Diketahui dan . Tentukan hasil dari matriks berikut

Pertanyaan

Diketahui $K equals open parentheses table row 4 12 row 9 cell negative 7 end cell end table close parentheses comma L equals open parentheses table row cell negative 3 end cell 2 row 0 cell negative 4 end cell end table close parentheses comma space$ dan $M equals open parentheses table row 0 3 row 4 1 end table close parentheses$ . Tentukan hasil dari matriks berikut

$left parenthesis K minus L right parenthesis M to the power of T$

Pembahasan Soal:

Diketahui matriks $K equals open parentheses table row 4 12 row 9 cell negative 7 end cell end table close parentheses comma L equals open parentheses table row cell negative 3 end cell 2 row 0 cell negative 4 end cell end table close parentheses comma space$ dan $M equals open parentheses table row 0 3 row 4 1 end table close parentheses$

Ingat kembali mengenai operasi perhitungan dalam matriks

Pengurangan Matriks

$A minus B equals open parentheses table row cell a subscript 1 end cell cell a subscript 2 end cell row cell a subscript 3 end cell cell a subscript 4 end cell end table close parentheses minus open parentheses table row cell b subscript 1 end cell cell b subscript 2 end cell row cell b subscript 3 end cell cell b subscript 4 end cell end table close parentheses equals open parentheses table row cell a subscript 1 minus b subscript 1 end cell cell a subscript 2 minus b subscript 2 end cell row cell a subscript 3 minus b subscript 3 end cell cell a subscript 4 minus b subscript 4 end cell end table close parentheses$

Perkalian Matriks

$A cross times B equals open parentheses table row cell a subscript 1 end cell cell a subscript 2 end cell row cell a subscript 3 end cell cell a subscript 4 end cell end table close parentheses cross times open parentheses table row cell b subscript 1 end cell cell b subscript 2 end cell row cell b subscript 3 end cell cell b subscript 4 end cell end table close parentheses equals open parentheses table row cell a subscript 1 b subscript 1 plus a subscript 2 b subscript 3 end cell cell a subscript 1 b subscript 2 minus a subscript 2 b subscript 4 end cell row cell a subscript 3 b subscript 1 minus a subscript 4 b subscript 3 end cell cell a subscript 3 b subscript 2 minus a subscript 4 b subscript 4 end cell end table close parentheses$

Transpose Matriks

$A equals open parentheses table row cell a subscript 1 end cell cell a subscript 2 end cell row cell a subscript 3 end cell cell a subscript 4 end cell end table close parentheses rightwards double arrow A to the power of T equals open parentheses table row cell a subscript 1 end cell cell a subscript 3 end cell row cell a subscript 2 end cell cell a subscript 4 end cell end table close parentheses$

Diperoleh perhitungan sebagai berikut

$table attributes columnalign right center left columnspacing 0px end attributes row cell K minus L end cell equals cell open parentheses table row 4 12 row 9 cell negative 7 end cell end table close parentheses minus open parentheses table row cell negative 3 end cell 2 row 0 cell negative 4 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell 4 minus left parenthesis negative 3 right parenthesis end cell cell 12 minus 2 end cell row cell 9 minus 0 end cell cell negative 7 minus left parenthesis negative 4 right parenthesis end cell end table close parentheses end cell row blank equals cell open parentheses table row 7 10 row 9 cell negative 3 end cell end table close parentheses space end cell end table$

$M equals open parentheses table row 0 3 row 4 1 end table close parentheses rightwards double arrow M to the power of T equals open parentheses table row 0 4 row 3 1 end table close parentheses$

Oleh karena itu, diperoleh

$table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis K minus L right parenthesis M to the power of T end cell equals cell open parentheses table row 7 10 row 9 cell negative 3 end cell end table close parentheses open parentheses table row 0 4 row 3 1 end table close parentheses end cell row blank equals cell open parentheses table row cell left parenthesis 7 cross times 0 right parenthesis plus left parenthesis 10 cross times 3 right parenthesis end cell cell left parenthesis 7 cross times 4 right parenthesis plus left parenthesis 10 cross times 1 right parenthesis end cell row cell left parenthesis 9 cross times 0 right parenthesis plus left parenthesis negative 3 cross times 3 right parenthesis end cell cell left parenthesis 9 cross times 4 right parenthesis plus left parenthesis negative 3 cross times 1 right parenthesis end cell end table close parentheses end cell row blank equals cell open parentheses table row cell 0 plus 30 end cell cell 28 plus 10 end cell row cell 0 minus 9 end cell cell 36 minus 3 end cell end table close parentheses end cell row blank equals cell open parentheses table row 30 38 row cell negative 9 end cell 33 end table close parentheses end cell end table$

Dengan demikian, diperoleh hasil dari $left parenthesis K minus L right parenthesis M to the power of T$ adalah $open parentheses table row 30 38 row cell negative 9 end cell 33 end table close parentheses$ .

