Diketahui matriks M=(cosxk1sinxksinx−cosx) dengan k konstanta bukan nol. Tentukan transpose dari M2.

Pertanyaan

Diketahui matriks $M=\begin{pmatrix}\cos\;x&k\;\sin\;x\\\frac1k\;\sin\;x&-\cos\;x\end{pmatrix}$ dengan $k$ konstanta bukan nol. Tentukan transpose dari $M^2$ .

Pembahasan Soal:

Perkalian kuadarat matriks $2 cross times 2$ yaitu:

$open parentheses table row a b row c d end table close parentheses open parentheses table row a b row c d end table close parentheses equals open parentheses table row cell a times a plus b times c end cell cell a times b plus b times d end cell row cell c times a plus d times c end cell cell c times b plus d times d end cell end table close parentheses$

Pada identitas trigonometri, berlaku:

$sin squared space x plus cos squared space x equals 1$

Penyelesaiannya yaitu:

$begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row M equals cell open parentheses table row cell cos space x end cell cell k space sin space x end cell row cell 1 over k space sin space x end cell cell negative cos space x end cell end table close parentheses end cell row cell M squared end cell equals cell open parentheses table row cell cos space x end cell cell k space sin space x end cell row cell 1 over k space sin space x end cell cell negative cos space x end cell end table close parentheses squared end cell row blank equals cell open parentheses table row cell cos space x end cell cell k space sin space x end cell row cell 1 over k space sin space x end cell cell negative cos space x end cell end table close parentheses open parentheses table row cell cos space x end cell cell k space sin space x end cell row cell 1 over k space sin space x end cell cell negative cos space x end cell end table close parentheses end cell row blank equals cell open parentheses table row cell cos space x open parentheses cos space x close parentheses plus down diagonal strike k space sin space x open parentheses fraction numerator 1 over denominator down diagonal strike k end fraction space sin space x close parentheses end cell cell cos space x open parentheses k space sin space x close parentheses plus k space sin space x open parentheses negative cos space x close parentheses end cell row cell 1 over k space sin space x open parentheses cos space x close parentheses plus open parentheses negative cos space x close parentheses open parentheses 1 over k space sin space x close parentheses end cell cell fraction numerator 1 over denominator down diagonal strike k end fraction space sin space x open parentheses down diagonal strike k space sin space x close parentheses plus open parentheses negative cos space x close parentheses open parentheses negative cos space x close parentheses end cell end table close parentheses end cell row blank equals cell open parentheses table row cell cos squared space x plus sin squared space x end cell cell k space cos space x times sin space x minus k space sin space x times cos space x end cell row cell fraction numerator sin space x times cos space x over denominator k end fraction minus fraction numerator cos space x times sin space x over denominator k end fraction end cell cell sin squared space x plus cos squared space x end cell end table close parentheses end cell row blank equals cell open parentheses table row 1 0 row 0 1 end table close parentheses end cell end table end style$

Kemudian tentukan transpose dari matriks $M squared$ .

Rumus transpose matriks $2 cross times 2$ yaitu:

Diperoleh hasilnya:

$table attributes columnalign right center left columnspacing 0px end attributes row cell M squared end cell equals cell open parentheses table row 1 0 row 0 1 end table close parentheses end cell row cell open parentheses M squared close parentheses to the power of T end cell equals cell open parentheses table row 1 0 row 0 1 end table close parentheses end cell end table$

Dengan demikian, transpose dari $M squared$ adalah $open parentheses table row 1 0 row 0 1 end table close parentheses$ .

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

S. Ayu

Mahasiswa/Alumni Universitas Muhammadiyah Prof. DR. Hamka

Terakhir diupdate 07 Oktober 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

Pertanyaan yang serupa

Diketahui At adalah transpose dari A. Jika C=(74−71−7172), B=(4228), dan A=C−1, maka determinan dari matriks AtB adalah ...

