Q74E B3Consider the regular tetrahedr... [FREE SOLUTION]

Chapter 3: Q74E (page 162)

$B^{3}$ Consider the regular tetrahedron in the accompanying sketch whose center is at the origin. Let ${\overset{鈫�}{v}}_{0}, {\overset{鈫�}{v}}_{1}, {\overset{鈫�}{v}}_{2}, {\overset{鈫�}{v}}_{3}$ be the position vectors of the four vertices of the tetrahedron:
${\overset{鈫�}{v}}_{0} = \overset{鈫�}{O} P_{0}, v_{1} = {\overset{鈫�}{OP}}_{1}, v_{2} = {\overset{鈫�}{OP}}_{2}, v_{3} = O \overset{鈫�}{P_{3}}$
a. Find the sum ${\overset{鈫�}{v}}_{0}, {\overset{鈫�}{v}}_{1}, {\overset{鈫�}{v}}_{2}, {\overset{鈫�}{v}}_{3}$
b. Find the coordinate vector of ${\overset{鈫�}{v}}_{0}$ with respect to the basis ${\overset{鈫�}{v}}_{1}, {\overset{鈫�}{v}}_{2}, {\overset{鈫�}{v}}_{3}$ .
c. Let T be the linear transformation with $T ({\overset{鈫�}{v}}_{0}) = {\overset{鈫�}{v}}_{3}, T ({\overset{鈫�}{v}}_{3}) = {\overset{鈫�}{v}}_{1} and T ({\overset{鈫�}{v}}_{1}) = {\overset{鈫�}{v}}_{1}$ . What is $T ({\overset{鈫�}{v}}_{2})$ Describe the transformation T geometrically (as a reflection, rotation, projection, or whatever). Find the matrix B of T with respect to the basis ${\overset{鈫�}{v}}_{1}, {\overset{鈫�}{v}}_{2}, {\overset{鈫�}{v}}_{3}$ .
What is Explain.

Short Answer

Expert verified

a. The sum ${\overset{鈫�}{v}}_{0}, {\overset{鈫�}{v}}_{1}, {\overset{鈫�}{v}}_{2}, {\overset{鈫�}{v}}_{3} = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]$

b. The coordinate vector of ${\overset{鈫�}{v}}_{0}$ with respect to the basis ${\overset{鈫�}{v}}_{1}, {\overset{鈫�}{v}}_{2}, {\overset{鈫�}{v}}_{3}$ is

$鈭� {[{\overset{鈫�}{v}}_{0}]}_{尉} = [\begin{matrix} 伪 \\ 尾 \\ 纬 \end{matrix}] = [\begin{matrix} - 1 \\ - 1 \\ - 1 \end{matrix}]$

c. The matrix B of T with respect to the basis ${\overset{鈫�}{v}}_{1}, {\overset{鈫�}{v}}_{2}, {\overset{鈫�}{v}}_{3}$ is

$B = [\begin{matrix} - 1 & 0 & 0 \\ - 1 & 1 & 0 \\ - 1 & 0 & 1 \end{matrix}]$

And $B^{3} = B$ .

Step by step solution

(a) Find the sum v→0,v→1,v→2,v→

We have given,

${\overset{鈫�}{v}}_{0} = O \overset{鈫�}{P_{0}} = [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}], v_{1} = O {\overset{鈫�}{P}}_{1} [\begin{matrix} - 1 \\ 1 \\ - 1 \end{matrix}], v_{3} = O {\overset{鈫�}{P}}_{3} [\begin{matrix} - 1 \\ - 1 \\ 1 \end{matrix}] {\overset{鈫�}{v}}_{0}, {\overset{鈫�}{v}}_{1}, {\overset{鈫�}{v}}_{2}, {\overset{鈫�}{v}}_{3} = [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}] + [\begin{matrix} 1 \\ - 1 \\ - 1 \end{matrix}] [\begin{matrix} - 1 \\ 1 \\ - 1 \end{matrix}] + [\begin{matrix} - 1 \\ - 1 \\ 1 \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]$

