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Soal Matematika, selesaikan dan dapatkan poin tertinggi​

Berikut ini adalah pertanyaan dari opparyeon92 pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

Soal Matematika, selesaikan dan dapatkan poin tertinggi​
Soal Matematika, selesaikan dan dapatkan poin tertinggi​

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Jawab:

1. \displaystyle \begin{pmatrix}-2 & 1 & 6\\ 6 & -3 & -17\\ 1 & 0 & -3\end{pmatrix}

2. 4 satuan luas

3. π satuan volume

4. \displaystyle -\frac{\cos(2x+1)}{2}+C

5. \displaystyle \begin{pmatrix}-6 & -5\\ 5 & 4\end{pmatrix}

Penjelasan dengan langkah-langkah:

Nomor 1.

Tentukan determinan nya

\displaystyle |M|=\begin{vmatrix}9 & 3 & 1\\ 1 & 0 & 2\\ 3 & 1 & 0\end{vmatrix}\\=9\begin{vmatrix}0 & 2\\ 1 & 0\end{vmatrix}-3\begin{vmatrix}1 & 2\\ 3 & 0\end{vmatrix}+1\begin{vmatrix}1 & 0\\ 3 & 1\end{vmatrix}\\=9[0(0)-2(1)]-3[1(0)-2(3)]+1[1(1)-0(3)]\\=9(-2)-3(-6)+1(1)\\=1

Tentukan minor nya

\displaystyle M_M=\begin{pmatrix}\begin{vmatrix}0 & 2\\ 1 & 0\end{vmatrix} & \begin{vmatrix}1 & 2\\ 3 & 0\end{vmatrix} & \begin{vmatrix}1 & 0\\ 3 & 1\end{vmatrix}\\ \begin{vmatrix}3 & 1\\ 1 & 0\end{vmatrix} & \begin{vmatrix}9 & 1\\ 3 & 0\end{vmatrix} & \begin{vmatrix}9 & 3\\ 3 & 1\end{vmatrix}\\ \begin{vmatrix}3 & 1\\ 0 & 2\end{vmatrix} & \begin{vmatrix}9 & 1\\ 1 & 2\end{vmatrix} & \begin{vmatrix}9 & 3\\ 1 & 0\end{vmatrix}\end{pmatrix}

\displaystyle =\begin{pmatrix}-2 & -6 & 1\\ -1 & -3 & 0\\ 6 & 17 & -3\end{pmatrix}

Tentukan kofaktor nya

\displaystyle M_C=\begin{pmatrix}+(-2) & -(-6) & +1\\ -(-1) & +(-3) & -0\\ +6 & -17 & +(-3)\end{pmatrix}\\=\begin{pmatrix}-2 & 6 & 1\\ 1 & -3 & 0\\ 6 & -17 & -3\end{pmatrix}

Tentukan adjoin nya

\displaystyle M_{\textrm{adj}}=\begin{pmatrix}-2 & 1 & 6\\ 6 & -3 & -17\\ 1 & 0 & -3\end{pmatrix}

Tentukan invers nya

\displaystyle M^{-1}=\frac{1}{|M|}M_{\textrm{adj}}\\=\frac{1}{1}\begin{pmatrix}-2 & 1 & 6\\ 6 & -3 & -17\\ 1 & 0 & -3\end{pmatrix}\\=\begin{pmatrix}-2 & 1 & 6\\ 6 & -3 & -17\\ 1 & 0 & -3\end{pmatrix}

Nomor 2

Gambar grafik dan batas-batas nya (terlampir):

Luas dibawah kurva L = ₐ∫ᵇ f(x) dx sedangkan luas diatas kurva L = -ₐ∫ᵇ f(x) dx

\displaystyle L=L_1+L_2\\=\int_{\frac{\pi}{2}}^{\pi}2\sin x~dx-\int_{\pi}^{\frac{3\pi}{2}}2\sin x~dx\\=2\left [ -\cos x \right ]_\frac{\pi}{2}^\pi-2\left [ -\cos x \right ]_\pi^\frac{3\pi}{2}\\=2\left [ -\cos \pi+\cos \frac{\pi}{2} \right ]-2\left [ -\cos \frac{3\pi}{2}+\cos \pi \right ]\\=2[-(-1)+0]-2(-0-1)\\=2+2\\=4

Nomor 3

Sketsa grafik y = x²

\displaystyle y=x^2\\\begin{matrix}x & -2 & -1 & 0 & 1 & 2\\ y & 4 & 1 & 0 & 1 & 4\end{matrix}

Sketsa grafik y = 2 - x²

\displaystyle y=2-x^2\\\begin{matrix}x & -2 & -1 & 0 & 1 & 2\\ y & -2 & 1 & 2 & 1 & -2\end{matrix}

Karena hanya pada kuadran I, maka grafik menjadi berikut (terlampir):

Diputar terhadap sumbu Y sehingga gambar nya menjadi (terlampir):

Metode kulit tabung

\displaystyle V=2\pi\int_{a}^{b}x(y_1-y_2)dx\\=2\pi\int_{0}^{1}x(2-x^2-x^2)dx\\=2\pi\int_{0}^{1}(2x-2x^3)dx\\=2\pi\left [ x^2-\frac{x^4}{2} \right ]_0^1\\=2\pi\left \[ \left ( 1^2-\frac{1^4}{2} \right )-\left ( 0^2-\frac{0^4}{2} \right ) \right \]\\=2\pi\left ( \frac{1}{2}-0 \right )\\=\pi