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

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Terakhir diupdate 13 September 2021

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Diketahui , dan . Tentukan hasil dari matriks berikut

Pembahasan Soal:

Diketahui $K equals open parentheses table row 4 12 row 9 cell negative 7 end cell end table close parentheses comma L equals open parentheses table row cell negative 3 end cell 2 row 0 cell negative 4 end cell end table close parentheses$ , dan $M equals open parentheses table row 0 3 row 4 1 end table close parentheses$ .

Akan ditentukan hasil dari $L to the power of T left parenthesis K plus M right parenthesis$

Ingat kembali mengenai operasi perhitungan dalam matriks

Penjumlahan Matriks

$A plus B equals open parentheses table row cell a subscript 1 end cell cell a subscript 2 end cell row cell a subscript 3 end cell cell a subscript 4 end cell end table close parentheses plus open parentheses table row cell b subscript 1 end cell cell b subscript 2 end cell row cell b subscript 3 end cell cell b subscript 4 end cell end table close parentheses equals open parentheses table row cell a subscript 1 plus b subscript 1 end cell cell a subscript 2 plus b subscript 2 end cell row cell a subscript 3 plus b subscript 3 end cell cell a subscript 4 plus b subscript 4 end cell end table close parentheses$

Perkalian Matriks

$A cross times B equals open parentheses table row cell a subscript 1 end cell cell a subscript 2 end cell row cell a subscript 3 end cell cell a subscript 4 end cell end table close parentheses cross times open parentheses table row cell b subscript 1 end cell cell b subscript 2 end cell row cell b subscript 3 end cell cell b subscript 4 end cell end table close parentheses equals open parentheses table row cell a subscript 1 b subscript 1 plus a subscript 2 b subscript 3 end cell cell a subscript 1 b subscript 2 plus a subscript 2 b subscript 4 end cell row cell a subscript 3 b subscript 1 plus a subscript 4 b subscript 3 end cell cell a subscript 3 b subscript 2 plus a subscript 4 b subscript 4 end cell end table close parentheses$

Transpose Matriks

Diperoleh perhitungan sebagai berikut

$table attributes columnalign right center left columnspacing 0px end attributes row cell K plus M end cell equals cell open parentheses table row 4 12 row 9 cell negative 7 end cell end table close parentheses plus open parentheses table row 0 3 row 4 1 end table close parentheses end cell row blank equals cell open parentheses table row cell 4 plus 0 end cell cell 12 plus 3 end cell row cell 9 plus 4 end cell cell negative 7 plus 1 end cell end table close parentheses end cell row blank equals cell open parentheses table row 4 15 row 13 cell negative 6 end cell end table close parentheses end cell end table$

$L equals open parentheses table row cell negative 3 end cell 2 row 0 cell negative 4 end cell end table close parentheses rightwards double arrow L to the power of T equals open parentheses table row cell negative 3 end cell 0 row 2 cell negative 4 end cell end table close parentheses$

Selanjutnya, diperoleh

$table attributes columnalign right center left columnspacing 0px end attributes row cell L to the power of T left parenthesis K plus M right parenthesis end cell equals cell open parentheses table row cell negative 3 end cell 0 row 2 cell negative 4 end cell end table close parentheses open parentheses table row 4 15 row 13 cell negative 6 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell left parenthesis negative 3 cross times 4 right parenthesis plus left parenthesis 0 cross times 13 right parenthesis end cell cell left parenthesis negative 3 cross times 15 right parenthesis plus left parenthesis 0 cross times left parenthesis negative 6 right parenthesis right parenthesis end cell row cell left parenthesis 2 cross times 4 right parenthesis plus left parenthesis negative 4 cross times 13 right parenthesis end cell cell left parenthesis 2 cross times 15 right parenthesis plus left parenthesis negative 4 cross times left parenthesis negative 6 right parenthesis right parenthesis end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 12 plus 0 end cell cell negative 45 plus 0 end cell row cell 8 minus 52 end cell cell 30 plus 24 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 12 end cell cell negative 45 end cell row cell negative 44 end cell 54 end table close parentheses end cell end table$

Dengan demikian, diperoleh nilai dari $L to the power of T left parenthesis K plus M right parenthesis$ adalah $open parentheses table row cell negative 12 end cell cell negative 45 end cell row cell negative 44 end cell 54 end table close parentheses$

Roboguru

Diketahui , , dan . Tentukan : b.

