Jawaban Soal Latihan 1 Perkalian Matriks

Soal Latihan 1

1. Diketahui matriks A dan B sebagai berikut :
   
   A = $\left(\left(,,\begin{array}{ccc}-1& 3& -4\\ 2& 0& 5\end{array},,\right)\right)$
   
   B = $\left(\left(,,\begin{array}{cc}1& 2\\ -2& 3\\ 3& -1\end{array},,\right)\right)$
   
   Tentukan :
   
   1. 3A
   2. 2A
   3. 5A^t
   4. 2A^t + 5B
   5. 2B^t + 3A


2. Diketahui matriks A dan B sebagai berikut :
   
   A = $\left(\left(,,\begin{array}{cc}\mathrm{a}& 4\\ \mathrm{2b}& \mathrm{3c}\end{array},,\right)\right)$
   
   B = $\left(\left(,,\begin{array}{cc}\mathrm{2c\; -\; 3b}& \mathrm{2a\; +\; 1}\\ \mathrm{a}& \mathrm{b\; +\; 7}\end{array},,\right)\right)$
   
   Tentukan nilai c jika A = 2B^t


3. Diketahui matriks A, B dan C sebagai berikut :
   
   A = $\left(\left(,,\begin{array}{cc}3& 2\\ 1& -1\end{array},,\right)\right)$
   
   B = $\left(\left(,,\begin{array}{cc}-1& 4\\ 5& 0\end{array},,\right)\right)$
   
   C = $\left(\left(,,\begin{array}{ccc}1& -3& 0\\ -2& 5& 1\end{array},,\right)\right)$
   
   Tentukan nilai dari :
   1. AB
   2. AC
   3. BC
   4. A^tC
   5. B^tC
   6. C^tA
   7. C^tB
   8. A^tB^t
   9. B^tA^t
   10. (AB)^t
   11. Apakah (AB)^t = B^tA^t


4. Tentukan nilai a dan b jika diketahui persamaan berikut :
   
   $\left(\left(,,\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ 3& -2\end{array},,\right)\right)\left(\left(,,\begin{array}{cc}6& -5\\ 2& 4\end{array},,\right)\right)=\left(\left(,,\begin{array}{cc}12& -27\\ 14& -23\end{array},,\right)\right)$
   


5. Tentukan nilai x dan y yang memenuhi persamaan berikut :
   
   $\left(\left(,,\begin{array}{cc}1& 5\\ 4& -6\end{array},,\right)\right)\left(\left(,,\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array},,\right)\right)=\left(\left(,,\begin{array}{c}-13\\ 26\end{array},,\right)\right)$
   
   

Jawab

1. 1. 3A = 3 $\left(\left(,,\begin{array}{ccc}-1& 3& -4\\ 2& 0& 5\end{array},,\right)\right)=\left(\left(,,\begin{array}{ccc}-3& 9& -12\\ 6& 0& 15\end{array},,\right)\right)$
      
   
   
   2. 2A = 2 $\left(\left(,,\begin{array}{ccc}-1& 3& -4\\ 2& 0& 5\end{array},,\right)\right)=\left(\left(,,\begin{array}{ccc}-2& 6& -8\\ 4& 0& 10\end{array},,\right)\right)$
      
   
   
   3. A^t = $\left(\left(,,\begin{array}{cc}-1& 2\\ 3& 0\\ -4& 5\end{array},,\right)\right)$
      
      5A^t = 5 $\left(\left(,,\begin{array}{cc}-1& 2\\ 3& 0\\ -4& 5\end{array},,\right)\right)=\left(\left(,,\begin{array}{cc}-5& 10\\ 15& 0\\ -20& 25\end{array},,\right)\right)$
      
   
   
   4. 2A^t + 5B = 2 $\left(\left(,,\begin{array}{cc}-1& 2\\ 3& 0\\ -4& 5\end{array},,\right)\right)$ + 5 $\left(\left(,,\begin{array}{cc}1& 2\\ -2& 3\\ 3& -1\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{cc}-2& 4\\ 6& 0\\ -8& 10\end{array},,\right)\right)+\left(\left(,,\begin{array}{cc}5& 10\\ -10& 15\\ 15& -5\end{array},,\right)\right)=\left(\left(,,\begin{array}{cc}3& 14\\ -4& 15\\ 7& 5\end{array},,\right)\right)$
   