Nomor 4

Metode substitusi

\displaystyle \int \sin (2x+1)dx\\=\int \sin u~\frac{du}{2}\\=\frac{1}{2}(-\cos u)+C\\=-\frac{\cos(2x+1)}{2}+C

Nomor 5

Persamaan matriks AX = B → X = A⁻¹B

\displaystyle \begin{pmatrix}3 & 4\\ 1 & 2\end{pmatrix}P=\begin{pmatrix}2 & 1\\ 4 & 3\end{pmatrix}\\P=\begin{pmatrix}3 & 4\\ 1 & 2\end{pmatrix}^{-1}\begin{pmatrix}2 & 1\\ 4 & 3\end{pmatrix}

\displaystyle =\frac{1}{3(2)-4(1)}\displaystyle \begin{pmatrix}2 & -4\\ -1 & 3\end{pmatrix}\begin{pmatrix}2 & 1\\ 4 & 3\end{pmatrix}\\=\begin{pmatrix}1 & -2\\ -\frac{1}{2} & \frac{3}{2}\end{pmatrix}\begin{pmatrix}2 & 1\\ 4 & 3\end{pmatrix}

\displaystyle =\begin{pmatrix}1(2)-2(4) & 1(1)-2(3)\\ -\frac{1}{2}(2)+\frac{3}{2}(4) & -\frac{1}{2}(1)+\frac{3}{2}(3)\end{pmatrix}\\=\begin{pmatrix}-6 & -5\\ 5 & 4\end{pmatrix}