Pembahasan Soal:

Invers matriks $2 cross times 2$ berlaku:

$table attributes columnalign right center left columnspacing 0px end attributes row A equals cell open parentheses table row a b row c d end table close parentheses end cell row cell A to the power of negative 1 end exponent end cell equals cell fraction numerator 1 over denominator open vertical bar A close vertical bar end fraction open parentheses table row d cell negative b end cell row cell negative c end cell a end table close parentheses end cell row blank equals cell fraction numerator 1 over denominator a d minus b c end fraction open parentheses table row d cell negative b end cell row cell negative c end cell a end table close parentheses end cell end table$

Lakukan operasi matriks untuk mengetahui matriks $A$ seperti berikut:

$table attributes columnalign right center left columnspacing 0px end attributes row A equals cell C to the power of negative 1 end exponent end cell row blank equals cell open parentheses table row cell 4 over 7 end cell cell negative 1 over 7 end cell row cell negative 1 over 7 end cell cell 2 over 7 end cell end table close parentheses to the power of negative 1 end exponent end cell row blank equals cell fraction numerator 1 over denominator begin display style 4 over 7 end style open parentheses begin display style 2 over 7 end style close parentheses minus open parentheses negative begin display style 1 over 7 end style close parentheses open parentheses negative begin display style 1 over 7 end style close parentheses end fraction open parentheses table row cell 2 over 7 end cell cell 1 over 7 end cell row cell 1 over 7 end cell cell 4 over 7 end cell end table close parentheses end cell row blank equals cell fraction numerator 1 over denominator begin display style 8 over 49 minus 1 over 49 end style end fraction open parentheses table row cell 2 over 7 end cell cell 1 over 7 end cell row cell 1 over 7 end cell cell 4 over 7 end cell end table close parentheses end cell row blank equals cell fraction numerator 1 over denominator begin display style 7 over 49 end style end fraction open parentheses table row cell 2 over 7 end cell cell 1 over 7 end cell row cell 1 over 7 end cell cell 4 over 7 end cell end table close parentheses end cell row blank equals cell fraction numerator 1 over denominator begin display style 1 over 7 end style end fraction open parentheses table row cell 2 over 7 end cell cell 1 over 7 end cell row cell 1 over 7 end cell cell 4 over 7 end cell end table close parentheses end cell row blank equals cell 7 times open parentheses table row cell 2 over 7 end cell cell 1 over 7 end cell row cell 1 over 7 end cell cell 4 over 7 end cell end table close parentheses end cell row blank equals cell open parentheses table row 2 1 row 1 4 end table close parentheses end cell end table$

Kemudian diperoleh transpose matriks $A$ .

Transpose matriks $2 cross times 2$ berlaku:

Sehingga transpose matriks $A$ yaitu:

$table attributes columnalign right center left columnspacing 0px end attributes row A equals cell open parentheses table row 2 1 row 1 4 end table close parentheses end cell row cell A to the power of t end cell equals cell open parentheses table row 2 1 row 1 4 end table close parentheses end cell end table$

Kemudian tentukan matriks $A to the power of t B$ dengan perkalian matriks.

$table attributes columnalign right center left columnspacing 0px end attributes row cell A to the power of t B end cell equals cell open parentheses table row 2 1 row 1 4 end table close parentheses open parentheses table row 4 2 row 2 8 end table close parentheses end cell row blank equals cell open parentheses table row cell 2 open parentheses 4 close parentheses plus 1 open parentheses 2 close parentheses end cell cell 2 open parentheses 2 close parentheses plus 1 open parentheses 8 close parentheses end cell row cell 1 open parentheses 4 close parentheses plus 4 open parentheses 2 close parentheses end cell cell 1 open parentheses 2 close parentheses plus 4 open parentheses 8 close parentheses end cell end table close parentheses end cell row blank equals cell open parentheses table row cell 8 plus 2 end cell cell 4 plus 8 end cell row cell 4 plus 8 end cell cell 2 plus 32 end cell end table close parentheses end cell row blank equals cell open parentheses table row 10 12 row 12 34 end table close parentheses end cell end table$

Kemudian tentukan determinan dari matriks $A to the power of t B$ .

