Jawaban Soal Latihan 1 Penjumlahan Dan Pengurangan Matriks

Soal Latihan 1

1. Diketahui matriks A, B dan C sebagai berikut :
   
   A = $\left(\left(,,\begin{array}{cc}-3& 2\\ 1& 3\end{array},,\right)\right)$
   
   B = $\left(\left(,,\begin{array}{cc}4& 1\\ 2& -5\end{array},,\right)\right)$
   
   C = $\left(\left(,,\begin{array}{cc}2& -2\\ 3& -1\end{array},,\right)\right)$
   
   Tentukan :
   1. A + B
   2. A - B
   3. B + C
   4. A - C
   5. C - B


2. Tentukan nilai x, y dan z dari persamaan matriks berikut :
   
   $\left(\left(,,\begin{array}{cc}\mathrm{x}& \mathrm{2y}\\ 8& \mathrm{3z}\end{array},,\right)\right)+\left(\left(,,\begin{array}{cc}5& 3\\ 1& 6\end{array},,\right)\right)=\left(\left(,,\begin{array}{cc}10& 11\\ 9& 15\end{array},,\right)\right)$
   
   


3. Diketahui matriks A dan B sebagai berikut :
   
   A = $\left(\left(,,\begin{array}{cc}6& \mathrm{z}\\ 10& \mathrm{x}\end{array},,\right)\right)$
   
   B = $\left(\left(,,\begin{array}{cc}\mathrm{2x}& \mathrm{2x}\\ 10& \frac{3}{2}\mathrm{y}\end{array},,\right)\right)$
   
   Tentukan nilai x, y dan z jika A - B = O

Jawab :

1. 1. A + B = $\left(\left(,,\begin{array}{cc}-3& 2\\ 1& 3\end{array},,\right)\right)+\left(\left(,,\begin{array}{cc}4& 1\\ 2& -5\end{array},,\right)\right)=\left(\left(,,\begin{array}{cc}1& 3\\ 3& -2\end{array},,\right)\right)$
      
   
   
   2. A - B = $\left(\left(,,\begin{array}{cc}-3& 2\\ 1& 3\end{array},,\right)\right)-\left(\left(,,\begin{array}{cc}4& 1\\ 2& -5\end{array},,\right)\right)=\left(\left(,,\begin{array}{cc}-7& 1\\ -1& 8\end{array},,\right)\right)$
      
      
   3. B + C = $\left(\left(,,\begin{array}{cc}4& 1\\ 2& -5\end{array},,\right)\right)+\left(\left(,,\begin{array}{cc}2& -2\\ 3& -1\end{array},,\right)\right)=\left(\left(,,\begin{array}{cc}6& -1\\ 5& -6\end{array},,\right)\right)$
      
      
   4. A - C = $\left(\left(,,\begin{array}{cc}-3& 2\\ 1& 3\end{array},,\right)\right)-\left(\left(,,\begin{array}{cc}2& -2\\ 3& -1\end{array},,\right)\right)=\left(\left(,,\begin{array}{cc}-5& 4\\ -2& 4\end{array},,\right)\right)$
      
      
   5. C - B = $\left(\left(,,\begin{array}{cc}2& -2\\ 3& -1\end{array},,\right)\right)-\left(\left(,,\begin{array}{cc}4& 1\\ 2& -5\end{array},,\right)\right)=\left(\left(,,\begin{array}{cc}-2& -3\\ 1& 4\end{array},,\right)\right)$
      
      