Jawab:1. [tex]\displaystyle \begin{pmatrix}-2 & 1 & 6\\ 6 & -3 & -17\\ 1 & 0 & -3\end{pmatrix}[/tex]2. 4 satuan luas3. π satuan volume4. [tex]\displaystyle -\frac{\cos(2x+1)}{2}+C[/tex]5. [tex]\displaystyle \begin{pmatrix}-6 & -5\\ 5 & 4\end{pmatrix}[/tex]Penjelasan dengan langkah-langkah:Nomor 1.Tentukan determinan nya[tex]\displaystyle |M|=\begin{vmatrix}9 & 3 & 1\\ 1 & 0 & 2\\ 3 & 1 & 0\end{vmatrix}\\=9\begin{vmatrix}0 & 2\\ 1 & 0\end{vmatrix}-3\begin{vmatrix}1 & 2\\ 3 & 0\end{vmatrix}+1\begin{vmatrix}1 & 0\\ 3 & 1\end{vmatrix}\\=9[0(0)-2(1)]-3[1(0)-2(3)]+1[1(1)-0(3)]\\=9(-2)-3(-6)+1(1)\\=1[/tex]Tentukan minor nya[tex]\displaystyle M_M=\begin{pmatrix}\begin{vmatrix}0 & 2\\ 1 & 0\end{vmatrix} & \begin{vmatrix}1 & 2\\ 3 & 0\end{vmatrix} & \begin{vmatrix}1 & 0\\ 3 & 1\end{vmatrix}\\ \begin{vmatrix}3 & 1\\ 1 & 0\end{vmatrix} & \begin{vmatrix}9 & 1\\ 3 & 0\end{vmatrix} & \begin{vmatrix}9 & 3\\ 3 & 1\end{vmatrix}\\ \begin{vmatrix}3 & 1\\ 0 & 2\end{vmatrix} & \begin{vmatrix}9 & 1\\ 1 & 2\end{vmatrix} & \begin{vmatrix}9 & 3\\ 1 & 0\end{vmatrix}\end{pmatrix}[/tex][tex]\displaystyle =\begin{pmatrix}-2 & -6 & 1\\ -1 & -3 & 0\\ 6 & 17 & -3\end{pmatrix}[/tex]Tentukan kofaktor nya[tex]\displaystyle M_C=\begin{pmatrix}+(-2) & -(-6) & +1\\ -(-1) & +(-3) & -0\\ +6 & -17 & +(-3)\end{pmatrix}\\=\begin{pmatrix}-2 & 6 & 1\\ 1 & -3 & 0\\ 6 & -17 & -3\end{pmatrix}[/tex]Tentukan adjoin nya[tex]\displaystyle M_{\textrm{adj}}=\begin{pmatrix}-2 & 1 & 6\\ 6 & -3 & -17\\ 1 & 0 & -3\end{pmatrix}[/tex]Tentukan invers nya[tex]\displaystyle M^{-1}=\frac{1}{|M|}M_{\textrm{adj}}\\=\frac{1}{1}\begin{pmatrix}-2 & 1 & 6\\ 6 & -3 & -17\\ 1 & 0 & -3\end{pmatrix}\\=\begin{pmatrix}-2 & 1 & 6\\ 6 & -3 & -17\\ 1 & 0 & -3\end{pmatrix}[/tex]Nomor 2Gambar grafik dan batas-batas nya (terlampir):Luas dibawah kurva L = ₐ∫ᵇ f(x) dx sedangkan luas diatas kurva L = -ₐ∫ᵇ f(x) dx[tex]\displaystyle L=L_1+L_2\\=\int_{\frac{\pi}{2}}^{\pi}2\sin x~dx-\int_{\pi}^{\frac{3\pi}{2}}2\sin x~dx\\=2\left [ -\cos x \right ]_\frac{\pi}{2}^\pi-2\left [ -\cos x \right ]_\pi^\frac{3\pi}{2}\\=2\left [ -\cos \pi+\cos \frac{\pi}{2} \right ]-2\left [ -\cos \frac{3\pi}{2}+\cos \pi \right ]\\=2[-(-1)+0]-2(-0-1)\\=2+2\\=4[/tex]Nomor 3Sketsa grafik y = x²[tex]\displaystyle y=x^2\\\begin{matrix}x & -2 & -1 & 0 & 1 & 2\\ y & 4 & 1 & 0 & 1 & 4\end{matrix}[/tex]Sketsa grafik y = 2 - x²[tex]\displaystyle y=2-x^2\\\begin{matrix}x & -2 & -1 & 0 & 1 & 2\\ y & -2 & 1 & 2 & 1 & -2\end{matrix}[/tex]Karena hanya pada kuadran I, maka grafik menjadi berikut (terlampir):Diputar terhadap sumbu Y sehingga gambar nya menjadi (terlampir):Metode kulit tabung[tex]\displaystyle V=2\pi\int_{a}^{b}x(y_1-y_2)dx\\=2\pi\int_{0}^{1}x(2-x^2-x^2)dx\\=2\pi\int_{0}^{1}(2x-2x^3)dx\\=2\pi\left [ x^2-\frac{x^4}{2} \right ]_0^1\\=2\pi\left \[ \left ( 1^2-\frac{1^4}{2} \right )-\left ( 0^2-\frac{0^4}{2} \right ) \right \]\\=2\pi\left ( \frac{1}{2}-0 \right )\\=\pi[/tex]Nomor 4Metode substitusi[tex]\displaystyle \int \sin (2x+1)dx\\=\int \sin u~\frac{du}{2}\\=\frac{1}{2}(-\cos u)+C\\=-\frac{\cos(2x+1)}{2}+C[/tex]Nomor 5Persamaan matriks AX = B → X = A⁻¹B[tex]\displaystyle \begin{pmatrix}3 & 