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 .
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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