Determinan matriks $2 cross times 2$ berlaku:

$open vertical bar table row a b row c d end table close vertical bar equals a d minus b c$

Maka,

$table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar A to the power of t B close vertical bar end cell equals cell open vertical bar table row 10 12 row 12 34 end table close vertical bar end cell row blank equals cell 10 open parentheses 34 close parentheses minus 12 open parentheses 12 close parentheses end cell row blank equals cell 340 minus 144 end cell row blank equals 196 end table$

Maka determinan dari matriks $A to the power of t B$ adalah $196$ .

Oleh karena itu, jawaban yang benar adalah D.

Roboguru

Jika X memenuhi persamaan (1121)X=(106) dan Y memenuhi persamaan Y=(21X)T, maka matriks adalah ...

Pembahasan Soal:

Persamaan matriks dengan $A$ adalah matriks persegi yang mempunyai invers atau $A to the power of negative 1 end exponent$ , berlaku:

$A X equals B$

$table attributes columnalign right center left columnspacing 0px end attributes row cell A X end cell equals B row cell A to the power of negative 1 end exponent A X end cell equals cell A to the power of negative 1 end exponent B end cell row cell I X end cell equals cell A to the power of negative 1 end exponent B end cell row X equals cell A to the power of negative 1 end exponent B end cell end table$

Dengan menggunakan rumus persamaan diatas, diperoleh penyelesaiannya yaitu:

$table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses table row 1 2 row 1 1 end table close parentheses X end cell equals cell open parentheses table row 10 row 6 end table close parentheses end cell row X equals cell open parentheses table row 1 2 row 1 1 end table close parentheses to the power of negative 1 end exponent times open parentheses table row 10 row 6 end table close parentheses end cell row X equals cell fraction numerator 1 over denominator 1 open parentheses 1 close parentheses minus 2 open parentheses 1 close parentheses end fraction open parentheses table row 1 cell negative 2 end cell row cell negative 1 end cell 1 end table close parentheses times open parentheses table row 10 row 6 end table close parentheses end cell row X equals cell fraction numerator 1 over denominator 1 minus 2 end fraction open parentheses table row cell 1 open parentheses 10 close parentheses plus open parentheses negative 2 close parentheses open parentheses 6 close parentheses end cell row cell negative 1 open parentheses 10 close parentheses plus 1 open parentheses 6 close parentheses end cell end table close parentheses end cell row X equals cell fraction numerator 1 over denominator negative 1 end fraction open parentheses table row cell 10 minus 12 end cell row cell negative 10 plus 6 end cell end table close parentheses end cell row X equals cell negative 1 times open parentheses table row cell 10 minus 12 end cell row cell negative 10 plus 6 end cell end table close parentheses end cell row X equals cell negative 1 times open parentheses table row cell negative 2 end cell row cell negative 4 end cell end table close parentheses end cell row X equals cell open parentheses table row 2 row 4 end table close parentheses end cell end table$

Kemudian lakukan operasi hitung matriks $X$ yang telah diperoleh.

Rumus transpose matriks $1 cross times 2$ yaitu:

$table attributes columnalign right center left columnspacing 0px end attributes row A equals cell open parentheses table row a row b end table close parentheses end cell row cell A to the power of T end cell equals cell open parentheses table row a b end table close parentheses end cell end table$

Diperoleh:

$table attributes columnalign right center left columnspacing 0px end attributes row Y equals cell open parentheses 1 half X close parentheses to the power of T end cell row blank equals cell open parentheses 1 half open parentheses table row 2 row 4 end table close parentheses close parentheses to the power of T end cell row blank equals cell open parentheses table row cell 2 over 2 end cell row cell 4 over 2 end cell end table close parentheses to the power of T end cell row blank equals cell open parentheses table row 1 row 2 end table close parentheses to the power of T end cell row blank equals cell open parentheses table row 1 2 end table close parentheses end cell end table$

Maka matriks $Y$ adalah $open parentheses table row 1 2 end table close parentheses$ .