4\\ 1 & 2\end{pmatrix}P=\begin{pmatrix}2 & 1\\ 4 & 3\end{pmatrix}\\P=\begin{pmatrix}3 & 4\\ 1 & 2\end{pmatrix}^{-1}\begin{pmatrix}2 & 1\\ 4 & 3\end{pmatrix}[/tex][tex]\displaystyle =\frac{1}{3(2)-4(1)}\displaystyle \begin{pmatrix}2 & -4\\ -1 & 3\end{pmatrix}\begin{pmatrix}2 & 1\\ 4 & 3\end{pmatrix}\\=\begin{pmatrix}1 & -2\\ -\frac{1}{2} & \frac{3}{2}\end{pmatrix}\begin{pmatrix}2 & 1\\ 4 & 3\end{pmatrix}[/tex][tex]\displaystyle =\begin{pmatrix}1(2)-2(4) & 1(1)-2(3)\\ -\frac{1}{2}(2)+\frac{3}{2}(4) & -\frac{1}{2}(1)+\frac{3}{2}(3)\end{pmatrix}\\=\begin{pmatrix}-6 & -5\\ 5 & 4\end{pmatrix}[/tex]Jawab:1. [tex]\displaystyle \begin{pmatrix}-2 & 1 & 6\\ 6 & -3 & -17\\ 1 & 0 & -3\end{pmatrix}[/tex]2. 4 satuan luas3. π satuan volume4. [tex]\displaystyle -\frac{\cos(2x+1)}{2}+C[/tex]5. [tex]\displaystyle \begin{pmatrix}-6 & -5\\ 5 & 4\end{pmatrix}[/tex]Penjelasan dengan langkah-langkah:Nomor 1.Tentukan determinan nya[tex]\displaystyle |M|=\begin{vmatrix}9 & 3 & 1\\ 1 & 0 & 2\\ 3 & 1 & 0\end{vmatrix}\\=9\begin{vmatrix}0 & 2\\ 1 & 0\end{vmatrix}-3\begin{vmatrix}1 & 2\\ 3 & 0\end{vmatrix}+1\begin{vmatrix}1 & 0\\ 3 & 1\end{vmatrix}\\=9[0(0)-2(1)]-3[1(0)-2(3)]+1[1(1)-0(3)]\\=9(-2)-3(-6)+1(1)\\=1[/tex]Tentukan minor nya[tex]\displaystyle M_M=\begin{pmatrix}\begin{vmatrix}0 & 2\\ 1 & 0\end{vmatrix} & \begin{vmatrix}1 & 2\\ 3 & 0\end{vmatrix} & \begin{vmatrix}1 & 0\\ 3 & 1\end{vmatrix}\\ \begin{vmatrix}3 & 1\\ 1 & 0\end{vmatrix} & \begin{vmatrix}9 & 1\\ 3 & 0\end{vmatrix} & \begin{vmatrix}9 & 3\\ 3 & 1\end{vmatrix}\\ \begin{vmatrix}3 & 1\\ 0 & 2\end{vmatrix} & \begin{vmatrix}9 & 1\\ 1 & 2\end{vmatrix} & \begin{vmatrix}9 & 3\\ 1 & 0\end{vmatrix}\end{pmatrix}[/tex][tex]\displaystyle =\begin{pmatrix}-2 & -6 & 1\\ -1 & -3 & 0\\ 6 & 17 & -3\end{pmatrix}[/tex]Tentukan kofaktor nya[tex]\displaystyle M_C=\begin{pmatrix}+(-2) & -(-6) & +1\\ -(-1) & +(-3) & -0\\ +6 & -17 & +(-3)\end{pmatrix}\\=\begin{pmatrix}-2 & 6 & 1\\ 1 & -3 & 0\\ 6 & -17 & -3\end{pmatrix}[/tex]Tentukan adjoin nya[tex]\displaystyle M_{\textrm{adj}}=\begin{pmatrix}-2 & 1 & 6\\ 6 & -3 & -17\\ 1 & 0 & -3\end{pmatrix}[/tex]Tentukan invers nya[tex]\displaystyle M^{-1}=\frac{1}{|M|}M_{\textrm{adj}}\\=\frac{1}{1}\begin{pmatrix}-2 & 1 & 6\\ 6 & -3 & -17\\ 1 & 0 & -3\end{pmatrix}\\=\begin{pmatrix}-2 & 1 & 6\\ 6 & -3 & -17\\ 1 & 0 & -3\end{pmatrix}[/tex]Nomor 2Gambar grafik dan batas-batas nya (terlampir):Luas dibawah kurva L = ₐ∫ᵇ f(x) dx sedangkan luas diatas kurva L = -ₐ∫ᵇ f(x) dx[tex]\displaystyle L=L_1+L_2\\=\int_{\frac{\pi}{2}}^{\pi}2\sin x~dx-\int_{\pi}^{\frac{3\pi}{2}}2\sin x~dx\\=2\left [ -\cos x \right ]_\frac{\pi}{2}^\pi-2\left [ -\cos x \right ]_\pi^\frac{3\pi}{2}\\=2\left [ -\cos \pi+\cos \frac{\pi}{2} \right ]-2\left [ -\cos \frac{3\pi}{2}+\cos \pi \right ]\\=2[-(-1)+0]-2(-0-1)\\=2+2\\=4[/tex]Nomor 3Sketsa grafik y = x²[tex]\displaystyle y=x^2\\\begin{matrix}x & -2 & -1 & 0 & 1 & 2\\ y & 4 & 1 & 0 & 1 & 4\end{matrix}[/tex]Sketsa grafik y = 2 - x²[tex]\displaystyle y=2-x^2\\\begin{matrix}x & -2 & -1 & 0 & 1 & 2\\ y & -2 & 1 & 2 & 1 & -2\end{matrix}[/tex]Karena hanya pada kuadran I, maka grafik menjadi berikut (terlampir):Diputar