   
   5. B^t = $\left(\left(,,\begin{array}{ccc}1& -2& 3\\ 2& 3& -1\end{array},,\right)\right)$
      
      2B^t + 3A = 2 $\left(\left(,,\begin{array}{ccc}1& -2& 3\\ 2& 3& -1\end{array},,\right)\right)$ + 3 $\left(\left(,,\begin{array}{ccc}-1& 3& -4\\ 2& 0& 5\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{ccc}2& -4& 6\\ 4& 6& -2\end{array},,\right)\right)$ + $\left(\left(,,\begin{array}{ccc}-3& 9& -12\\ 6& 0& 15\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{ccc}-1& 5& -6\\ 10& 6& 13\end{array},,\right)\right)$


2. B^t = $\left(\left(,,\begin{array}{cc}\mathrm{2c\; -\; 3b}& \mathrm{a}\\ \mathrm{2a\; +\; 1}& \mathrm{b\; +\; 7}\end{array},,\right)\right)$
   
   A = 2B^t
   
   $\left(\left(,,\begin{array}{cc}\mathrm{a}& 4\\ \mathrm{2b}& \mathrm{3c}\end{array},,\right)\right)$ = 2 $\left(\left(,,\begin{array}{cc}\mathrm{2c\; -\; 3b}& \mathrm{a}\\ \mathrm{2a\; +\; 1}& \mathrm{b\; +\; 7}\end{array},,\right)\right)$
   
   $\left(\left(,,\begin{array}{cc}\mathrm{a}& 4\\ \mathrm{2b}& \mathrm{3c}\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{cc}\mathrm{4c\; -\; 6b}& \mathrm{2a}\\ \mathrm{4a\; +\; 2}& \mathrm{2b\; +\; 14}\end{array},,\right)\right)$
   
   2a = 4
   a = 2
   
   2b = 4a + 2
   2b = 4(2) + 2
   2b = 8 + 2
   2b = 10
   b = 5
   
   3c = 2b + 14
   3c = 2(5) + 14
   3c = 10 + 14
   3c = 24
   c = 8
   


3. 1. AB = $\left(\left(,,\begin{array}{cc}3& 2\\ 1& -1\end{array},,\right)\right)$ $\left(\left(,,\begin{array}{cc}-1& 4\\ 5& 0\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{cc}\mathrm{3\; .\; -1\; +\; 2\; .\; 5}& \mathrm{3\; .\; 4\; +\; 2\; .\; 0}\\ \mathrm{1\; .\; -1\; +\; -1\; .\; 5}& \mathrm{1\; .\; 4\; +\; -1\; .\; 0}\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{cc}7& 12\\ -6& 4\end{array},,\right)\right)$
   
   
   2. AC = $\left(\left(,,\begin{array}{cc}3& 2\\ 1& -1\end{array},,\right)\right)$ $\left(\left(,,\begin{array}{ccc}1& -3& 0\\ -2& 5& 1\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{ccc}\mathrm{3\; .\; 1\; +\; 2\; .\; -2}& \mathrm{3\; .\; -3\; +\; 2\; .\; 5}& \mathrm{3\; .\; 0\; +\; 2\; .\; 1}\\ \mathrm{1\; .\; 1\; +\; -1\; .\; -2}& \mathrm{1\; .\; -3\; +\; -1\; .\; 5}& \mathrm{1\; .\; 0\; +\; -1\; .\; 1}\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{ccc}-1& 1& 2\\ 3& -8& -1\end{array},,\right)\right)$
   
   
   3. BC = $\left(\left(,,\begin{array}{cc}-1& 4\\ 5& 0\end{array},,\right)\right)$ $\left(\left(,,\begin{array}{ccc}1& -3& 0\\ -2& 5& 1\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{ccc}\mathrm{-1\; .\; 1\; +\; 4\; .\; -2}& \mathrm{-1\; .\; -3\; +\; 4\; .\; 5}& \mathrm{-1\; .\; 0\; +\; 4\; .\; 1}\\ \mathrm{5\; .\; 1\; +\; 0\; .\; -2}& \mathrm{5\; .\; -3\; +\; 0\; .\; 5}& \mathrm{5\; .\; 0\; +\; 0\; .\; 1}\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{ccc}-9& 23& 4\\ 5& -15& 0\end{array},,\right)\right)$
   
