Exercise Zone : Matriks

Berikut ini adalah kumpulan soal mengenai Matriks tingkat dasar. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.
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No. 1

Jika matriks AB=(5623){A\cdot B=\begin{pmatrix}5&6\\2&3\end{pmatrix}} dan detA=3{\det A=3}, maka det(3BA1){\det\left(3BA^{-1}\right)} adalah....
  1. 11
  2. 22
  3. 33
  1. 44
  2. 55
AB=(5623)AB=(5)(3)(2)(6)AB=15123B=3B=1\begin{aligned}A\cdot B&=\begin{pmatrix}5&6\\2&3\end{pmatrix}\\[8pt]|AB|&=(5)(3)-(2)(6)\\|A||B|&=15-12\\3|B|&=3\\|B|&=1\end{aligned}

3BA1=32BA1=9(1)(1A)=9(13)=3\begin{aligned}\left|3BA^{-1}\right|&=3^2|B|\left|A^{-1}\right|\\&=9(1)\left(\dfrac1{|A|}\right)\\&=9\left(\dfrac13\right)\\&=3\end{aligned}

No. 2

Jika A=(110112){A=\begin{pmatrix}-1&-1&0\\-1&1&2\end{pmatrix}}, B=(1x1y0z){B=\begin{pmatrix}-1&x\\1&y\\0&z\end{pmatrix}} dan AB=(0224){AB=\begin{pmatrix}0&2\\2&4\end{pmatrix}}, maka nilai zx{z-x} adalah ....
AB=(0224)(110112)(1x1y0z)=(0224)(0xy2x+y+2z)=(0224)\begin{aligned}AB&=\begin{pmatrix}0&2\\2&4\end{pmatrix}\\\begin{pmatrix}-1&-1&0\\-1&1&2\end{pmatrix}\begin{pmatrix}-1&x\\1&y\\0&z\end{pmatrix}&=\begin{pmatrix}0&2\\2&4\end{pmatrix}\\\begin{pmatrix}0&-x-y\\2&-x+y+2z\end{pmatrix}&=\begin{pmatrix}0&2\\2&4\end{pmatrix}\end{aligned}

xy=2x+y+2z=4+2x+2z=62z2x=6zx=3\begin{aligned}-x-y&=2\\-x+y+2z&=4&\qquad+\\\hline-2x+2z&=6\\2z-2x&=6\\z-x&=3\end{aligned}

No. 3

Diketahui matriks A=(1234){A=\begin{pmatrix}1&2\\3&4\end{pmatrix}} dan B=(1yx3){B=\begin{pmatrix}1&y\\x&3\end{pmatrix}}. Jika determinan ABAB adalah 1010 maka xy=xy= ....
AB=10AB=10(2)(3xy)=103xy=5xy=8\begin{aligned}|AB|&=10\\|A||B|&=10\\(-2)(3-xy)&=10\\3-xy&=-5\\xy&=8\end{aligned}

No. 4

Diberikan matriks P=(2143){P=\begin{pmatrix}2&-1\\4&3\end{pmatrix}} dan Q=(2r1rp+1){Q=\begin{pmatrix}2r&1\\r&p+1\end{pmatrix}} dengan r0{r\neq0} dan p0{p\neq0}. Matriks PQPQ tidak mempunyai invers apabila nilai p=p= ....
Matriks PQPQ tidak mempunyai invers maka:
PQ=0PQ=0((2)(3)(1)(4))((2r)(p+1)(1)(r))=0(10)(2pr+2rr)=02pr+r=0r(2p+1)=02p+1=02p=1p=12\begin{aligned}|PQ|&=0\\|P||Q|&=0\\((2)(3)-(-1)(4))((2r)(p+1)-(1)(r))&=0\\(10)(2pr+2r-r)&=0\\2pr+r&=0\\r(2p+1)&=0\\2p+1&=0\\2p&=-1\\p&=-\dfrac12\end{aligned}

No. 5

Jika P=(1213){P=\begin{pmatrix}1&2\\1&3\end{pmatrix}} dan (xyzz)=2P1{\begin{pmatrix}x&y\\-z&z\end{pmatrix}=2P^{-1}} dengan P1P^{-1} menyatakan invers matriks PP, maka x+y={x+y=} ....
P=(1)(3)(2)(1)=1\begin{aligned}|P|&=(1)(3)-(2)(1)\\&=1\end{aligned}