2. $\left(\left(,,\begin{array}{cc}\mathrm{x}& \mathrm{2y}\\ 8& \mathrm{3z}\end{array},,\right)\right)+\left(\left(,,\begin{array}{cc}5& 3\\ 1& 6\end{array},,\right)\right)=\left(\left(,,\begin{array}{cc}10& 11\\ 9& 15\end{array},,\right)\right)$
   
   x + 5 = 10
   x = 10 - 5
   x = 5
   
   2y + 3 = 11
   2y = 11 - 3
   2y = 8
   y = 4
   
   3z + 6 = 15
   3z = 15 - 6
   3z = 9
   z = 3


3. A - B = O
   
   $\left(\left(,,\begin{array}{cc}6& \mathrm{z}\\ 10& \mathrm{x}\end{array},,\right)\right)-\left(\left(,,\begin{array}{cc}\mathrm{2x}& \mathrm{2x}\\ 10& \frac{3}{2}\mathrm{y}\end{array},,\right)\right)=\left(\left(,,\begin{array}{cc}0& 0\\ 0& 0\end{array},,\right)\right)$
   
   6 - 2x = 0
   6 = 2x
   x = 3
   
   z - 2x = 0
   z - 2(3) = 0
   z = 6
   
   $\mathrm{x}-\frac{3}{2}\mathrm{y}=0$
   $3-\frac{3}{2}\mathrm{y}=0$
   $3=\frac{3}{2}\mathrm{y}$
   $\mathrm{y}=3.\frac{2}{3}$
   $\mathrm{y}=2$
   

Jawaban Soal Latihan 1 Matriks

Soal Latihan 1

1. Diketahui matriks A sebagai berikut :
   
   A = $\left(\left(,,\begin{array}{ccccc}9& 0& 3& -7& 6\\ -5& 6& 8& 1& 4\\ 5& 7& 0& 2& 8\end{array},,\right)\right)$
   
   
   1. Sebutkan ordo dari matriks A
   2. Sebutkan elemen-elemen kolom ke 2
   3. Sebutkan elemen-elemen baris ke 3
   4. Jika a_ij menyatakan elemen baris ke i dan kolom ke j maka tentukan a₂₃, a₁₄ dan a₃₅


2. Berikan contoh matriks berordo 3x3 untuk :
   1. Matriks diagonal
   2. Matriks identitas
   3. Matriks segitiga atas
   4. Matriks segitiga bawah


3. Tulislah transpose dari matriks berikut :
   
   
   1. A = $\left(\left(,,\begin{array}{cc}4& 3\\ 5& 6\end{array},,\right)\right)$
   
   
   2. B = $\left(\left(,,\begin{array}{cc}2& 4\\ -3& 1\\ 0& 5\end{array},,\right)\right)$
   
   
   3. C = $\left(\left(,,\begin{array}{ccc}4& 5& -1\\ 2& 3& 0\end{array},,\right)\right)$
   
   
   4. D = $\left(\left(,,\begin{array}{ccc}4& 0& 5\\ 2& 4& 7\\ -1& 3& -2\end{array},,\right)\right)$


4. Tentukan nilai x dan y dari persamaan berikut :
   
   
   1. $\left(\left(,,\begin{array}{cc}\mathrm{x}& \mathrm{2y}\\ 0& 3\end{array},,\right)\right)=\left(\left(,,\begin{array}{cc}1& 8\\ 0& 3\end{array},,\right)\right)$
   
   
   2. $\left(\left(,,\begin{array}{cc}\mathrm{x\; +\; y}& 5\\ 4& \mathrm{x\; -\; y}\end{array},,\right)\right)=\left(\left(,,\begin{array}{cc}5& 5\\ 4& 1\end{array},,\right)\right)$
   
   
   3. $\left(\left(,,\begin{array}{cc}\mathrm{x}& 4\\ \mathrm{2y}& \mathrm{3z}\end{array},,\right)\right)=\left(\left(,,\begin{array}{cc}\mathrm{4z\; -\; 6y}& \mathrm{2x}\\ \mathrm{4x\; +\; 2}& \mathrm{2y\; +\; 14}\end{array},,\right)\right)$


5. Tentukan x dan y jika A = B^t
   
   
   1. A = $\left(\left(,,\begin{array}{cc}\mathrm{2x}& 0\\ 1& \mathrm{-4y}\end{array},,\right)\right)$ dan B = $\left(\left(,,\begin{array}{cc}8& 1\\ 0& 40\end{array},,\right)\right)$
   