Thus

(b) Finding the coordinate vector of v→0 with respect to the basis v→1,v→2,v→3

$Let 貜 = \{{\overset{鈫�}{v}}_{1}, {\overset{鈫�}{v}}_{2}, {\overset{鈫�}{v}}_{3}\} and {[{\overset{鈫�}{v}}_{0}]}_{貜} = [\begin{matrix} 伪 \\ 尾 \\ 纬 \end{matrix}]$

Then

${\overset{鈫�}{v}}_{0} = 伪 {\overset{鈫�}{v}}_{1} + 尾 {\overset{鈫�}{v}}_{2} + 纬 {\overset{鈫�}{v}}_{3} [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}] = 伪 [\begin{matrix} 1 \\ - 1 \\ - 1 \end{matrix}] + 尾 [\begin{matrix} - 1 \\ 1 \\ - 1 \end{matrix}] + 纬 [\begin{matrix} - 1 \\ - 1 \\ 1 \end{matrix}] 鈬� 1 = 伪 - 尾 - 纬 1 = - 伪 + 尾 - 纬 1 = - 伪 - 尾 + 纬鈬� 伪 - 1, 尾 = - 1, 纬 = - 1 鈭� {[{\overset{鈫�}{v}}_{0}]}_{貜} = [\begin{matrix} 伪 \\ 尾 \\ 纬 \end{matrix}] = [\begin{matrix} - 1 \\ - 1 \\ - 1 \end{matrix}]$

(c) Finding T(v→2)

We have given

$T ({\overset{鈫�}{v}}_{0}) = \overset{鈫�}{v_{3}}, T ({\overset{鈫�}{v}}_{3}) = {\overset{鈫�}{v}}_{1} and T ({\overset{鈫�}{v}}_{1}) = {\overset{鈫�}{v}}_{0}$ $鈬� T ([\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}]) = [\begin{matrix} 1 \\ - 1 \\ - 1 \end{matrix}], T ([\begin{matrix} - 1 \\ 1 \\ - 1 \end{matrix}]) = [\begin{matrix} - 1 \\ - 1 \\ 1 \end{matrix}] a n d T ([\begin{matrix} 1 \\ - 1 \\ - 1 \end{matrix}]) = [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}]$

Now,

$[\begin{matrix} - 1 \\ 1 \\ - 1 \end{matrix}] = c_{1} [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}] + c_{2} [\begin{matrix} - 1 \\ - 1 \\ 1 \end{matrix}] + c_{3} [\begin{matrix} 1 \\ - 1 \\ - 1 \end{matrix}] 鈬� - 1 = c_{1} - c_{2} + c_{3} 1 = c_{1} - c_{2} - c_{3} - 1 = c_{1} + c_{2} - c_{3} 鈬� c_{1} = c_{2} = c_{3} = - 1$

$鈬� [\begin{matrix} - 1 \\ 1 \\ - 1 \end{matrix}] = (- 1) [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}] + (- 1) [\begin{matrix} - 1 \\ 1 \\ - 1 \end{matrix}] + (- 1) [\begin{matrix} 1 \\ 1 \\ - 1 \end{matrix}] 鈬� T ([\begin{matrix} - 1 \\ 1 \\ - 1 \end{matrix}]) = (- 1) T ([\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}]) + (- 1) T ([\begin{matrix} - 1 \\ - 1 \\ 1 \end{matrix}]) + (- 1) T ([\begin{matrix} 1 \\ - 1 \\ - 1 \end{matrix}]) 鈬� T ([\begin{matrix} - 1 \\ 1 \\ - 1 \end{matrix}]) = (- 1) ([\begin{matrix} - 1 \\ 1 \\ - 1 \end{matrix}]) + (- 1) [\begin{matrix} 1 \\ - 1 \\ - 1 \end{matrix}] + (- 1) [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}] 鈬� T ([\begin{matrix} - 1 \\ 1 \\ - 1 \end{matrix}]) = [\begin{matrix} 1 \\ 1 \\ - 1 \end{matrix}] + [\begin{matrix} - 1 \\ 1 \\ 1 \end{matrix}] + [\begin{matrix} - 1 \\ - 1 \\ - 1 \end{matrix}] 鈬� T ([\begin{matrix} - 1 \\ 1 \\ - 1 \end{matrix}]) = [\begin{matrix} - 1 \\ 1 \\ - 1 \end{matrix}] 鈬� T ({\overset{鈫�}{v}}_{2}) = [\begin{matrix} - 1 \\ 1 \\ - 1 \end{matrix}]$

