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컴퓨터과학[2-2]/[2-2]선형대수

선형대수 11강 연습문제 풀이

by boolean 2015. 10. 31.
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선형대수 11강 연습문제 풀이

Question 1. 벡터공간 $V=R^3$와 기저$ Ã = {(1,1,0),(1,0,1),(0,1,1)}$에 대한 벡터 $(-1,4,1)$을 기저 $Ã$로 표현한 좌표는 ?

  1. $(1,-2,3)_Ã$
  2. $(-2,3,1)_Ã$
  3. $(1,3,-2)_Ã$
  4. $(1,2,-3)_Ã$
Solution
  1. $(-1, 4, 1) = k_1(1, 1, 0) + k_2(1, 0, 1) + k_3(0, 1, 1) \\ = (k_1, k_1, 0) + (k_2, 0, k_2) + (0, k_3, k_3) \\ = (k_1 + k_2, k_1 + k_3, k_2 + k_3) \\ \longrightarrow k_1 + k_2 = -1, \quad k_1 + k_3 = 4, \quad k_2 + k_3 = 1 \\ k_1 = -k_2 - 1, \quad k_3 =k_2 + 1 + 4, \quad k_2 + k_2 + 1 + 4 = 1 \\ k_1 = 1, \quad k_3 =3, \quad k_2 = -2 \\ \color{red}{(1, 3, -2)_Ã} $

Question 2. $T:R3->R3가 T(x,y,z)=(y,x-y,z)$를 만족하고, $R^3$의 주어진 기저$ Ã = {(1,1,0),(0,2,1),(0,-1,2)}$와 표준기저$\tilde{E}$ 를 기저로 할때$(T)_ \tilde{E}^\tilde{A}$는?

  1. $\left( \begin{array}{ccc} 1 & 2 & -1 \\ 0 & -2 & 1 \\0 & 1 & 2\end{array}\right)$
  2. $\left( \begin{array}{ccc} 1 & 2 & 1 \\ 0 & 0 & -2 \\0 & 0 & 0\end{array}\right)$
  3. $\left( \begin{array}{ccc} 1 & 1 & \frac{1}{2} \\ 1& 0 & 0 \\0 & 1 & 0\end{array}\right)$
  4. $\left( \begin{array}{ccc} \frac{1}{2}& 1 & 0 \\ 0 & 1 & 0 \\1 & 1 & 1\end{array}\right)$
Solution
  1. $T(1, 1, 0) = (1, 0, 0), \quad T(0, 2, 1) = (2, -2, 1), \quad T(0, -1, 2) = (-1, 1, 2) \\$표준기저$ \longrightarrow \color{red}{\left( \begin{array}{ccc} 1 & 2 & -1 \\ 0 & -2 & 1 \\0 & 1 & 2\end{array}\right)}$

Question 3. $T:R2→R2, T(x,y)=(2y,-x), \quad U:R2→R2, \quad U:(x,y)=(0,y) $일 때, 다음 물음에 답하여라.

$(T+U)(x,y)$의 값은?
  1. $(x, y - x)$
  2. $(2y, y + x)$
  3. $(x, y + x)$
  4. $(2y, y - x)$
Solution
  1. $(2y, -x) + (0, y) = (2y + 0, -x + y) = \color{red}{ (2y, y - x)}$

Question 4. $T:R2→R2, T(x,y)=(2y,-x), \quad U:R2→R2, \quad U:(x,y)=(0,y) $일 때, 다음 물음에 답하여라.

표준기저에 의한 $T+U$의 행렬은 ?
  1. $\left( \begin{array}{cc} 1 &0 \\ -1&2 \end{array}\right)$
  2. $\left( \begin{array}{cc} 0 &2 \\ -1&1 \end{array}\right)$
  3. $\left( \begin{array}{cc} 2 &1 \\ -1&0 \end{array}\right)$
  4. $\left( \begin{array}{cc} -1 &2 \\ 0&1 \end{array}\right)$
Solution
  1. $(T+U)(1, 0) = (0, -1) \\ (T + U)(0, 1) = (2, 1)$ 그러므로 행렬로 표현하면
    $\color{red}{\left( \begin{array}{cc} 0 &2 \\ -1&1 \end{array}\right)}$

Question 5. $T:R2→R2, T(x,y)=(2y,-x), \quad U:R2→R2, \quad U:(x,y)=(0,y) $일 때, 다음 물음에 답하여라.

