Question

Soal Kelas XI Matriks Sertakan Cara ​

Answer 1

Jawab:

Penjelasan dengan langkah-langkah:

1. [tex]A = \left[\begin{array}{cc}5&3\\2&6\\\end{array}\right][/tex] ; [tex]B = \left[\begin{array}{cc}2&-1\\0&-3\\\end{array}\right][/tex]

a. A + B = [tex]\left[\begin{array}{cc}5&3\\2&6\\\end{array}\right] + \left[\begin{array}{cc}2&-1\\0&-3\\\end{array}\right] = \left[\begin{array}{cc}5+2&3+(-1)\\2+0&6+(-3)\\\end{array}\right] = \left[\begin{array}{cc}7&2\\2&3\\\end{array}\right][/tex]

b. [tex]A^{T} = \left[\begin{array}{cc}5&2\\3&6\end{array}\right][/tex]

[tex]A^{T} + 3B = \left[\begin{array}{cc}5&2\\3&6\\\end{array}\right] + 3\left[\begin{array}{cc}2&-1\\0&-3\\\end{array}\right] = \left[\begin{array}{cc}5+6&2+(-3)\\3+0&6+(-9)\\\end{array}\right] = \left[\begin{array}{cc}11&-1\\3&-3\\\end{array}\right][/tex]

c. A x B = [tex]\left[\begin{array}{cc}5&3\\2&6\\\end{array}\right]\left[\begin{array}{cc}2&-1\\0&-3\\\end{array}\right] = \left[\begin{array}{cc}10+0&-5+(-9)\\4+0&-2+(-18)\\\end{array}\right] = \left[\begin{array}{cc}10&-14\\4&-20\\\end{array}\right][/tex]

d. B² = [tex]\left[\begin{array}{cc}2&-1\\0&-3\\\end{array}\right]^2 = \left[\begin{array}{cc}2&-1\\0&-3\\\end{array}\right] \left[\begin{array}{cc}2&-1\\0&-3\\\end{array}\right] = \left[\begin{array}{cc}4+0&-2+3\\0+0&0+9\\\end{array}\right] = \left[\begin{array}{cc}4&1\\0&9\\\end{array}\right][/tex]

2. [tex]C=\left[\begin{array}{cc}2&-2\\5&2\\1&2\end{array}\right][/tex] ; [tex]D = \left[\begin{array}{ccc}-2&3&1\\0&-1&0\\\end{array}\right][/tex]

a. [tex]C^{T} - D = \left[\begin{array}{ccc}2&5&1\\-2&2&2\\\end{array}\right] - \left[\begin{array}{ccc}-2&3&1\\0&-1&0\\\end{array}\right] = \left[\begin{array}{ccc}2-(-2)&5-3&1-1\\-2-0&2-(-1)&2-0\\\end{array}\right]= \left[\begin{array}{ccc}4&2&0\\-2&3&2\\\end{array}\right][/tex]

b. [tex]C x D = \left[\begin{array}{cc}2&-2\\5&2\\1&2\end{array}\right] \left[\begin{array}{ccc}-2&3&1\\0&-1&0\end{array}\right] = \left[\begin{array}{ccc}-4+0&6+2&2+0\\-10+0&15-2&5+0\\-2+0&3+(-2)&1+0\end{array}\right] =\left[\begin{array}{ccc}-4&8&2\\-10&13&5\\-2&1&1\end{array}\right][/tex]

[tex](CxD)^{T} =\left[\begin{array}{ccc}-4&-10&-2\\8&13&1\\2&5&1\end{array}\right][/tex]

c. [tex]D^{T} =\left[\begin{array}{cc}-2&0\\3&-1\\1&0\end{array}\right][/tex] ; [tex]C^{T} = \left[\begin{array}{ccc}2&5&1\\-2&2&2\\\end{array}\right][/tex][tex]C^{T} x D^{T} = \left[\begin{array}{ccc}2&5&1\\-2&2&2\\\end{array}\right] \left[\begin{array}{cc}-2&0\\3&-1\\1&0\end{array}\right] = \left[\begin{array}{cc}-4+15+1&0+(-5)+0\\4+6+2&0+(-2)+0\end{array}\right] = \left[\begin{array}{cc}12&-5\\12&-2\end{array}\right][/tex]

d. [tex]C^{T} + D = \left[\begin{array}{ccc}2&5&1\\-2&2&2\\\end{array}\right]+\left[\begin{array}{ccc}-2&3&1\\0&-1&0\end{array}\right] = \left[\begin{array}{ccc}0&8&2\\-2&1&2\\\end{array}\right][/tex]

3. K = L

a = 6

b = 2a = 2(6) = 12

3c = 4b = 4(12) = 48

c = 48/3 = 16