terhadap sumbu Y sehingga gambar nya menjadi (terlampir):Metode kulit tabung[tex]\displaystyle V=2\pi\int_{a}^{b}x(y_1-y_2)dx\\=2\pi\int_{0}^{1}x(2-x^2-x^2)dx\\=2\pi\int_{0}^{1}(2x-2x^3)dx\\=2\pi\left [ x^2-\frac{x^4}{2} \right ]_0^1\\=2\pi\left \[ \left ( 1^2-\frac{1^4}{2} \right )-\left ( 0^2-\frac{0^4}{2} \right ) \right \]\\=2\pi\left ( \frac{1}{2}-0 \right )\\=\pi[/tex]Nomor 4Metode substitusi[tex]\displaystyle \int \sin (2x+1)dx\\=\int \sin u~\frac{du}{2}\\=\frac{1}{2}(-\cos u)+C\\=-\frac{\cos(2x+1)}{2}+C[/tex]Nomor 5Persamaan matriks AX = B → X = A⁻¹B[tex]\displaystyle \begin{pmatrix}3 & 4\\ 1 & 2\end{pmatrix}P=\begin{pmatrix}2 & 1\\ 4 & 3\end{pmatrix}\\P=\begin{pmatrix}3 & 4\\ 1 & 2\end{pmatrix}^{-1}\begin{pmatrix}2 & 1\\ 4 & 3\end{pmatrix}[/tex][tex]\displaystyle =\frac{1}{3(2)-4(1)}\displaystyle \begin{pmatrix}2 & -4\\ -1 & 3\end{pmatrix}\begin{pmatrix}2 & 1\\ 4 & 3\end{pmatrix}\\=\begin{pmatrix}1 & -2\\ -\frac{1}{2} & \frac{3}{2}\end{pmatrix}\begin{pmatrix}2 & 1\\ 4 & 3\end{pmatrix}[/tex][tex]\displaystyle =\begin{pmatrix}1(2)-2(4) & 1(1)-2(3)\\ -\frac{1}{2}(2)+\frac{3}{2}(4) & -\frac{1}{2}(1)+\frac{3}{2}(3)\end{pmatrix}\\=\begin{pmatrix}-6 & -5\\ 5 & 4\end{pmatrix}[/tex]Jawab:1. [tex]\displaystyle \begin{pmatrix}-2 & 1 & 6\\ 6 & -3 & -17\\ 1 & 0 & -3\end{pmatrix}[/tex]2. 4 satuan luas3. π satuan volume4. [tex]\displaystyle -\frac{\cos(2x+1)}{2}+C[/tex]5. [tex]\displaystyle \begin{pmatrix}-6 & -5\\ 5 & 4\end{pmatrix}[/tex]Penjelasan dengan langkah-langkah:Nomor 1.Tentukan determinan nya[tex]\displaystyle |M|=\begin{vmatrix}9 & 3 & 1\\ 1 & 0 & 2\\ 3 & 1 & 0\end{vmatrix}\\=9\begin{vmatrix}0 & 2\\ 1 & 0\end{vmatrix}-3\begin{vmatrix}1 & 2\\ 3 & 0\end{vmatrix}+1\begin{vmatrix}1 & 0\\ 3 & 1\end{vmatrix}\\=9[0(0)-2(1)]-3[1(0)-2(3)]+1[1(1)-0(3)]\\=9(-2)-3(-6)+1(1)\\=1[/tex]Tentukan minor nya[tex]\displaystyle M_M=\begin{pmatrix}\begin{vmatrix}0 & 2\\ 1 & 0\end{vmatrix} & \begin{vmatrix}1 & 2\\ 3 & 0\end{vmatrix} & \begin{vmatrix}1 & 0\\ 3 & 1\end{vmatrix}\\ \begin{vmatrix}3 & 1\\ 1 & 0\end{vmatrix} & \begin{vmatrix}9 & 1\\ 3 & 0\end{vmatrix} & \begin{vmatrix}9 & 3\\ 3 & 1\end{vmatrix}\\ \begin{vmatrix}3 & 1\\ 0 & 2\end{vmatrix} & \begin{vmatrix}9 & 1\\ 1 & 2\end{vmatrix} & \begin{vmatrix}9 & 3\\ 1 & 0\end{vmatrix}\end{pmatrix}[/tex][tex]\displaystyle =\begin{pmatrix}-2 & -6 & 1\\ -1 & -3 & 0\\ 6 & 17 & -3\end{pmatrix}[/tex]Tentukan kofaktor nya[tex]\displaystyle M_C=\begin{pmatrix}+(-2) & -(-6) & +1\\ -(-1) & +(-3) & -0\\ +6 & -17 & +(-3)\end{pmatrix}\\=\begin{pmatrix}-2 & 6 & 1\\ 1 & -3 & 0\\ 6 & -17 & -3\end{pmatrix}[/tex]Tentukan adjoin nya[tex]\displaystyle M_{\textrm{adj}}=\begin{pmatrix}-2 & 1 & 6\\ 6 & -3 & -17\\ 1 & 0 & -3\end{pmatrix}[/tex]Tentukan invers nya[tex]\displaystyle M^{-1}=\frac{1}{|M|}M_{\textrm{adj}}\\=\frac{1}{1}\begin{pmatrix}-2 & 1 & 6\\ 6 & -3 & -17\\ 1 & 0 & -3\end{pmatrix}\\=\begin{pmatrix}-2 & 1 & 6\\ 6 & -3 & -17\\ 1 & 0 & -3\end{pmatrix}[/tex]Nomor 2Gambar grafik dan batas-batas nya (terlampir):Luas dibawah kurva L = ₐ∫ᵇ f(x) dx sedangkan luas diatas kurva L = -ₐ∫ᵇ f(x) dx[tex]\displaystyle L=L_1+L_2\\=\int_{\frac{\pi}{2}}^{\pi}2\sin