기저$ Ã = {(1,0), (1, -1)}$에 의한$(T \circ U)_\tilde{A}^\tilde{A}$와$(U \circ T)_\tilde{A}^\tilde{A}$를 바르게 묶은 것은?
  1. $\left( \begin{array}{cc} 0 &0 \\ -2&0 \end{array}\right) \left( \begin{array}{cc}- 1 &-1 \\ 0 &0 \end{array}\right)$
  2. $\left( \begin{array}{cc} 0 &-2 \\ 0&0 \end{array}\right) \left( \begin{array}{cc}0 &0 \\ -1&-1 \end{array}\right)$
  3. $\left( \begin{array}{cc} 0 &0 \\ 0&-2 \end{array}\right) \left( \begin{array}{cc} 0 &0 \\ -1&-1 \end{array}\right)$
  4. $\left( \begin{array}{cc} 0 &-2 \\ 0&0 \end{array}\right) \left( \begin{array}{cc} -1 &-1 \\ 0&0 \end{array}\right)$
Solution
  1. $(T(U)_\tilde{A}^\tilde{A})_\tilde{A}^\tilde{A}\\ U(1, 0) = (0, y) \rightarrow (0, 0) \rightarrow T(0. 0) = (2y, -x) \rightarrow (0, 0) \\ U(1, -1) = (0, y) \rightarrow (0, -1) \rightarrow T(0. 0) = (2y, -x) \rightarrow (-2, 0) \longrightarrow \color{red}{\left( \begin{array}{cc} 0 &-2 \\ 0&0 \end{array}\right)} \\ (U(T)_\tilde{A}^\tilde{A})_\tilde{A}^\tilde{A}\\ T(1, 0) = (2y,-x) \rightarrow (0, -1) \rightarrow U(0. -1) = (0, y) \rightarrow (0, -1) \\ T(1, -1) = (-2y, -x) \rightarrow (-2, -1) \rightarrow U(-2. -1) = (0, y) \rightarrow (0, -1) \longrightarrow \color{red}{\left( \begin{array}{cc} 0 &-1 \\ 0&-1 \end{array}\right)}$

Question 6. 주어진 기저로 주어진 벡터의 좌표를 구하여라.

$V=R2, Ã = {(2, 3), (-1, 4)}$
  1. $(3, 10)$
  2. $(-7, 6$
  3. $(11, 11)$
Solution
  1. $(3, 10) = k_1(2, 3) + k_2(-1, 4) = (2k_1 + (-k_2), 3k_1 + 4k_2) \rightarrow 2k_1 - k_2 = 3, 3k_1 + 4k_2 = 10 \\ k_1 = 2, \quad k_2 = 1 \longrightarrow \color{red}{(2, 1)}$
  2. $(-7, 6) = k_1(2, 3) + k_2(-1, 4) = (2k_1 + (-k_2), 3k_1 + 4k_2) \rightarrow 2k_1 - k_2 = -7, 3k_1 + 4k_2 = 6 \\ k_1 = -1, \quad k_2 = 5 \longrightarrow \color{red}{(1, 5)}$
  3. $(11, 11) = k_1(2, 3) + k_2(-1, 4) = (2k_1 + (-k_2), 3k_1 + 4k_2) \rightarrow 2k_1 - k_2 = 11, 3k_1 + 4k_2 =11 \\ k_1 = 5, \quad k_2 = -1 \longrightarrow \color{red}{(5, -1)}$

Question 7. 주어진 기저로 주어진 벡터의 좌표를 구하여라.