Finding the B matrix of T with respect to the basis ؏=(v→1,v→2,v→3)

The B matrix of T with respect to the basis $貜 = ({\overset{鈫�}{v}}_{1}, {\overset{鈫�}{v}}_{2}, {\overset{鈫�}{v}}_{3})$ is given by

$B = [{[T ({\overset{鈫�}{v}}_{1})]}_{貜} {[T ({\overset{鈫�}{v}}_{2})]}_{貜} {[T ({\overset{鈫�}{v}}_{3})]}_{貜}]$

We have given,

role="math" localid="1664275553613" ${\overset{鈫�}{v}}_{1} = [\begin{matrix} 1 \\ - 1 \\ - 1 \end{matrix}], {\overset{鈫�}{v}}_{2} = [\begin{matrix} - 1 \\ 1 \\ - 1 \end{matrix}], {\overset{鈫�}{v}}_{3} = [\begin{matrix} - 1 \\ - 1 \\ 1 \end{matrix}] and T ({\overset{鈫�}{v}}_{1}) = [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}], T ({\overset{鈫�}{v}}_{2}) = [\begin{matrix} - 1 \\ 1 \\ - 1 \end{matrix}], T ({\overset{鈫�}{v}}_{3}) = [\begin{matrix} 1 \\ - 1 \\ - 1 \end{matrix}]$

Now,

$T ({\overset{鈫�}{v}}_{1}) = a_{1} {\overset{鈫�}{v}}_{1} + a_{2} {\overset{鈫�}{v}}_{2} + a_{3} {\overset{鈫�}{v}}_{3} 鈬� [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}] = a_{1} [\begin{array}{c} 1 \\ 鈭� 1 \\ 鈭� 1 \end{array}] + a_{2} [\begin{array}{c} 鈭� 1 \\ 1 \\ 鈭� 1 \end{array}] + a_{3} [\begin{array}{c} 鈭� 1 \\ 鈭� 1 \\ 1 \end{array}] 鈬� 1 = a_{1} 鈭� a_{2} 鈭� a_{3} 1 = 鈭� a_{1} + a_{2} 鈭� a_{3} 1 = 鈭� a_{1} 鈭� a_{2} + a_{3} 鈬� a_{1} = a_{2} = a_{3} = 鈭� 1 and T ({\overset{鈫�}{v}}_{2}) = b_{1} {\overset{鈫�}{v}}_{1} + b_{2} {\overset{鈫�}{v}}_{2} + b_{3} {\overset{鈫�}{v}}_{3} 鈬� [\begin{array}{c} 鈭� 1 \\ 1 \\ 鈭� 1 \end{array}] = b_{1} [\begin{array}{c} 1 \\ 鈭� 1 \\ 鈭� 1 \end{array}] + b_{2} [\begin{array}{c} 鈭� 1 \\ 1 \\ 鈭� 1 \end{array}] + b_{3} [\begin{array}{c} 鈭� 1 \\ 鈭� 1 \\ 1 \end{array}] 鈬� 鈭� 1 = b_{1} 鈭� b_{2} 鈭� b_{3} 1 = 鈭� b_{1} + b_{2} 鈭� b_{3} 鈭� 1 = 鈭� b_{1} 鈭� b_{2} + b_{3} 鈬� b_{1} = 0, b_{2} = 1, b_{3} = 0 T ({\overset{鈫�}{v}}_{3}) = c_{1} {\overset{鈫�}{v}}_{1} + c_{2} {\overset{鈫�}{v}}_{2} + c_{3} {\overset{鈫�}{v}}_{3} 鈬� [\begin{array}{c} 鈭� 1 \\ 鈭� 1 \\ 1 \end{array}] = c_{1} [\begin{array}{c} 1 \\ 鈭� 1 \\ 鈭� 1 \end{array}] + c_{2} [\begin{array}{c} 鈭� 1 \\ 1 \\ 鈭� 1 \end{array}] + c_{3} [\begin{array}{c} 鈭� 1 \\ 鈭� 1 \\ 1 \end{array}]$