x~dx-\int_{\pi}^{\frac{3\pi}{2}}2\sin x~dx\\=2\left [ -\cos x \right ]_\frac{\pi}{2}^\pi-2\left [ -\cos x \right ]_\pi^\frac{3\pi}{2}\\=2\left [ -\cos \pi+\cos \frac{\pi}{2} \right ]-2\left [ -\cos \frac{3\pi}{2}+\cos \pi \right ]\\=2[-(-1)+0]-2(-0-1)\\=2+2\\=4[/tex]Nomor 3Sketsa grafik y = x²[tex]\displaystyle y=x^2\\\begin{matrix}x & -2 & -1 & 0 & 1 & 2\\ y & 4 & 1 & 0 & 1 & 4\end{matrix}[/tex]Sketsa grafik y = 2 - x²[tex]\displaystyle y=2-x^2\\\begin{matrix}x & -2 & -1 & 0 & 1 & 2\\ y & -2 & 1 & 2 & 1 & -2\end{matrix}[/tex]Karena hanya pada kuadran I, maka grafik menjadi berikut (terlampir):Diputar terhadap sumbu Y sehingga gambar nya menjadi (terlampir):Metode kulit tabung[tex]\displaystyle V=2\pi\int_{a}^{b}x(y_1-y_2)dx\\=2\pi\int_{0}^{1}x(2-x^2-x^2)dx\\=2\pi\int_{0}^{1}(2x-2x^3)dx\\=2\pi\left [ x^2-\frac{x^4}{2} \right ]_0^1\\=2\pi\left \[ \left ( 1^2-\frac{1^4}{2} \right )-\left ( 0^2-\frac{0^4}{2} \right ) \right \]\\=2\pi\left ( \frac{1}{2}-0 \right )\\=\pi[/tex]Nomor 4Metode substitusi[tex]\displaystyle \int \sin (2x+1)dx\\=\int \sin u~\frac{du}{2}\\=\frac{1}{2}(-\cos u)+C\\=-\frac{\cos(2x+1)}{2}+C[/tex]Nomor 5Persamaan matriks AX = B → X = A⁻¹B[tex]\displaystyle \begin{pmatrix}3 & 4\\ 1 & 2\end{pmatrix}P=\begin{pmatrix}2 & 1\\ 4 & 3\end{pmatrix}\\P=\begin{pmatrix}3 & 4\\ 1 & 2\end{pmatrix}^{-1}\begin{pmatrix}2 & 1\\ 4 & 3\end{pmatrix}[/tex][tex]\displaystyle =\frac{1}{3(2)-4(1)}\displaystyle \begin{pmatrix}2 & -4\\ -1 & 3\end{pmatrix}\begin{pmatrix}2 & 1\\ 4 & 3\end{pmatrix}\\=\begin{pmatrix}1 & -2\\ -\frac{1}{2} & \frac{3}{2}\end{pmatrix}\begin{pmatrix}2 & 1\\ 4 & 3\end{pmatrix}[/tex][tex]\displaystyle =\begin{pmatrix}1(2)-2(4) & 1(1)-2(3)\\ -\frac{1}{2}(2)+\frac{3}{2}(4) & -\frac{1}{2}(1)+\frac{3}{2}(3)\end{pmatrix}\\=\begin{pmatrix}-6 & -5\\ 5 & 4\end{pmatrix}[/tex]Jawab:1. [tex]\displaystyle \begin{pmatrix}-2 & 1 & 6\\ 6 & -3 & -17\\ 1 & 0 & -3\end{pmatrix}[/tex]2. 4 satuan luas3. π satuan volume4. [tex]\displaystyle -\frac{\cos(2x+1)}{2}+C[/tex]5. [tex]\displaystyle \begin{pmatrix}-6 & -5\\ 5 & 4\end{pmatrix}[/tex]Penjelasan dengan langkah-langkah:Nomor 1.Tentukan determinan nya[tex]\displaystyle |M|=\begin{vmatrix}9 & 3 & 1\\ 1 & 0 & 2\\ 3 & 1 & 0\end{vmatrix}\\=9\begin{vmatrix}0 & 2\\ 1 & 0\end{vmatrix}-3\begin{vmatrix}1 & 2\\ 3 & 0\end{vmatrix}+1\begin{vmatrix}1 & 0\\ 3 & 1\end{vmatrix}\\=9[0(0)-2(1)]-3[1(0)-2(3)]+1[1(1)-0(3)]\\=9(-2)-3(-6)+1(1)\\=1[/tex]Tentukan minor nya[tex]\displaystyle M_M=\begin{pmatrix}\begin{vmatrix}0 & 2\\ 1 & 0\end{vmatrix} & \begin{vmatrix}1 & 2\\ 3 & 0\end{vmatrix} & \begin{vmatrix}1 & 0\\ 3 & 1\end{vmatrix}\\ \begin{vmatrix}3 & 1\\ 1 & 0\end{vmatrix} & \begin{vmatrix}9 & 1\\ 3 & 0\end{vmatrix} & \begin{vmatrix}9 & 3\\ 3 & 1\end{vmatrix}\\ \begin{vmatrix}3 & 1\\ 0 & 2\end{vmatrix} & \begin{vmatrix}9 & 1\\ 1 & 2\end{vmatrix} & \begin{vmatrix}9 & 3\\ 1 & 0\end{vmatrix}\end{pmatrix}[/tex][tex]\displaystyle =\begin{pmatrix}-2 & -6 & 1\\ -1 & -3 & 0\\ 6 & 17 & -3\end{pmatrix}[/tex]Tentukan kofaktor nya[tex]\displaystyle M_C=\begin{pmatrix}+(-2) & -(-6) & +1\\ -(-1) & +(-3) & -0\\ +6 & -17 & +(-3)\end{pmatrix}\\=\begin{pmatrix}-2 & 6 & 1\\ 1 & -3 & 0\\ 6 & -17 & -3\end{pmatrix}[/tex]Tentukan adjoin nya[tex]\displaystyle M_{\textrm{adj}}=\begin{pmatrix}-2 & 1 & 6\\ 6 & -3 & -17\\ 1 & 0 & -3\end{pmatrix}[/tex]Tentukan invers nya[tex]\displaystyle M^{-1}=\frac{1}{|M|}M_{\textrm{adj}}\\=\frac{1}{1}\begin{pmatrix}-2 & 1 & 6\\ 6 & -3 & -17\\ 1 & 0 & -3\end{pmatrix}\\=\begin{pmatrix}-2 & 1 & 6\\ 6 & -3 & -17\\ 1 & 0 & -3\end{pmatrix}[/tex]Nomor 2Gambar grafik dan batas-batas nya (terlampir):Luas dibawah kurva L = ₐ∫ᵇ f(x) dx sedangkan luas diatas kurva L = -ₐ∫ᵇ f(x) dx[tex]\displaystyle L=L_1+L_2\\=\int_{\frac{\pi}{2}}^{\pi}2\sin x~dx-\int_{\pi}^{\frac{3\pi}{2}}2\sin x~dx\\=2\left [ -\cos x \right ]_\frac{\pi}{2}^\pi-2\left [ -\cos x \right ]_\pi^\frac{3\pi}{2}\\=2\left [ -\cos \pi+\cos \frac{\pi}{2} \right ]-2\left [ -\cos \frac{3\pi}{2}+\cos \pi \right ]\\=2[-(-1)+0]-2(-0-1)\\=2+2\\=4[/tex]Nomor 3Sketsa grafik y = x²[tex]\displaystyle y=x^2\\\begin{matrix}x & -2 & -1 & 0 & 1 & 2\\ y & 4 & 1 & 0 & 1 & 4\end{matrix}[/tex]Sketsa grafik y = 2 - x²[tex]\displaystyle y=2-x^2\\\begin{matrix}x & -2 & -1 & 0 & 1 & 2\\ y & -2 & 1 & 2 & 1 & -2\end{matrix}[/tex]Karena hanya pada kuadran I, maka grafik menjadi berikut (terlampir):Diputar terhadap sumbu Y sehingga gambar nya menjadi (terlampir):Metode kulit tabung[tex]\displaystyle V=2\pi\int_{a}^{b}x(y_1-y_2)dx\\=2\pi\int_{0}^{1}x(2-x^2-x^2)dx\\=2\pi\int_{0}^{1}(2x-2x^3)dx\\=2\pi\left [ x^2-\frac{x^4}{2} \right ]_0^1\\=2\pi\left \[ \left ( 1^2-\frac{1^4}{2} \right )-\left ( 0^2-\frac{0^4}{2} \right ) \right \]\\=2\pi\left ( \frac{1}{2}-0 \right )\\=\pi[/tex]Nomor 4Metode substitusi[tex]\displaystyle \int \sin (2x+1)dx\\=\int \sin u~\frac{du}{2}\\=\frac{1}{2}(-\cos u)+C\\=-\frac{\cos(2x+1)}{2}+C[/tex]Nomor 5Persamaan matriks AX = B → X = A⁻¹B[tex]\displaystyle \begin{pmatrix}3 & 4\\ 1 & 2\end{pmatrix}P=\begin{pmatrix}2 & 1\\ 4 & 3\end{pmatrix}\\P=\begin{pmatrix}3 & 4\\ 1 & 2\end{pmatrix}^{-1}\begin{pmatrix}2 & 1\\ 4 & 3\end{pmatrix}[/tex][tex]\displaystyle =\frac{1}{3(2)-4(1)}\displaystyle \begin{pmatrix}2 & -4\\ -1 & 3\end{pmatrix}\begin{pmatrix}2 & 1\\ 4 & 3\end{pmatrix}\\=\begin{pmatrix}1 & -2\\ -\frac{1}{2} & \frac{3}{2}\end{pmatrix}\begin{pmatrix}2 & 1\\ 4 & 3\end{pmatrix}[/tex][tex]\displaystyle =\begin{pmatrix}1(2)-2(4) & 1(1)-2(3)\\ -\frac{1}{2}(2)+\frac{3}{2}(4) & -\frac{1}{2}(1)+\frac{3}{2}(3)\end{pmatrix}\\=\begin{pmatrix}-6 & -5\\ 5 & 4\end{pmatrix}[/tex]Jawab:1. [tex]\displaystyle \begin{pmatrix}-2 & 1 & 6\\ 6 & -3 & -17\\ 1 & 0 & -3\end{pmatrix}[/tex]2. 4 satuan luas3. π satuan volume4. [tex]\displaystyle -\frac{\cos(2x+1)}{2}+C[/tex]5. [tex]\displaystyle \begin{pmatrix}-6 & -5\\ 5 & 4\end{pmatrix}[/tex]Penjelasan dengan langkah-langkah:Nomor 1.Tentukan determinan nya[tex]\displaystyle |M|=\begin{vmatrix}9 & 3 & 1\\ 1 & 0 & 2\\ 3 & 1 & 0\end{vmatrix}\\=9\begin{vmatrix}0 & 2\\ 1 & 0\end{vmatrix}-3\begin{vmatrix}1 & 2\\ 3 & 0\end{vmatrix}+1\begin{vmatrix}1 & 0\\ 3 & 1\end{vmatrix}\\=9[0(0)-2(1)]-3[1(0)-2(3)]+1[1(1)-0(3)]\\=9(-2)-3(-6)+1(1)\\=1[/tex]Tentukan minor nya[tex]\displaystyle M_M=\begin{pmatrix}\begin{vmatrix}0 & 2\\ 1 & 0\end{vmatrix} & \begin{vmatrix}1 & 2\\ 3 & 