   
   2. A = $\left(\left(,,\begin{array}{cc}\mathrm{x\; -\; 2y}& 1\\ 0& \mathrm{-3y}\end{array},,\right)\right)$ dan B = $\left(\left(,,\begin{array}{cc}8& 0\\ 1& 40\end{array},,\right)\right)$

Jawab :

1. 1. ordo mariks A = 3 x 5
   2. elemen-elemen kolom ke 2 = 0, 6, 7
   3. elemen-elemen baris ke 3 = 5, 7, 0, 2, 8
   4. a₂₃ = 8
      a₁₄ = -7
      a₃₅ = 8


2. 1. matriks diagonal = $\left(\left(,,\begin{array}{ccc}1& 0& 0\\ 0& 2& 0\\ 0& 0& 3\end{array},,\right)\right)$
   
   
   2. matriks identitas = $\left(\left(,,\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array},,\right)\right)$
   
   
   3. matriks segitiga atas = $\left(\left(,,\begin{array}{ccc}4& 1& 5\\ 0& 2& 7\\ 0& 0& 3\end{array},,\right)\right)$
   
   
   4. matriks segitiga bawah = $\left(\left(,,\begin{array}{ccc}4& 0& 0\\ 2& 6& 0\\ 5& 3& 1\end{array},,\right)\right)$


3. 1. A^t = $\left(\left(,,\begin{array}{cc}4& 5\\ 3& 6\end{array},,\right)\right)$
   
   
   2. B^t = $\left(\left(,,\begin{array}{ccc}2& -3& 0\\ 4& 1& 5\end{array},,\right)\right)$
   
   
   3. C^t = $\left(\left(,,\begin{array}{cc}4& 2\\ 5& 3\\ -1& 0\end{array},,\right)\right)$
   
   
   4. D^t = $\left(\left(,,\begin{array}{ccc}4& 2& -1\\ 0& 4& 3\\ 5& 7& -2\end{array},,\right)\right)$


4. 1. x = 1    2y = 8
             y = 4
   
   
   2. x + y = 5
      x = 5 - y
      
      x - y = 1
      (5 - y) - y = 1
      5 - 2y = 1
      -2y = 1 - 5
      -2y = -4
      y = 2
      
      x = 5 - y
      x = 5 - 2
      x = 3
   
   
   3. 4 = 2x
      x = 2
      
      2y = 4x + 2
      2y = 4(2) + 2
      2y = 8 + 2
      2y = 10
      y = 5
      
      3z = 2y + 14
      3z = 2(5) + 14
      3z = 10 + 14
      3z = 24
      z = 8


5. 1. B^t = $\left(\left(,,\begin{array}{cc}8& 0\\ 1& 40\end{array},,\right)\right)$
      
      A = B^t
      $\left(\left(,,\begin{array}{cc}\mathrm{2x}& 0\\ 1& \mathrm{-4y}\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{cc}8& 0\\ 1& 40\end{array},,\right)\right)$
      
      2x = 8
      x = 4
      
      -4y = 40
      y = -10
   
   
   2. B^t = $\left(\left(,,\begin{array}{cc}8& 1\\ 0& 40\end{array},,\right)\right)$
      
      A = B^t
      $\left(\left(,,\begin{array}{cc}\mathrm{x\; -\; 2y}& 1\\ 0& \mathrm{-3y}\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{cc}8& 1\\ 0& 40\end{array},,\right)\right)$
      
      -3y = 40
      $\mathrm{y}=-\frac{40}{3}$
      
      x - 2y = 8
      $\mathrm{x}-2(-\frac{40}{3})=8$
      $\mathrm{x}+\frac{80}{3}=8$
      $\mathrm{x}=8-\frac{80}{3}$
      $\mathrm{x}=\frac{24}{3}-\frac{80}{3}$
      $\mathrm{x}=-\frac{56}{3}$
      

Jawaban Soal Latihan 1 Perbandingan Trigonometri Suatu Sudut Dalam Segitiga Siku-siku