Pembahasan Soal:

Diberikan matriks $K equals open parentheses table row 4 12 row 9 cell negative 7 end cell end table close parentheses$ , $L equals open parentheses table row cell negative 3 end cell 2 row 0 cell negative 4 end cell end table close parentheses$ , dan $M equals open parentheses table row 0 3 row 4 1 end table close parentheses$ . Penjumlahan matriks K dan M dapat dihitung seperti betikut :

Hasil perkalian antara transpose dari matriks L dengan penjumlahan matriks K dan M adalah

$table attributes columnalign right center left columnspacing 0px end attributes row cell L to the power of T left parenthesis K plus M right parenthesis end cell equals cell open parentheses table row cell negative 3 end cell 0 row 2 cell negative 4 end cell end table close parentheses open parentheses table row 4 15 row 13 cell negative 6 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 3 left parenthesis 4 right parenthesis plus 0 left parenthesis 13 right parenthesis end cell cell negative 3 left parenthesis 15 right parenthesis plus 0 left parenthesis negative 6 right parenthesis end cell row cell 2 left parenthesis 4 right parenthesis minus 4 left parenthesis 13 right parenthesis end cell cell 2 left parenthesis 15 right parenthesis minus 4 left parenthesis negative 6 right parenthesis end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 12 plus 0 end cell cell negative 45 plus 0 end cell row cell 8 minus 52 end cell cell 30 plus 24 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 12 end cell cell negative 45 end cell row cell negative 46 end cell 54 end table close parentheses end cell end table$

Jadi, hasil dari operasi matriks $L to the power of T left parenthesis K plus M right parenthesis$ adalah $open parentheses table row cell negative 12 end cell cell negative 45 end cell row cell negative 46 end cell 54 end table close parentheses$ .

Roboguru

Diketahui matriks adalah transpos matriks S, maka nilai m adalah ....

Pembahasan Soal:

$begin mathsize 14px style straight P equals open parentheses table row cell negative 2 end cell 3 row 1 5 end table close parentheses comma space straight Q equals open parentheses table row 4 cell negative 6 end cell row 2 1 end table close parentheses comma space straight R equals open parentheses table row straight n cell negative 3 end cell row 1 cell negative 5 end cell end table close parentheses comma space straight S equals open parentheses table row straight m 3 row cell 2 straight n end cell 4 end table close parentheses. end style$

$PQ space equals space straight R space plus space straight S to the power of straight T open parentheses table row cell negative 2 end cell 3 row 1 5 end table close parentheses open parentheses table row 4 cell negative 6 end cell row 2 1 end table close parentheses equals open parentheses table row straight n cell negative 3 end cell row 1 cell negative 5 end cell end table close parentheses plus open parentheses table row straight m cell 2 straight n end cell row 3 4 end table close parentheses open parentheses table row cell negative 8 plus 6 end cell cell 12 plus 3 end cell row cell 4 plus 10 end cell cell negative 6 plus 5 end cell end table close parentheses equals open parentheses table row cell straight n plus straight m end cell cell negative 3 plus 2 straight n end cell row 14 cell negative 1 end cell end table close parentheses$

Roboguru

3. Diketahui matriks adalah transpose matriks A. Jika matriks ...

Pembahasan Soal:

Jika diketahui matriks $A equals open parentheses table row a b row c d end table close parentheses$ dan $B equals open parentheses table row e f row g h end table close parentheses$ , maka dapat ditentukan transpose dan operasi penjumlahan matriks sebagai berikut.

$A to the power of T equals open parentheses table row a c row b d end table close parentheses$

$A plus B equals open parentheses table row a b row c d end table close parentheses plus open parentheses table row e f row g h end table close parentheses equals open parentheses table row cell a plus e end cell cell b plus f end cell row cell c plus g end cell cell d plus h end cell end table close parentheses$

Operasi perkalian dua buah matriks dapat dilakukan dengan syarat banyak kolom pada matriks pertama sama dengan banyak baris pada matriks kedua.

Jika diketahui matriks $A equals open parentheses table row a b row c d end table close parentheses$ , maka dapat ditentukan determinan dan invers matriks sebagai berikut.