$V=R3, Ã = {(1, 1, 1),(0, 1, 1),(0, 0, -1)}$
  1. $(1, 2, 3)$
  2. $(1, 3, 0)$
  3. $(8, 9, 4)$
Solution
  1. $(1, 2, 3) = k_1(1, 1, 1) + k_2(0, 1, 1) + k_3(0, 0, -1)= (k_1, k -1 + k_2, k_1 + k_2 -k_3) \rightarrow k_1 = 1, k_1 + k_2 = 2, k_1 + k_2 - k_3 = 3 \\ k_1 = 1, \quad k_2 = 1, \quad k_3 = -1 \longrightarrow \color{red}{(1, 1, -1)}$
  2. $(1, 3, 0) = k_1(1, 1, 1) + k_2(0, 1, 1) + k_3(0, 0, -1)= (k_1, k -1 + k_2, k_1 + k_2 -k_3) \rightarrow k_1 = 1, k_1 + k_2 = 3, k_1 + k_2 - k_3 = 0 \\ k_1 = 1, \quad k_2 = 2, \quad k_3 = 3 \longrightarrow \color{red}{(1, 2, 3)}$
  3. $(8, 9, 4) = k_1(1, 1, 1) + k_2(0, 1, 1) + k_3(0, 0, -1)= (k_1, k -1 + k_2, k_1 + k_2 -k_3) \rightarrow k_1 = 8, k_1 + k_2 = 9, k_1 + k_2 - k_3 = 4 \\ k_1 = 8, \quad k_2 = 1, \quad k_3 = 5 \longrightarrow \color{red}{(8, 1, 5)}$

Question 8. T를 V에서 V로의 선형변환이라 하자. 그리고 벡터공간 $V$의 기저를$ Ã =\{A1, A2, A3 \} $이라 하고 선형변환 $T$가 $T(A1) = A1,\quad T(A2) = A1+A2, \quad T(A3) = A1+A2+A3$를 만족한다고 할 때, 다음 물음에 답하라.