$鈬� 鈭� 1 = c_{1} 鈭� c_{2} 鈭� c_{3} 鈭� 1 = 鈭� c_{1} + c_{2} 鈭� c_{3} 1 = 鈭� c_{1} 鈭� c_{2} + c_{3} 鈬� c_{1} = 0, c_{2} = 0, c_{3} = 1 鈭� B = [\begin{array}{lll} {T ({\overset{鈫�}{v}}_{1})]}_{I} & {[T ({\overset{鈫�}{v}}_{2})]}_{I} {[T ({\overset{鈫�}{v}}_{3})]}_{鉁�} \end{array}] 鈬� B = [\begin{array}{lll} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array}] 鈬� B = [\begin{array}{lll} 鈭� 1 & 0 & 0 \\ 鈭� 1 & 1 & 0 \\ 鈭� 1 & 0 & 1 \end{array}]$

Also

$B^{2} = [\begin{array}{lll} 鈭� 1 & 0 & 0 \\ 鈭� 1 & 1 & 0 \\ 鈭� 1 & 0 & 1 \end{array}] [\begin{array}{lll} 鈭� 1 & 0 & 0 \\ 鈭� 1 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{lll} 鈭� 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] 鈬� B^{3} = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] [\begin{array}{lll} 鈭� 1 & 0 & 0 \\ 鈭� 1 & 1 & 0 \\ 鈭� 1 & 0 & 1 \end{array}] = [\begin{array}{lll} 鈭� 1 & 0 & 0 \\ 鈭� 1 & 1 & 0 \\ 鈭� 1 & 0 & 1 \end{array}] 鈬� B^{3} = [\begin{array}{lll} 鈭� 1 & 0 & 0 \\ 鈭� 1 & 1 & 0 \\ 鈭� 1 & 0 & 1 \end{array}] = B 鈬� B^{3} = B$

Final Answer

a. The sum ${\overset{鈫�}{v}}_{0} + {\overset{鈫�}{v}}_{1} + {\overset{鈫�}{v}}_{2} + {\overset{鈫�}{v}}_{3} = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]$

b. The coordinate vector of ${\overset{鈫�}{v}}_{0}$ with respect to the basis ${\overset{鈫�}{v}}_{1}, {\overset{鈫�}{v}}_{2}, {\overset{鈫�}{v}}_{3}$ is

$鈭� {[{\overset{鈫�}{v}}_{0}]}_{貜} = [\begin{matrix} 伪 \\ 尾 \\ 纬 \end{matrix}] = [\begin{matrix} - 1 \\ - 1 \\ - 1 \end{matrix}]$

c. The matrix B of T with respect to the basis ${\overset{鈫�}{v}}_{1} + {\overset{鈫�}{v}}_{2} + {\overset{鈫�}{v}}_{3}$ is

$B [\begin{matrix} - 1 & 0 & 0 \\ - 1 & 1 & 0 \\ - 1 & 0 & 1 \end{matrix}]$

And $B^{3} = B$ .

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Short Answer

Step by step solution

(a) Find the sum v→0,v→1,v→2,v→

(b) Finding the coordinate vector of v→0 with respect to the basis v→1,v→2,v→3

(c) Finding T(v→2)

Finding the B matrix of T with respect to the basis ؏=(v→1,v→2,v→3)

Final Answer

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