   
   4. A^t = $\left(\left(,,\begin{array}{cc}3& 1\\ 2& -1\end{array},,\right)\right)$
      
      A^tC = $\left(\left(,,\begin{array}{cc}3& 1\\ 2& -1\end{array},,\right)\right)$ $\left(\left(,,\begin{array}{ccc}1& -3& 0\\ -2& 5& 1\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{ccc}\mathrm{3\; .\; 1\; +\; 1\; .\; -2}& \mathrm{3\; .\; -3\; +\; 1\; .\; 5}& \mathrm{3\; .\; 0\; +\; 1\; .\; 1}\\ \mathrm{2\; .\; 1\; +\; -1\; .\; -2}& \mathrm{2\; .\; -3\; +\; -1\; .\; 5}& \mathrm{2\; .\; 0\; +\; -1\; .\; 1}\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{ccc}1& -4& 1\\ 4& -11& -1\end{array},,\right)\right)$
   
   
   5. B^t = $\left(\left(,,\begin{array}{cc}-1& 5\\ 4& 0\end{array},,\right)\right)$
      
      B^tC = $\left(\left(,,\begin{array}{cc}-1& 5\\ 4& 0\end{array},,\right)\right)$ $\left(\left(,,\begin{array}{ccc}1& -3& 0\\ -2& 5& 1\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{ccc}\mathrm{-1\; .\; 1\; +\; 5\; .\; -2}& \mathrm{-1\; .\; -3\; +\; 5\; .\; 5}& \mathrm{-1\; .\; 0\; +\; 5\; .\; 1}\\ \mathrm{4\; .\; 1\; +\; 0\; .\; -2}& \mathrm{4\; .\; -3\; +\; 0\; .\; 5}& \mathrm{4\; .\; 0\; +\; 0\; .\; 1}\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{ccc}-11& 28& 5\\ 4& -12& 0\end{array},,\right)\right)$
   
   
   6. C^t = $\left(\left(,,\begin{array}{cc}1& -2\\ -3& 5\\ 0& 1\end{array},,\right)\right)$
      
      C^tA = $\left(\left(,,\begin{array}{cc}1& -2\\ -3& 5\\ 0& 1\end{array},,\right)\right)$ $\left(\left(,,\begin{array}{cc}3& 2\\ 1& -1\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{cc}\mathrm{1\; .\; 3\; +\; -2\; .\; 1}& \mathrm{1\; .\; 2\; +\; -2\; .\; -1}\\ \mathrm{-3\; .\; 3\; +\; 5\; .\; 1}& \mathrm{-3\; .\; 2\; +\; 5\; .\; -1}\\ \mathrm{0\; .\; 3\; +\; 1\; .\; 1}& \mathrm{0\; .\; 2\; +\; 1\; .\; -1}\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{cc}1& 4\\ -4& -11\\ 1& -1\end{array},,\right)\right)$
   
   
   7. C^tB = $\left(\left(,,\begin{array}{cc}1& -2\\ -3& 5\\ 0& 1\end{array},,\right)\right)$ $\left(\left(,,\begin{array}{cc}-1& 4\\ 5& 0\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{cc}\mathrm{1\; .\; -1\; +\; -2\; .\; 5}& \mathrm{1\; .\; 4\; +\; -2\; .\; 0}\\ \mathrm{-3\; .\; -1\; +\; 5\; .\; 5}& \mathrm{-3\; .\; 4\; +\; 5\; .\; 0}\\ \mathrm{0\; .\; -1\; +\; 1\; .\; 5}& \mathrm{0\; .\; 4\; +\; 1\; .\; 0}\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{cc}-11& 4\\ 28& -12\\ 5& 0\end{array},,\right)\right)$
   