  1. $T$에 의한 기저 $Ã$의 상을 기저 $Ã$로 표현한 행렬을 구하시오.
  2. $\tilde{B}=\{A1, -A1-A2, A1+A2+A3\}$이라 할 때 　　$T$에 의한 $\tilde{B}$기저 의 상을 기저 $\tilde{B}$로 표현한 행렬을 구하시오.
  3. 벡터공간 $V$의 기저 $Ã$에서 $(2)$의 $\tilde{B}$로의 기저 변환 행렬을 구하시오.
  4. 벡터공간 $V$의 기저 (2)의 $\tilde{B}$에서 Ã로의 기저 변환 행렬을 구하시오.
Solution
  1. $T(A_1) =A_1 = k_1(A_1) + k_2(A_2) + k_3(A_3) = 1(A_1) + 0(A_2) + 0(A_3) \longrightarrow (1, 0, 0)_\tilde{A} \\ T(A_2) =A_1 + A_2 = k_1(A_1) + k_2(A_2) + k_3(A_3) = 1(A_1) + 1(A_2) + 0(A_3) \longrightarrow (1, 1, 0)_\tilde{A} \\ T(A_1) =A_1 + A_2 + A_3 = k_1(A_1) + k_2(A_2) + k_3(A_3) = 1(A_1) + 1(A_2) + 1(A_3) \longrightarrow (1, 1, 1)_\tilde{A} \\ \longrightarrow \color{red}{(T)_\tilde{A}^\tilde{A} = \left( \begin{array}{ccc} 1 & 1 & 1 \\ 0 & 1 &1 \\0 & 0 & 1\end{array}\right)}$
  2. $T(A_1) = A_1 = k_1(A_1) + k_2(-A_1-A_2) + k_3(A_1 +A_2 +A_3) \\ = k_1(A_1) + k_2(-A_1-A_2) + k_3(A_1 + A_2 + A_3) \\ = k_1A_1 -k_2A_1 -k_2A_2 + k_3A_1 + k_3A_2 + k_3A_3 \\ =A_1(k_1-k_2 + k_3) + A_2(-k_2 - k_3) + A_3(k_3) \longrightarrow k_3 = 0, \quad k_2 = 0, \quad k_1 = 1 \rightarrow (1, 0, 0)_\tilde{B} \\ T(A_2) =A_1 + A_2 = k_1(A_1) + k_2(-A_1-A_2) + k_3(A_1 +A_2 +A_3) \\ = k_1(A_1) + k_2(-A_1-A_2) + k_3(A_1 + A_2 + A_3) \\ = k_1A_1 -k_2A_1 -k_2A_2 + k_3A_1 + k_3A_2 + k_3A_3 \\ =A_1(k_1-k_2 + k_3) + A_2(-k_2 - k_3) + A_3(k_3) \longrightarrow k_3 = 0, \quad k_2 = -1, \quad k_1 = 0 \rightarrow (0, -1, 0)_\tilde{B} \\ T(A_3) =A_1 + A_2 + A_3= k_1(A_1) + k_2(-A_1-A_2) + k_3(A_1 +A_2 +A_3) \\ = k_1(A_1) + k_2(-A_1-A_2) + k_3(A_1 + A_2 + A_3) \\ = k_1A_1 -k_2A_1 -k_2A_2 + k_3A_1 + k_3A_2 + k_3A_3 \\ =A_1(k_1-k_2 + k_3) + A_2(-k_2 - k_3) + A_3(k_3) \longrightarrow k_3 = 1, \quad k_2 = -2, \quad k_1 = -2 \rightarrow (1, -2, -2)_\tilde{B} \\ \longrightarrow \color{red}{(T)_\tilde{B}^\tilde{B} = \left( \begin{array}{ccc} 1 & 0 & 1 \\ 0 & -1 &-2 \\0 & 0 & -2\end{array}\right)}$
  3. $ I(A_1) = A_1 = k_1(A_1) + k_2(-A_1-A_2) + k_3(A_1 +A_2 +A_3) \\ = k_1(A_1) + k_2(-A_1-A_2) + k_3(A_1 + A_2 + A_3) \\ = k_1A_1 -k_2A_1 -k_2A_2 + k_3A_1 + k_3A_2 + k_3A_3 \\ =A_1(k_1-k_2 + k_3) + A_2(-k_2 - k_3) + A_3(k_3) \longrightarrow k_3 = 0, \quad k_2 = 0, \quad k_1 = 1 \rightarrow (1, 0, 0)_\tilde{B} \\ I(A_2) =A_2 = k_1(A_1) + k_2(-A_1-A_2) + k_3(A_1 +A_2 +A_3) \\ = k_1(A_1) + k_2(-A_1-A_2) + k_3(A_1 + A_2 + A_3) \\ = k_1A_1 -k_2A_1 -k_2A_2 + k_3A_1 + k_3A_2 + k_3A_3 \\ =A_1(k_1-k_2 + k_3) + A_2(-k_2 - k_3) + A_3(k_3) \longrightarrow k_3 = 0, \quad k_2 = -1, \quad k_1 = -1 \rightarrow (0, -1, -1)_\tilde{B} \\ T(A_3) = A_3= k_1(A_1) + k_2(-A_1-A_2) + k_3(A_1 +A_2 +A_3) \\ = k_1(A_1) + k_2(-A_1-A_2) + k_3(A_1 + A_2 + A_3) \\ = k_1A_1 -k_2A_1 -k_2A_2 + k_3A_1 + k_3A_2 + k_3A_3 \\ =A_1(k_1-k_2 + k_3) + A_2(-k_2 - k_3) + A_3(k_3) \longrightarrow k_3 = 1, \quad k_2 = 1, \quad k_1 = 0 \rightarrow (1, 1, 0)_\tilde{B} \\ \longrightarrow \color{red}{ (I)_\tilde{B}^\tilde{A} = \left( \begin{array}{ccc} 1 & 0 & 1 \\ 0 & -1 &1 \\0 & -1 & 0\end{array}\right)}$
  4. $I(A_1) =A_1 = k_1(A_1) + k_2(A_2) + k_3(A_3) = 1(A_1) + 0(A_2) + 0(A_3) \longrightarrow (1, 0, 0)_\tilde{A} \\ I(-A_1-A_2) =-A_1 - A_2 = k_1(A_1) + k_2(A_2) + k_3(A_3) = -1(A_1) - 1(A_2) + 0(A_3) \longrightarrow (-1, -1, 0)_\tilde{A} \\ T(A_1) =A_1 + A_2 + A_3 = k_1(A_1) + k_2(A_2) + k_3(A_3) = 1(A_1) + 1(A_2) + 1(A_3) \longrightarrow (1, 1, 1)_\tilde{A} \\ \longrightarrow \color{red}{(I)_\tilde{A}^\tilde{B} = \left( \begin{array}{ccc} 1 & -1 & 1 \\ 0 & -1 &1 \\0 & 0 & 1\end{array}\right)}$

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