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Dear Students,
Welcome to today’s quiz! This is your opportunity to demonstrate what you’ve learned so far, so do your best. Please keep in mind that you have a maximum of 60 minutes to complete all the questions. Make sure to manage your time wisely and answer each question thoughtfully.
Good luck!
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Pertanyaan 1 dari 6
1. Pertanyaan
1 pointsJika \(\textbf{B} = \begin{bmatrix}2 &-1\\3& 4 \end{bmatrix}\) dan I adalah matriks identitas, maka determinan dari \(\textbf{B}^2 + \textbf{BI} \:-\:3\textbf{B}\) adalah…
Benar
\(\textbf{B}^2 + \textbf{BI} \:-\:3\textbf{B}\)
\(\begin{bmatrix}2 &-1\\3& 4 \end{bmatrix}\cdot \begin{bmatrix}2 &-1\\3& 4 \end{bmatrix} + \begin{bmatrix}2 &-1\\3& 4 \end{bmatrix} \cdot \begin{bmatrix}1 &0\\0& 1\end{bmatrix} \:-\:3\begin{bmatrix}2 &-1\\3& 4 \end{bmatrix}\)
\(\begin{bmatrix}1 &-6\\18& 13 \end{bmatrix} + \begin{bmatrix}2 &-1\\3& 4 \end{bmatrix}\:-\:\begin{bmatrix}6 &-3\\9& 12 \end{bmatrix}\)
\(\begin{bmatrix}3&-7\\21& 17 \end{bmatrix}\:-\:\begin{bmatrix}6 &-3\\9& 12 \end{bmatrix}\)
\(\begin{bmatrix}-3 &-4\\12& 5 \end{bmatrix}\)
Salah
\(\textbf{B}^2 + \textbf{BI} \:-\:3\textbf{B}\)
\(\begin{bmatrix}2 &-1\\3& 4 \end{bmatrix}\cdot \begin{bmatrix}2 &-1\\3& 4 \end{bmatrix} + \begin{bmatrix}2 &-1\\3& 4 \end{bmatrix} \cdot \begin{bmatrix}1 &0\\0& 1\end{bmatrix} \:-\:3\begin{bmatrix}2 &-1\\3& 4 \end{bmatrix}\)
\(\begin{bmatrix}1 &-6\\18& 13 \end{bmatrix} + \begin{bmatrix}2 &-1\\3& 4 \end{bmatrix}\:-\:\begin{bmatrix}6 &-3\\9& 12 \end{bmatrix}\)
\(\begin{bmatrix}3&-7\\21& 17 \end{bmatrix}\:-\:\begin{bmatrix}6 &-3\\9& 12 \end{bmatrix}\)
\(\begin{bmatrix}-3 &-4\\12& 5 \end{bmatrix}\)
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Pertanyaan 2 dari 6
2. Pertanyaan
1 pointsDiketahui matriks \(\textbf{A} = \begin{bmatrix}4 & 2 & 8 \\ 2 & x & 5 \\ 3 & 2 & 4 \end{bmatrix}\) dan \(\left|\textbf{A}\right| = -2\). Nilai dari \(x\) adalah…
Benar
\(+4\cdot \text{ det}\begin{bmatrix}x& 5\\ 2 &4 \end{bmatrix}\:-\:2\cdot \text{ det}\begin{bmatrix}2& 5\\ 3 &4 \end{bmatrix} + 8\cdot \text{ det}\begin{bmatrix}2& x\\ 3 &2 \end{bmatrix}= -2\)
\(4(4x\:-\:10)\:-\:2(8\:-\:15) + 8(4\:-\:3x) = -2\)
\(16x\:-\:40 \:-\:16 + 30 + 32\:-\:24x = -2\)
\(-8x + 6 = -2\)
\(-8x = -8\)
\(x = 1\)
Salah
\(+4\cdot \text{ det}\begin{bmatrix}x& 5\\ 2 &4 \end{bmatrix}\:-\:2\cdot \text{ det}\begin{bmatrix}2& 5\\ 3 &4 \end{bmatrix} + 8\cdot \text{ det}\begin{bmatrix}2& x\\ 3 &2 \end{bmatrix}= -2\)
\(4(4x\:-\:10)\:-\:2(8\:-\:15) + 8(4\:-\:3x) = -2\)
\(16x\:-\:40 \:-\:16 + 30 + 32\:-\:24x = -2\)
\(-8x + 6 = -2\)
\(-8x = -8\)
\(x = 1\)
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Pertanyaan 3 dari 6
3. Pertanyaan
1 pointsDiketahui matriks \(\textbf{A} = \begin{bmatrix}0 & -2 \\ 1 & 4 \end{bmatrix}\), \(\textbf{B} = \begin{bmatrix}2x + 2y & 2 \\ x\:-\:4y & -1 \end{bmatrix}\), dan \(\textbf{C} = \begin{bmatrix}5 & -2 \\ 3 & -1 \end{bmatrix}\).