   
   8. A^tB^t = $\left(\left(,,\begin{array}{cc}3& 1\\ 2& -1\end{array},,\right)\right)$ $\left(\left(,,\begin{array}{cc}-1& 5\\ 4& 0\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{cc}\mathrm{3\; .\; -1\; +\; 1\; .\; 4}& \mathrm{3\; .\; 5\; +\; 1\; .\; 0}\\ \mathrm{2\; .\; -1\; +\; -1\; .\; 4}& \mathrm{2\; .\; 5\; +\; -1\; .\; 0}\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{cc}1& 15\\ -6& 10\end{array},,\right)\right)$
   
   
   9. B^tA^t = $\left(\left(,,\begin{array}{cc}-1& 5\\ 4& 0\end{array},,\right)\right)$ $\left(\left(,,\begin{array}{cc}3& 1\\ 2& -1\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{cc}\mathrm{-1\; .\; 3\; +\; 5\; .\; 2}& \mathrm{-1\; .\; 1\; +\; 5\; .\; -1}\\ \mathrm{4\; .\; 3\; +\; 0\; .\; 2}& \mathrm{4\; .\; 1\; +\; 0\; .\; -1}\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{cc}7& -6\\ 12& 4\end{array},,\right)\right)$
   
   
   10. AB = $\left(\left(,,\begin{array}{cc}7& 12\\ -6& 4\end{array},,\right)\right)$
       
       (AB)^t = $\left(\left(,,\begin{array}{cc}7& -6\\ 12& 4\end{array},,\right)\right)$
   
   
   11. Apakah (AB)^t = B^tA^t ?
       Iya


4. $\left(\left(,,\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ 3& -2\end{array},,\right)\right)\left(\left(,,\begin{array}{cc}6& -5\\ 2& 4\end{array},,\right)\right)=\left(\left(,,\begin{array}{cc}12& -27\\ 14& -23\end{array},,\right)\right)$
   
   $\left(\left(,,\begin{array}{cc}\mathrm{a\; .\; 6\; +\; b\; .\; 2}& \mathrm{a\; .\; -5\; +\; b\; .\; 4}\\ \mathrm{3\; .\; 6\; +\; -2\; .\; 2}& \mathrm{3\; .\; -5\; +\; -2\; .\; 4}\end{array},,\right)\right)=\left(\left(,,\begin{array}{cc}12& -27\\ 14& -23\end{array},,\right)\right)$
   
   $\left(\left(,,\begin{array}{cc}\mathrm{6a\; +\; 2b}& \mathrm{-5a\; +\; 4b}\\ 14& -23\end{array},,\right)\right)=\left(\left(,,\begin{array}{cc}12& -27\\ 14& -23\end{array},,\right)\right)$
   
   6a + 2b = 12
   2b = 12 - 6a
   b = 6 - 3a
   
   -5a + 4b = -27
   -5a + 4(6 - 3a) = -27
   -5a + 24 - 12a = -27
   -17a = -27 - 24
   -17a = -51
   a = 3
   
   b = 6 - 3a
   b = 6 - 3(3)
   b = 6 - 9
   b = -3


5. $\left(\left(,,\begin{array}{cc}1& 5\\ 4& -6\end{array},,\right)\right)\left(\left(,,\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array},,\right)\right)=\left(\left(,,\begin{array}{c}-13\\ 26\end{array},,\right)\right)$
   
   $\left(\left(,,\begin{array}{c}\mathrm{1\; .\; x\; +\; 5\; .\; y}\\ \mathrm{4\; .\; x\; +\; -6\; .\; y}\end{array},,\right)\right)=\left(\left(,,\begin{array}{c}-13\\ 26\end{array},,\right)\right)$
   
   $\left(\left(,,\begin{array}{c}\mathrm{x\; +\; 5y}\\ \mathrm{4x\; -\; 6y}\end{array},,\right)\right)=\left(\left(,,\begin{array}{c}-13\\ 26\end{array},,\right)\right)$
   
   x + 5y = -13
   x = -13 - 5y
   
   4x - 6y = 26
   4(-13 - 5y) - 6y = 26
   -52 - 20y - 6y = 26
   -26y = 26 + 52
   -26y = 78
   y = -3
   
   x = -13 - 5y
   x = -13 - 5(-3)
   x = -13 + 15
   x = 2