Soal Latihan 1

1. Diketahui segitiga ABC sebagai berikut :
   
   
   
   
   Tentukan :
   
   
   1. Panjang AC
   2. Nilai perbandingan : sin α, cos α, tan α, cot α, sec α, cosec α






2. Pada segitiga KLM siku-siku di K, diketahui nilai cos M = $\frac{1}{5}\sqrt{5}$. Tentukan nilai tan M dan cosec M
3. Ditentukan bahwa a° adalah besar sudut XOP. Hitunglah nilai dari sin a°, cos a°, tan a°, cot a°, sec a° dan cosec a°, jika koordinat titik P adalah :
   1. (8, 6)
   2. (-4, 6)
   3. (12, -5)
   4. (-3, -4)
   5. (1, 3)

Jawab :

1. 1. AB = 12
      BC = 5
      $AC=\sqrt{{AB}^2+{BC}^2}$
      $=\sqrt{{12}^2+5^2}$
      $=\sqrt{144+25}$
      $=\sqrt{169}$
      $=13$
      
   2. sin α = $\frac{sisi\; hadap}{sisi\; miring}=\frac{5}{13}$
      cos α = $\frac{sisi\; dekat}{sisi\; miring}=\frac{12}{13}$
      tan α = $\frac{sisi\; hadap}{sisi\; dekat}=\frac{5}{12}$
      cot α = $\frac{sisi\; dekat}{sisi\; hadap}=\frac{12}{5}$
      sec α = $\frac{sisi\; miring}{sisi\; dekat}=\frac{13}{12}$
      cosec α = $\frac{sisi\; miring}{sisi\; hadap}=\frac{13}{5}$
2. cos M = $\frac{1}{5}\sqrt{5}$
   
   
   
   
   tan M = $\frac{sisi\; hadap}{sisi\; dekat}=\frac{2\sqrt{5}}{\sqrt{5}}=2$
   
   cosec M = $\frac{sisi\; miring}{sisi\; hadap}=\frac{5}{2\sqrt{5}}=\frac{5}{2\sqrt{5}}.\frac{\sqrt{5}}{\sqrt{5}}=\frac{5\sqrt{5}}{10}=\frac{1}{2}\sqrt{5}$







3. 1. 
      
      sisi hadap = 6
      sisi dekat = 8
      sisi miring = $\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10$
      
      sin a° = $\frac{sisi\; hadap}{sisi\; miring}=\frac{6}{10}=\frac{3}{5}$
      cos a° = $\frac{sisi\; dekat}{sisi\; miring}=\frac{8}{10}=\frac{4}{5}$
      tan a° = $\frac{sisi\; hadap}{sisi\; dekat}=\frac{6}{8}=\frac{3}{4}$
      cot a° = $\frac{sisi\; dekat}{sisi\; hadap}=\frac{8}{6}=\frac{4}{3}$
      sec a° = $\frac{sisi\; miring}{sisi\; dekat}=\frac{10}{8}=\frac{5}{4}$
      cosec a° = $\frac{sisi\; miring}{sisi\; hadap}=\frac{10}{6}=\frac{5}{3}$
      
   
   
   
   2. 
      
      sisi hadap = 6
      sisi dekat = -4
      sisi miring = $\sqrt{6^2+{\mathrm{(-4)}}^2}=\sqrt{36+16}=\sqrt{52}=\sqrt{4.13}=2\sqrt{13}$
      
      sin a° = $\frac{sisi\; hadap}{sisi\; miring}=\frac{6}{2\sqrt{13}}=\frac{3}{\sqrt{13}}.\frac{\sqrt{13}}{\sqrt{13}}=\frac{3\sqrt{13}}{13}$
      cos a° = $\frac{sisi\; dekat}{sisi\; miring}=\frac{-4}{2\sqrt{13}}=\frac{-2}{\sqrt{13}}.\frac{\sqrt{13}}{\sqrt{13}}=-\frac{2\sqrt{13}}{13}$
      tan a° = $\frac{sisi\; hadap}{sisi\; dekat}=\frac{6}{-4}=-\frac{3}{2}$
      cot a° = $\frac{sisi\; dekat}{sisi\; hadap}=\frac{-4}{6}=-\frac{2}{3}$
      sec a° = $\frac{sisi\; miring}{sisi\; dekat}=\frac{2\sqrt{13}}{-4}=-\frac{1}{2}\sqrt{13}$
      cosec a° = $\frac{sisi\; miring}{sisi\; hadap}=\frac{2\sqrt{13}}{6}=\frac{1}{3}\sqrt{13}$
      