Jika \(\textbf{A}^{-1}\cdot \textbf{B} = \textbf{C}^{\text{T}}\), maka nilai \(2x + 3y\) adalah…
Benar
\(\textbf{A}^{-1}\cdot \textbf{B} = \textbf{C}^{\text{T}}\)
\(\textbf{A}\cdot\textbf{A}^{-1}\cdot \textbf{B} = \textbf{A}\cdot\textbf{C}^{\text{T}}\)
\(\textbf{B} = \textbf{A}\cdot\textbf{C}^{\text{T}}\)
\(\begin{bmatrix}2x + 2y & 2 \\ x\:-\:4y & -1 \end{bmatrix} = \begin{bmatrix}0 & -2 \\ 1 & 4 \end{bmatrix} \cdot \begin{bmatrix}5 & 3 \\ -2 & -1 \end{bmatrix}\)
\(\begin{bmatrix}2x + 2y & 2 \\ x\:-\:4y & -1 \end{bmatrix} = \begin{bmatrix}4 & 2 \\ -3 & -1 \end{bmatrix}\)
\(2x + 2y = 4\)
\(x + y = 2 \dotso \color{red} (1)\)
\(x\:-\:4y = -3 \dotso \color{red} (2)\)
Kurangkan persamaan (1) dengan persamaan (2)
\(5y = 5 \rightarrow y = 1\)
Substitusikan \(y = 1\) ke dalam persamaan (1)
\(x + 1 = 2\)
\(x = 1\)
Jadi, nilai \(2x + 3y = 2(1) + 3(1) = 5\)
Salah
\(\textbf{A}^{-1}\cdot \textbf{B} = \textbf{C}^{\text{T}}\)
\(\textbf{A}\cdot\textbf{A}^{-1}\cdot \textbf{B} = \textbf{A}\cdot\textbf{C}^{\text{T}}\)
\(\textbf{B} = \textbf{A}\cdot\textbf{C}^{\text{T}}\)
\(\begin{bmatrix}2x + 2y & 2 \\ x\:-\:4y & -1 \end{bmatrix} = \begin{bmatrix}0 & -2 \\ 1 & 4 \end{bmatrix} \cdot \begin{bmatrix}5 & 3 \\ -2 & -1 \end{bmatrix}\)
\(\begin{bmatrix}2x + 2y & 2 \\ x\:-\:4y & -1 \end{bmatrix} = \begin{bmatrix}4 & 2 \\ -3 & -1 \end{bmatrix}\)
\(2x + 2y = 4\)
\(x + y = 2 \dotso \color{red} (1)\)
\(x\:-\:4y = -3 \dotso \color{red} (2)\)
Kurangkan persamaan (1) dengan persamaan (2)
\(5y = 5 \rightarrow y = 1\)
Substitusikan \(y = 1\) ke dalam persamaan (1)
\(x + 1 = 2\)
\(x = 1\)
Jadi, nilai \(2x + 3y = 2(1) + 3(1) = 5\)
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Pertanyaan 4 dari 6
4. Pertanyaan
1 pointsDiketahui matriks \(\textbf{A} = \begin{bmatrix}x & 4 \\ 2 & x\:-\:2 \end{bmatrix}\) adalah matriks singular. Jika penyelesaian dari matriks \(\textbf{A}\) adalah \(x_1\) dan \(x_2\) dengan \(x_1 > x_2\), maka nilai dari \(x_1^2 + 8x_2\) adalah…
Benar
Matriks singular adalah matriks yang determinannya sama dengan nol.
\(\text{ Det A} = x(x\:-\:2) \:-\:4(2)\)
\(0 = x^2\:-\:2x\:-\:8\)
\(0 = (x \:-\:4)(x + 2)\)
\(x \:-\:4 = 0 \rightarrow x = 4\)
\(x + 2 = 0 \rightarrow x = -2\)
Karena \(x_1 > x_2\) maka \(x_1 = 4\) dan \(x_2 = -2\)
\(x_1^2 + 8x_2 = 4^2 + 8(-2)\)
\(x_1^2 + 8x_2 = 16 \:-\:16\)
\(x_1^2 + 8x_2 = 0\)
Salah
Matriks singular adalah matriks yang determinannya sama dengan nol.
