1&1 \\\\\n 2 & 3&-1\\\\3&-5&-7\n\\end{bmatrix}-3\\begin{bmatrix}\n 1 & 1&1 \\\\\n 2 & 3&-1\\\\3&-5&-7\n\\end{bmatrix}\\\\[9pt]=\\begin{bmatrix}\n -2 & -2&-2 \\\\\n -4 & -6&2\\\\-6&10&14\n\\end{bmatrix}-\\begin{bmatrix}\n 3 & 3&3 \\\\\n 6 & 9&-3\\\\9&-15&-21\n\\end{bmatrix}\\\\[9pt]=\\begin{bmatrix}\n -5 & -5&-5 \\\\\n -10 & -15&5\\\\-15&25&49\n\\end{bmatrix}"


Hence, RHS=LHS


"(iii) (A^T)^T=\\begin{bmatrix}\n 3 & 0&2 \\\\\n 4 & -6&3\\\\-2&1&8\n\\end{bmatrix}''=\\begin{bmatrix}\n 3 & 4&-2 \\\\\n 0& -6&1\\\\2&3&8\n\\end{bmatrix}'=\\begin{bmatrix}\n 3 & 0&2 \\\\\n 4 & -6&3\\\\-2&1&8\n\\end{bmatrix}=A"


Hence, "(A^T)^T=A"


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