P1=1P(3211)=11(3211)=(3211)\begin{aligned}P^{-1}&=\dfrac1{|P|}\begin{pmatrix}3&-2\\-1&1\end{pmatrix}\\&=\dfrac11\begin{pmatrix}3&-2\\-1&1\end{pmatrix}\\&=\begin{pmatrix}3&-2\\-1&1\end{pmatrix}\end{aligned}

(xyzz)=2P1(xyzz)=2(3211)(xyzz)=(6422)\begin{aligned}\begin{pmatrix}x&y\\-z&z\end{pmatrix}&=2P^{-1}\\\begin{pmatrix}x&y\\-z&z\end{pmatrix}&=2\begin{pmatrix}3&-2\\-1&1\end{pmatrix}\\\begin{pmatrix}x&y\\-z&z\end{pmatrix}&=\begin{pmatrix}6&-4\\-2&2\end{pmatrix}\end{aligned}
x=6x=6 dan y=4y=-4

x+y=6+(4)=2\begin{aligned}x+y&=6+(-4)\\&=2\end{aligned}

No. 6

Matriks AA yang memenuhi (2k10)A=(24k10){\begin{pmatrix}2&k\\1&0\end{pmatrix}A=\begin{pmatrix}2&4k\\1&0\end{pmatrix}} adalah ....
(2k10)A=(24k10)A=(2k10)1(24k10)=120k1(0k12)(24k10)=1k(k004k)=(1004)

No. 7

Jika AA, BB, CC, DD, EE, dan MM adalah matriks persegi yang berukuran sama dan memenuhi A1B1C1D1E1=M{A^{-1}B^{-1}C^{-1}D^{-1}E^{-1}=M}, maka matriks CC adalah ....
A1B1C1D1E1=MEDCBA=M1DCBA=E1M1CBA=D1E1M1CB=D1E1M1A1C=D1E1M1A1B1

No. 8

Diketahui matriks K=(1432)K=\begin{pmatrix}-1&4\\3&-2\end{pmatrix}, L=(5132)L=\begin{pmatrix}5&1\\3&-2\end{pmatrix} dan M=(9061)M=\begin{pmatrix}9&0\\6&-1\end{pmatrix}. Matriks 2KL+M{2K-L+M} adalah....
  1. (2883)\begin{pmatrix}-2&8\\-8&3\end{pmatrix}
  2. (1763)\begin{pmatrix}1&7\\-6&3\end{pmatrix}
  3. (2793)\begin{pmatrix}2&7\\9&-3\end{pmatrix}
  1. (2793)\begin{pmatrix}2&7\\9&3\end{pmatrix}
  2. (2973)\begin{pmatrix}2&9\\7&-3\end{pmatrix}
2KL+M=2(1432)(5132)+(9061)=(2864)(5132)+(9061)=(2793)

No. 9

Diketahui matriks A=(1232)A=\begin{pmatrix}-1&2\\3&-2\end{pmatrix} dan B=(1230)B=\begin{pmatrix}1&-2\\3&0\end{pmatrix}. Jika matriks C=AB{C=AB}, maka determinan matriks C=C=
  1. 1818
  2. 1212
  3. 24-24
  1. 30-30
  2. 36-36
\(\eqalign{ C&=AB\ |C|&=|A||B|\ &=\left[(-1)(-2)-(2)(3)\right]\cdot[(1)(0)-(-2)(3)]\ &=[2-6][0-(-6)]\ &=(-4)(6)\ &=-24 }$

No. 10

Diketahui matriks A=(1x133x)A=\begin{pmatrix}1&x&-1\\3&3&x\end{pmatrix}, B=(2y345x)B=\begin{pmatrix}2&y&3\\4&5&x\end{pmatrix}, dan C=(3x2682)C=\begin{pmatrix}3&x&2\\6&8&2\end{pmatrix}. Jika A+B=C{A+B=C}, maka x+y={x+y=} ....
  1. 2-2
  2. 1-1
  3. 00
  1. 11
  2. 44
A+B=C(1x133x)+(2y345x)=(3x2682)(3x+y2782x)=(3x2682)

x+y=xy=0

2x=2x=1

x+y=1+0=1

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