$text det end text A equals a d minus b c$

$A to the power of negative 1 end exponent equals fraction numerator 1 over denominator text det end text space A end fraction open parentheses table row d cell negative b end cell row cell negative c end cell a end table close parentheses$

Pada persamaan matriks berlaku rumus berikut.

$A X equals B space rightwards double arrow space X equals A to the power of negative 1 end exponent B X A equals B space rightwards double arrow space X equals B A to the power of negative 1 end exponent$

Penyelesaian soal di atas adalah sebagai berikut.

$table attributes columnalign right center left columnspacing 0px end attributes row cell A X end cell equals cell B plus A to the power of straight T end cell row X equals cell A to the power of negative 1 end exponent left parenthesis straight B plus straight A to the power of straight T right parenthesis end cell row X equals cell fraction numerator 1 over denominator 3 left parenthesis 5 right parenthesis minus 7 left parenthesis 2 right parenthesis end fraction open parentheses table row 5 cell negative 7 end cell row cell negative 2 end cell 3 end table close parentheses open parentheses open parentheses table row 1 3 row cell negative 4 end cell cell negative 3 end cell end table close parentheses plus open parentheses table row 3 2 row 7 5 end table close parentheses close parentheses end cell row X equals cell open parentheses 1 close parentheses open parentheses table row 5 cell negative 7 end cell row cell negative 2 end cell 3 end table close parentheses open parentheses open parentheses table row 1 3 row cell negative 4 end cell cell negative 3 end cell end table close parentheses plus open parentheses table row 3 2 row 7 5 end table close parentheses close parentheses end cell row X equals cell open parentheses table row 5 cell negative 7 end cell row cell negative 2 end cell 3 end table close parentheses open parentheses table row 4 5 row 3 2 end table close parentheses end cell row X equals cell open parentheses table row cell negative 1 end cell 11 row 1 cell negative 4 end cell end table close parentheses end cell end table$

Jadi, jawaban yang tepat adalah B. $space$

Roboguru

Tentukan : d.

Pembahasan Soal:

Pertama, tentukan penjumlahan/pengurangan matriks dengan cara menjumlahkan/mengurangkan elemen matriks yang seletak,

$table attributes columnalign right center left columnspacing 0px end attributes row cell A plus B plus C end cell equals cell open parentheses table row 1 2 row 3 4 end table close parentheses plus open parentheses table row 2 1 row 4 3 end table close parentheses plus open parentheses table row 2 0 row 2 1 end table close parentheses end cell row blank equals cell open parentheses table row cell 1 plus 2 plus 2 end cell cell 2 plus 1 plus 0 end cell row cell 3 plus 4 plus 2 end cell cell 4 plus 3 plus 1 end cell end table close parentheses end cell row blank equals cell open parentheses table row 5 3 row 9 8 end table close parentheses end cell end table$

Selanjutnya tentukan transpose matriks (baris menjadi kolom) seperti $M equals open parentheses table row straight a straight b row straight c straight d end table close parentheses rightwards arrow M apostrophe equals open parentheses table row straight a straight c row straight b straight d end table close parentheses$ , maka

$table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis A plus B plus C right parenthesis apostrophe end cell equals cell open parentheses table row 5 9 row 3 8 end table close parentheses end cell row cell D apostrophe end cell equals cell open parentheses table row 1 3 4 row 2 2 5 end table close parentheses end cell end table$

Ingat operasi perkalian matriks berikut ini:

$open parentheses table row straight a straight b row straight c straight d end table close parentheses table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses table row straight e straight f straight g row straight h straight i straight j end table close parentheses end cell end table equals table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses table row cell ae plus bh end cell cell af plus bi end cell cell ag plus bj end cell row cell ce plus dh end cell cell cf plus di end cell cell cg plus dj end cell end table close parentheses end cell end table$

Maka,

$table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis A plus B plus C right parenthesis apostrophe D apostrophe end cell equals cell open parentheses table row 5 9 row 3 8 end table close parentheses open parentheses table row 1 3 4 row 2 2 5 end table close parentheses end cell row blank equals cell open parentheses table row cell 5 plus 18 end cell cell 15 plus 18 end cell cell 20 plus 45 end cell row cell 3 plus 16 end cell cell 9 plus 16 end cell cell 12 plus 40 end cell end table close parentheses end cell row blank equals cell open parentheses table row 23 33 65 row 19 25 52 end table close parentheses end cell end table$

Jadi, hasil operasi tersebut adalah $table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses table row 23 33 65 row 19 25 52 end table close parentheses end cell end table$ .

Roboguru