0\end{vmatrix} & \begin{vmatrix}1 & 0\\ 3 & 1\end{vmatrix}\\ \begin{vmatrix}3 & 1\\ 1 & 0\end{vmatrix} & \begin{vmatrix}9 & 1\\ 3 & 0\end{vmatrix} & \begin{vmatrix}9 & 3\\ 3 & 1\end{vmatrix}\\ \begin{vmatrix}3 & 1\\ 0 & 2\end{vmatrix} & \begin{vmatrix}9 & 1\\ 1 & 2\end{vmatrix} & \begin{vmatrix}9 & 3\\ 1 & 0\end{vmatrix}\end{pmatrix}[/tex][tex]\displaystyle =\begin{pmatrix}-2 & -6 & 1\\ -1 & -3 & 0\\ 6 & 17 & -3\end{pmatrix}[/tex]Tentukan kofaktor nya[tex]\displaystyle M_C=\begin{pmatrix}+(-2) & -(-6) & +1\\ -(-1) & +(-3) & -0\\ +6 & -17 & +(-3)\end{pmatrix}\\=\begin{pmatrix}-2 & 6 & 1\\ 1 & -3 & 0\\ 6 & -17 & -3\end{pmatrix}[/tex]Tentukan adjoin nya[tex]\displaystyle M_{\textrm{adj}}=\begin{pmatrix}-2 & 1 & 6\\ 6 & -3 & -17\\ 1 & 0 & -3\end{pmatrix}[/tex]Tentukan invers nya[tex]\displaystyle M^{-1}=\frac{1}{|M|}M_{\textrm{adj}}\\=\frac{1}{1}\begin{pmatrix}-2 & 1 & 6\\ 6 & -3 & -17\\ 1 & 0 & -3\end{pmatrix}\\=\begin{pmatrix}-2 & 1 & 6\\ 6 & -3 & -17\\ 1 & 0 & -3\end{pmatrix}[/tex]Nomor 2Gambar grafik dan batas-batas nya (terlampir):Luas dibawah kurva L = ₐ∫ᵇ f(x) dx sedangkan luas diatas kurva L = -ₐ∫ᵇ f(x) dx[tex]\displaystyle L=L_1+L_2\\=\int_{\frac{\pi}{2}}^{\pi}2\sin x~dx-\int_{\pi}^{\frac{3\pi}{2}}2\sin x~dx\\=2\left [ -\cos x \right ]_\frac{\pi}{2}^\pi-2\left [ -\cos x \right ]_\pi^\frac{3\pi}{2}\\=2\left [ -\cos \pi+\cos \frac{\pi}{2} \right ]-2\left [ -\cos \frac{3\pi}{2}+\cos \pi \right ]\\=2[-(-1)+0]-2(-0-1)\\=2+2\\=4[/tex]Nomor 3Sketsa grafik y = x²[tex]\displaystyle y=x^2\\\begin{matrix}x & -2 & -1 & 0 & 1 & 2\\ y & 4 & 1 & 0 & 1 & 4\end{matrix}[/tex]Sketsa grafik y = 2 - x²[tex]\displaystyle y=2-x^2\\\begin{matrix}x & -2 & -1 & 0 & 1 & 2\\ y & -2 & 1 & 2 & 1 & -2\end{matrix}[/tex]Karena hanya pada kuadran I, maka grafik menjadi berikut (terlampir):Diputar terhadap sumbu Y sehingga gambar nya menjadi (terlampir):Metode kulit tabung[tex]\displaystyle V=2\pi\int_{a}^{b}x(y_1-y_2)dx\\=2\pi\int_{0}^{1}x(2-x^2-x^2)dx\\=2\pi\int_{0}^{1}(2x-2x^3)dx\\=2\pi\left [ x^2-\frac{x^4}{2} \right ]_0^1\\=2\pi\left \[ \left ( 1^2-\frac{1^4}{2} \right )-\left ( 0^2-\frac{0^4}{2} \right ) \right \]\\=2\pi\left ( \frac{1}{2}-0 \right )\\=\pi[/tex]Nomor 4Metode substitusi[tex]\displaystyle \int \sin (2x+1)dx\\=\int \sin u~\frac{du}{2}\\=\frac{1}{2}(-\cos u)+C\\=-\frac{\cos(2x+1)}{2}+C[/tex]Nomor 5Persamaan matriks AX = B → X = A⁻¹B[tex]\displaystyle \begin{pmatrix}3 & 4\\ 1 & 2\end{pmatrix}P=\begin{pmatrix}2 & 1\\ 4 & 3\end{pmatrix}\\P=\begin{pmatrix}3 & 4\\ 1 & 2\end{pmatrix}^{-1}\begin{pmatrix}2 & 1\\ 4 & 3\end{pmatrix}[/tex][tex]\displaystyle =\frac{1}{3(2)-4(1)}\displaystyle \begin{pmatrix}2 & -4\\ -1 & 3\end{pmatrix}\begin{pmatrix}2 & 1\\ 4 & 3\end{pmatrix}\\=\begin{pmatrix}1 & -2\\ -\frac{1}{2} & \frac{3}{2}\end{pmatrix}\begin{pmatrix}2 & 1\\ 4 & 3\end{pmatrix}[/tex][tex]\displaystyle =\begin{pmatrix}1(2)-2(4) & 1(1)-2(3)\\ -\frac{1}{2}(2)+\frac{3}{2}(4) & -\frac{1}{2}(1)+\frac{3}{2}(3)\end{pmatrix}\\=\begin{pmatrix}-6 & -5\\ 5 & 4\end{pmatrix}[/tex]

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Last Update: Thu, 16 Feb 23
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