Oleh karena itu, jawaban yang benar adalah D.

Roboguru

Diketahui , , dan . Tentukan nilai dari CT×B.

Pembahasan Soal:

Ingat transpose dari suatu matrik $C$ dengan ordo $m cross times n$ adalah matrik $C subscript T$ dengan ordo $n cross times m$ yang didapatkan dengan mempertukarkan baris dan kolom dari matriks $C$ . Karena ordo dari matrik $C$ pada soal $2 cross times 3$ maka matriks transposenya juga berordo $3 cross times 2$ .

$table attributes columnalign right center left columnspacing 0px end attributes row C equals cell open parentheses table row cell negative 1 end cell 1 2 row 2 0 cell negative 1 end cell end table close parentheses end cell row cell C subscript T end cell equals cell open parentheses table row cell negative 1 end cell 2 row 1 0 row 2 cell negative 1 end cell end table close parentheses end cell end table$

Ingat operasi hitung perkalian pada 2 matriks dengan mengalikan baris dan kolom yang sesuai dengan peraturan perkalian matriks.

$table attributes columnalign right center left columnspacing 0px end attributes row cell C to the power of T cross times B end cell equals cell open parentheses table row cell negative 1 end cell 2 row 1 0 row 2 cell negative 1 end cell end table close parentheses cross times open parentheses table row 2 1 row 1 cell negative 1 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 2 plus 2 end cell cell negative 1 plus left parenthesis negative 2 right parenthesis end cell row cell 2 plus 0 end cell cell 1 plus 0 end cell row cell 4 plus left parenthesis negative 1 right parenthesis end cell cell 2 plus 1 end cell end table close parentheses end cell row blank equals cell open parentheses table row 0 cell negative 3 end cell row 2 1 row 3 3 end table close parentheses end cell end table$

Jadi, dapat disimpulkan bahwa nilai dari $C to the power of T cross times B$ adalah $open parentheses table row 0 cell negative 3 end cell row 2 1 row 3 3 end table close parentheses$ .

Roboguru

Sebuah matriks disebut matriks ortogonal jika A−1=AT. Diketahui matriks A=⎝⎛73b72727376a72c⎠⎞ adalah matriks ortogonal, maka nilai dari 7a2+b2+c2 adalah ...

Pembahasan Soal:

Suatu matriks $A$ disebut matriks ortogonal jika berlaku $A to the power of negative 1 end exponent equals A to the power of T$ (Invers matriks sama dengan transpose matriks). Persamaan ini ekuivalen dengan $A times A to the power of T equals I$ setelah kedua ruas dikalikan dengan $A$ .

Matriks $A equals open parentheses table row cell 3 over 7 end cell cell 2 over 7 end cell a row b cell 3 over 7 end cell cell 2 over 7 end cell row cell 2 over 7 end cell cell 6 over 7 end cell c end table close parentheses$ maka $A to the power of T equals open parentheses table row cell 3 over 7 end cell b cell 2 over 7 end cell row cell 2 over 7 end cell cell 3 over 7 end cell cell 6 over 7 end cell row a cell 2 over 7 end cell c end table close parentheses$ , dengan matriks identitas atau $I equals open parentheses table row 1 0 0 row 0 1 0 row 0 0 1 end table close parentheses$ .

Sehingga diperoleh

$table attributes columnalign right center left columnspacing 0px end attributes row cell A times A to the power of T end cell equals I row cell open parentheses table row cell 3 over 7 end cell cell 2 over 7 end cell a row b cell 3 over 7 end cell cell 2 over 7 end cell row cell 2 over 7 end cell cell 6 over 7 end cell c end table close parentheses open parentheses table row cell 3 over 7 end cell b cell 2 over 7 end cell row cell 2 over 7 end cell cell 3 over 7 end cell cell 6 over 7 end cell row a cell 2 over 7 end cell c end table close parentheses end cell equals cell open parentheses table row 1 0 0 row 0 1 0 row 0 0 1 end table close parentheses end cell end table$

