Answer to Question #200538 in Linear Algebra for Snakho

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Answer to Question #200538 in Linear Algebra for Snakho

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Linear Algebra

Question #200538

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

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

Expert's answer

A = "\\begin{bmatrix}\n 3 & 0 & 2 \\\\\n 4 & -6 & 3\\\\\n -2 & 1 & 8\\\\ \n\\end{bmatrix}"

B = "\\begin{bmatrix}\n -5 & 1 &1 \\\\\n 0 & 3 & 0 \\\\\n7 & 6 & 2 \\\\\n\\end{bmatrix}"

C = "\\begin{bmatrix}\n 1 & 1 &1\\\\\n 2 &3&-1\\\\\n3&-5&-7 \n\\end{bmatrix}"

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

a) A+(B+C) =(A+B) +C

LHS:

= A+(B+C)

="\\begin{bmatrix}\n 3 & 0 & 2 \\\\\n 4 & -6 & 3\\\\\n -2 & 1 & 8\\\\ \n\\end{bmatrix}" + "\\begin{bmatrix}\n -4 & 2 &2 \\\\\n 2 & 6 & -1 \\\\\n10 & 1 & -5 \\\\\n\\end{bmatrix}"

= "\\begin{bmatrix}\n -1 & 2 &4\\\\\n 6 & 0 &2\\\\\n8&2&3\n\\end{bmatrix}"

RHS:

(A+B) +C

= "\\begin{bmatrix}\n-2&1&3\\\\\n4&-3&3\\\\\n5&7&10\n\\end{bmatrix}" + "\\begin{bmatrix}\n 1 & 1 &1\\\\\n 2 &3&-1\\\\\n3&-5&-7 \n\\end{bmatrix}"

= "\\begin{bmatrix}\n -1 & 2 &4\\\\\n 6 & 0 &2\\\\\n8&2&3\n\\end{bmatrix}"

LHS = RHS

b) A(BC) =(AB) C

LHS:

= "\\begin{bmatrix}\n 3 & 0 & 2 \\\\\n 4 & -6 & 3\\\\\n -2 & 1 & 8\\\\ \n\\end{bmatrix}" * "\\begin{bmatrix}\n0&-7&-13\\\\\n6&9&-3\\\\\n25&15&-13\\\\\n\n\n\n\n\n\n\n\n\\end{bmatrix}"

= "\\begin{bmatrix}\n50&9&-65\\\\\n39&-37&-73\\\\\n206&143&-81\\\\\n\\end{bmatrix}"

RHS:

= "\\begin{bmatrix}\n -1 & 15&7 \\\\\n 1 & 4&10\\\\\n66&49&14\\\\\n\\end{bmatrix}" * "\\begin{bmatrix}\n 1 & 1 &1\\\\\n 2 &3&-1\\\\\n3&-5&-7 \n\\end{bmatrix}"

= "\\begin{bmatrix}\n50&9&-65\\\\\n39&-37&-73\\\\\n206&143&-81\\\\\n\\end{bmatrix}"

LHS = RHS

(ii) (a-b) C=aC - bC and a(B - C) =aB - aC

a) (a-b) C=aC - bC

LHS:

= (a - b) * "\\begin{bmatrix}\n 1 & 1 &1\\\\\n 2 &3&-1\\\\\n3&-5&-7 \n\\end{bmatrix}"

= "\\begin{bmatrix}\n a-b & a-b &a-b\\\\\n 2(a - b) &3(a - b)&-1(a - b)\\\\\n3(a - b)&-5(a - b)&-7(a - b)\\\\ \n\\end{bmatrix}"

RHS:

= a * "\\begin{bmatrix}\n 1 & 1 &1\\\\\n 2 &3&-1\\\\\n3&-5&-7 \n\\end{bmatrix}" - b * "\\begin{bmatrix}\n 1 & 1 &1\\\\\n 2 &3&-1\\\\\n3&-5&-7 \n\\end{bmatrix}"

= "\\begin{bmatrix}\n a & a &a\\\\\n 2a &3a&-a\\\\\n3a&-5a&-7a \n\\end{bmatrix}" - "\\begin{bmatrix}\n b & b &b\\\\\n 2b &3b&-b\\\\\n3b&-5b&-7b \n\\end{bmatrix}"

= "\\begin{bmatrix}\n a-b & a-b &a-b\\\\\n 2(a - b) &3(a - b)&-1(a - b)\\\\\n3(a - b)&-5(a - b)&-7(a - b)\\\\ \n\\end{bmatrix}"

LHS = RHS

b) a(B - C) =aB - aC

LHS:

= a * "\\begin{bmatrix}\n-6&0&0\\\\\n-2&0&1\\\\\n4&11&9\n\\end{bmatrix}"

= "\\begin{bmatrix}\n-6a&0&0\\\\\n-2a&0&a\\\\\n4a&11a&9a\n\\end{bmatrix}"

RHS:

= aB - aC

= a * "\\begin{bmatrix}\n -5 & 1 &1 \\\\\n 0 & 3 & 0 \\\\\n7 & 6 & 2 \\\\\n\\end{bmatrix}" - a * "\\begin{bmatrix}\n 1 & 1 &1\\\\\n 2 &3&-1\\\\\n3&-5&-7 \n\\end{bmatrix}"

= "\\begin{bmatrix}\n-6a&0&0\\\\\n-2a&0&a\\\\\n4a&11a&9a\n\\end{bmatrix}"

LHS = RHS

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

a) (A^T) ^T=A

LHS:

= ( A^T) ^T

= "(\\begin{bmatrix}\n 3 & 4 & -2 \\\\\n 4 & -6 & 1\\\\\n -2 & 1 & 8\\\\ \n\\end{bmatrix} )^T"

= "\\begin{bmatrix}\n 3 & 4 & -2 \\\\\n 4 & -6 & 1\\\\\n -2 & 1 & 8\\\\ \n\\end{bmatrix}" = A

LHS = RHS

(b) (A - B) ^T=A^T - B^T

LHS:

= "(\\begin{bmatrix}\n 8 & -1 & 1 \\\\\n 4 & -9 & 3\\\\\n -9 & -5 & 6\\\\ \n\\end{bmatrix} )^T"

= "\\begin{bmatrix}\n 8 & 4 & -9 \\\\\n -1 & -9 & -5\\\\\n 1 & 3 & 6\\\\ \n\\end{bmatrix}"

RHS:

= A^T - B^T

= "(\\begin{bmatrix}\n 3 & 0 & 2 \\\\\n 4 & -6 & 3\\\\\n -2 & 1 & 8\\\\ \n\\end{bmatrix} ) ^T" - "(\\begin{bmatrix}\n -5 & 1 &1 \\\\\n 0 & 3 & 0 \\\\\n7 & 6 & 2 \\\\\n\\end{bmatrix})^T"

= "\\begin{bmatrix}\n 8 & 4 & -9 \\\\\n -1 & -9 & -5\\\\\n 1 & 3 & 6\\\\ \n\\end{bmatrix}"

LHS = RHS