\(\text{ Det A} = x(x\:-\:2) \:-\:4(2)\)
\(0 = x^2\:-\:2x\:-\:8\)
\(0 = (x \:-\:4)(x + 2)\)
\(x \:-\:4 = 0 \rightarrow x = 4\)
\(x + 2 = 0 \rightarrow x = -2\)
Karena \(x_1 > x_2\) maka \(x_1 = 4\) dan \(x_2 = -2\)
\(x_1^2 + 8x_2 = 4^2 + 8(-2)\)
\(x_1^2 + 8x_2 = 16 \:-\:16\)
\(x_1^2 + 8x_2 = 0\)
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Pertanyaan 5 dari 6
5. Pertanyaan
1 pointsTentukan nilai dari \(x + 2y\) yang memenuhi persamaan \(\begin{bmatrix}2 & -5 \\ 3& 4 \end{bmatrix} \begin{bmatrix}x \\ y \end{bmatrix} = \begin{bmatrix}-19 \\ 29 \end{bmatrix} \)
Benar
\(\begin{bmatrix}2 & -5 \\ 3& 4 \end{bmatrix} \begin{bmatrix}x \\ y \end{bmatrix} = \begin{bmatrix}-19 \\ 29 \end{bmatrix} \)
\(\begin{bmatrix}x \\ y \end{bmatrix} = \begin{bmatrix}2 & -5 \\ 3& 4 \end{bmatrix}^{-1} \begin{bmatrix}-19 \\ 29 \end{bmatrix} \)
\(\begin{bmatrix}x \\ y \end{bmatrix} = \dfrac{1}{8\:-\:(-15)}\begin{bmatrix}4 & 5 \\ -3 & 2 \end{bmatrix} \begin{bmatrix}-19 \\ 29 \end{bmatrix}\)
\(\begin{bmatrix}x \\ y \end{bmatrix} = \dfrac{1}{23}\begin{bmatrix}4 & 5 \\ -3 & 2 \end{bmatrix} \begin{bmatrix}-19 \\ 29 \end{bmatrix}\)
\(\begin{bmatrix}x \\ y \end{bmatrix} = \dfrac{1}{23}\begin{bmatrix}69\\ 115 \end{bmatrix}\)
\(\begin{bmatrix}x \\ y \end{bmatrix} = \begin{bmatrix}3\\ 5 \end{bmatrix}\)
\(x = 3\)
\(y = 5\)
\(x + 2y = 3 + 2(5) = 13\)
Salah
\(\begin{bmatrix}2 & -5 \\ 3& 4 \end{bmatrix} \begin{bmatrix}x \\ y \end{bmatrix} = \begin{bmatrix}-19 \\ 29 \end{bmatrix} \)
\(\begin{bmatrix}x \\ y \end{bmatrix} = \begin{bmatrix}2 & -5 \\ 3& 4 \end{bmatrix}^{-1} \begin{bmatrix}-19 \\ 29 \end{bmatrix} \)
\(\begin{bmatrix}x \\ y \end{bmatrix} = \dfrac{1}{8\:-\:(-15)}\begin{bmatrix}4 & 5 \\ -3 & 2 \end{bmatrix} \begin{bmatrix}-19 \\ 29 \end{bmatrix}\)
\(\begin{bmatrix}x \\ y \end{bmatrix} = \dfrac{1}{23}\begin{bmatrix}4 & 5 \\ -3 & 2 \end{bmatrix} \begin{bmatrix}-19 \\ 29 \end{bmatrix}\)
\(\begin{bmatrix}x \\ y \end{bmatrix} = \dfrac{1}{23}\begin{bmatrix}69\\ 115 \end{bmatrix}\)
\(\begin{bmatrix}x \\ y \end{bmatrix} = \begin{bmatrix}3\\ 5 \end{bmatrix}\)
\(x = 3\)
\(y = 5\)
\(x + 2y = 3 + 2(5) = 13\)
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Pertanyaan 6 dari 6
6. Pertanyaan
1 pointsAgar sistem persamaan linear:
\(ax + by \:-\:3z = -3\)
\(-2x \:-\:by + cz = -1\)
\(ax + 3y \:-\:cz = -5\)
Mempunyai penyelesaian \(x = 3, y = -2\) dan \(z = 1\). Nilai dari \(a + 2b + c\) adalah…
Benar
Salah
Substitusikan \(x = 3, y = -2\) dan \(z = 1\) ke dalam setiap persamaan
\(3a \:-\:2b = 0\)
\(2b + c = 5\)
\(3a \:-\:c = 1\)
Persamaan di atas dapat ditulis:
\(\begin{bmatrix}3 & -2 & 0 \\ 0& 2 & 1 \\ 3 & 0 & -1 \end{bmatrix} \begin{bmatrix}a \\ b \\c \end{bmatrix} = \begin{bmatrix}0 \\ 5 \\1 \end{bmatrix} \)
Gunakan metode Cramer untuk menentukan nilai \(a, b, \text{ dan } c\)
\(a = \dfrac{\text{det} \begin{bmatrix}0 & -2 & 0\\ 5& 2 & 1 \\ 1 & 0 & -1 \end{bmatrix}}{\text{det} \begin{bmatrix}3 & -2 & 0 \\ 0& 2 & 1 \\ 3 & 0 & -1 \end{bmatrix}}\)
\(a = 1\)
\(b = \dfrac{\text{det} \begin{bmatrix}3 & 0 & 0\\ 0& 5 & 1 \\ 3 & 1 & -1 \end{bmatrix}}{\text{det} \begin{bmatrix}3 & -2 & 0 \\ 0& 2 & 1 \\ 3 & 0 & -1 \end{bmatrix}}\)
\(b = \dfrac{3}{2}\)
\(c = \dfrac{\text{det} \begin{bmatrix}3 & -2 & 0\\ 0& 2 & 5 \\ 3 & 0 & 1 \end{bmatrix}}{\text{det} \begin{bmatrix}3 & -2 & 0 \\ 0& 2 & 1 \\ 3 & 0 & -1 \end{bmatrix}}\)
\(c = 2 \)
\(a + 2b + c = 1 + 2\cdot \dfrac{3}{2} + 2 = 6\)