Untuk mendapat entri baris pertama kolom pertama, diperoleh:

$table attributes columnalign right center left columnspacing 0px end attributes row cell 3 over 7 open parentheses 3 over 7 close parentheses plus 2 over 7 open parentheses 2 over 7 close parentheses plus a open parentheses a close parentheses end cell equals 1 row cell 9 over 49 plus 4 over 49 plus a squared end cell equals 1 row cell 13 over 49 plus a squared end cell equals 1 row cell a squared end cell equals cell 1 minus 13 over 49 end cell row cell a squared end cell equals cell 49 over 49 minus 13 over 49 end cell row cell a squared end cell equals cell 36 over 49 end cell end table$

Untuk mendapat entri baris kedua kolom kedua, diperoleh:

$table attributes columnalign right center left columnspacing 0px end attributes row cell b open parentheses b close parentheses plus 3 over 7 open parentheses 3 over 7 close parentheses plus 2 over 7 open parentheses 2 over 7 close parentheses end cell equals 1 row cell b squared plus 9 over 49 plus 4 over 49 end cell equals 1 row cell b squared plus 13 over 49 end cell equals 1 row cell b squared end cell equals cell 1 minus 13 over 49 end cell row cell b squared end cell equals cell 49 over 49 minus 13 over 49 end cell row cell b squared end cell equals cell 36 over 49 end cell end table$

Untuk mendapat entri baris ketiga kolom ketiga, diperoleh:

$table attributes columnalign right center left columnspacing 0px end attributes row cell 2 over 7 open parentheses 2 over 7 close parentheses plus 6 over 7 open parentheses 6 over 7 close parentheses plus c open parentheses c close parentheses end cell equals 1 row cell 4 over 49 plus 36 over 49 plus c squared end cell equals 1 row cell 40 over 49 plus c squared end cell equals 1 row cell c squared end cell equals cell 1 minus 40 over 49 end cell row cell c squared end cell equals cell 49 over 49 minus 40 over 49 end cell row cell c squared end cell equals cell 9 over 49 end cell end table$

Sehingga diperoleh:

$table attributes columnalign right center left columnspacing 0px end attributes row cell 7 square root of a squared plus b squared plus c squared end root end cell equals cell 7 square root of 36 over 49 plus 36 over 49 plus 9 over 49 end root end cell row blank equals cell 7 square root of 81 over 49 end root end cell row blank equals cell 7 open parentheses 9 over 7 close parentheses end cell row blank equals 9 end table$

Maka nilai dari $7 square root of a squared plus b squared plus c squared end root$ adalah $9$ .

Jadi, jawaban yang tepat adalah B.

Roboguru

Diketahui A=(4−320) dan B=(−15). Tentukan : b. BT⋅A

Pembahasan Soal:

Diberikan matriks $A equals open parentheses table row 4 2 row cell negative 3 end cell 0 end table close parentheses$ dan $B equals open parentheses table row cell negative 1 end cell row 5 end table close parentheses$ . Perkalian dari transpose matriks B dengan matriks A dapat dihitung seperti berikut :

$table attributes columnalign right center left columnspacing 0px end attributes row cell B to the power of T times A end cell equals cell open parentheses table row cell negative 1 end cell 5 end table close parentheses open parentheses table row 4 2 row cell negative 3 end cell 0 end table close parentheses end cell row blank equals cell open parentheses table row cell negative 1 left parenthesis 4 right parenthesis plus 5 left parenthesis negative 3 right parenthesis end cell cell negative 1 left parenthesis 2 right parenthesis plus 5 left parenthesis 0 right parenthesis end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 4 minus 15 end cell cell negative 2 plus 0 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 19 end cell cell negative 2 end cell end table close parentheses end cell end table$

Jadi, hasil dari operasi matriks $B to the power of T times A$ adalah $open parentheses table row cell negative 19 end cell cell negative 2 end cell end table close parentheses$ .

Roboguru