   
   
   
   
   3. 
      sisi hadap = -5
      sisi dekat = 12
      sisi miring = $\sqrt{{\mathrm{(-5)}}^2+{12}^2}=\sqrt{25+144}=\sqrt{169}=13$
      
      sin a° = $\frac{sisi\; hadap}{sisi\; miring}=\frac{-5}{13}=-\frac{5}{13}$
      cos a° = $\frac{sisi\; dekat}{sisi\; miring}=\frac{12}{13}$
      tan a° = $\frac{sisi\; hadap}{sisi\; dekat}=\frac{-5}{12}=-\frac{5}{12}$
      cot a° = $\frac{sisi\; dekat}{sisi\; hadap}=\frac{12}{-5}=-\frac{12}{5}$
      sec a° = $\frac{sisi\; miring}{sisi\; dekat}=\frac{13}{12}$
      cosec a° = $\frac{sisi\; miring}{sisi\; hadap}=\frac{13}{-5}=-\frac{13}{5}$
      
   
   
   4. 
      
      
      sisi hadap = -4
      sisi dekat = -3
      sisi miring = $\sqrt{{\mathrm{(-4)}}^2+{\mathrm{(-3)}}^2}=\sqrt{16+9}=\sqrt{25}=5$
      
      sin a° = $\frac{sisi\; hadap}{sisi\; miring}=\frac{-4}{5}=-\frac{4}{5}$
      cos a° = $\frac{sisi\; dekat}{sisi\; miring}=\frac{-3}{5}=-\frac{3}{5}$
      tan a° = $\frac{sisi\; hadap}{sisi\; dekat}=\frac{-4}{-3}=\frac{4}{3}$
      cot a° = $\frac{sisi\; dekat}{sisi\; hadap}=\frac{-3}{-4}=\frac{3}{4}$
      sec a° = $\frac{sisi\; miring}{sisi\; dekat}=\frac{5}{-3}=-\frac{5}{3}$
      cosec a° = $\frac{sisi\; miring}{sisi\; hadap}=\frac{5}{-4}=-\frac{5}{4}$
      
   
   
   
   
   
   5. 
      
      
      sisi hadap = 3
      sisi dekat = 1
      sisi miring = $\sqrt{3^2+1^2}=\sqrt{9+1}=\sqrt{10}$
      
      sin a° = $\frac{sisi\; hadap}{sisi\; miring}=\frac{3}{\sqrt{10}}=\frac{3}{\sqrt{10}}.\frac{\sqrt{10}}{\sqrt{10}}=\frac{3\sqrt{10}}{10}=\frac{3}{10}\sqrt{10}$
      cos a° = $\frac{sisi\; dekat}{sisi\; miring}=\frac{1}{\sqrt{10}}=\frac{1}{\sqrt{10}}.\frac{\sqrt{10}}{\sqrt{10}}=\frac{1}{10}\sqrt{10}$
      tan a° = $\frac{sisi\; hadap}{sisi\; dekat}=\frac{3}{1}=3$
      cot a° = $\frac{sisi\; dekat}{sisi\; hadap}=\frac{1}{3}$
      sec a° = $\frac{sisi\; miring}{sisi\; dekat}=\frac{\sqrt{10}}{1}=\sqrt{10}$
      cosec a° = $\frac{sisi\; miring}{sisi\; hadap}=\frac{\sqrt{10}}{3}=\frac{1}{3}\sqrt{10}$