soal matriks kelas 11 ,dengan penjelasan
1. soal matriks kelas 11 ,dengan penjelasan
Jawab:
2 3
-1 4
Penjelasan dengan langkah-langkah:
At = baris jadi kolom
kolom jadi baris
At = 2 3
-1 4
2. Ini jawabannya apa ya? Soal matriks kelas 11
Penjelasan dengan langkah-langkah:
Jawabannyaadadiatassemogabisamembantu
3. Mohon penjelasan dari Soal ini, materi matriks kelas 11
(matrik A) x = (matrik B)
x= (matrik A invers) (matrikB)
X (matrikA) = (matrik B)
X=(mateik B) (matrik A invers)
4. Soal matriks kelas 11
Jawab: D
Penjelasan :
B.A=C
[tex]B. \left[\begin{array}{ccc}-5&7\\1&-3\end{array}\right] = \left[\begin{array}{ccc}13&-23\\-26&46\end{array}\right][/tex]
[tex]\left[\begin{array}{ccc}a&b\\c&d\end{array}\right].\left[\begin{array}{ccc}-5&7\\1&-3\end{array}\right] = \left[\begin{array}{ccc}13&-23\\-26&46\end{array}\right][/tex]
[tex]\left[\begin{array}{ccc}a.(-5)+b.(1) &a.(7)+b.(-3)\\c.(-5)+d.(1)&c.(7)+d(-3)\end{array}\right] = \left[\begin{array}{ccc}13&-23\\-26&46\end{array}\right][/tex]
[tex]a(-5)+b(1)=13\\a(7)+b(-3)=-23[/tex] [tex]|kali 3\\|kali (-1)[/tex] [tex]|\\|[/tex] [tex]-15a+3b=39\\-7a+3b=23[/tex] (kemudian dikurangi)
[tex]-8a=16\\a=-2[/tex]
[tex]-7a+3b=23\\-7(-2)+3b=23\\b=3\\[/tex]
[tex]c(-5)+d(1)=-26\\
c(7)+d(-3)=46[/tex] [tex]|\\|[/tex] [tex]kali 3\\kali (-1)[/tex] [tex]|\\|[/tex] [tex]-15c+3d=-78\\-7c+3d=-46[/tex] (kemudian dikurangi)
[tex]-8c=-32\\c=4[/tex]
[tex]-7c+3d=-46\\-7(4)+3d=-46\\d=-6\\[/tex]
sehingga dapat disimpulkan matrix B = [tex]\left[\begin{array}{ccc}-2&3\\4&-6\end{array}\right][/tex]
5. 1 soal ini aja (matriks kelas 11)
Materi : Matriks
Jawaban terlampir
Semoga membantu
jawaban di lampiran ya
6. Soal matriks matematika peminatan kelas 11
Jawab :
Berdasarkan matriks tersebut, diperoleh persamaan :
• a + 3 = 5 – a
a + a = 5 – 3
2a = 2
a = 2/2
a = 1
• 8 + b = 4 – b
b + b = 4 – 8
2b = -4
b = -4/2
b = -2
• -1 + c = c – c
-1 + c = 0
c = 0 + 1
c = 1
• d + (-9) = -13 – d
d – 9 = -13 – d
d + d = -13 + 9
2d = -4
d = -4/2
d = -2
Maka, nilai dari a+b+c+d adalah
a + b + c + d = 1 + (-2) + 1 + (-2)
a + b + c + d = 1 – 2 + 1 – 2
a + b + c + d = -1 – 1 = -2
Jawaban : tidak ada di opsi
vin
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Mapel : Matematika
Kelas : X SMA
Materi : Matriks
Kata Kunci : Persamaan Matriks
Kode Soal : 2 (Matematika)
Kode Kategorisasi : 10.2.8
7. Soal Matriks kelas 11. Sekalian cara jawabnya ya
Jawab:
E. 17
Penjelasan dengan langkah-langkah:
[tex]PQ^{T} = R\\\\\left[\begin{array}{ccc}12&4\\0&-11\end{array}\right] \left[\begin{array}{ccc}x&-3\\2y&4\end{array}\right] = \left[\begin{array}{ccc}96&-20\\66&-44\end{array}\right] \\\\\left[\begin{array}{ccc}12x+8y&-20\\-22y&-44\end{array}\right] = \left[\begin{array}{ccc}96&-20\\66&-44\end{array}\right] \\[/tex]
[tex]-22y = 66\\y = 66 : -22 = -3\\\\12x + 8y = 96\\12x – 24 = 96\\12x = 96 + 24\\x = 120 : 12 = 10[/tex]
[tex]2x + y = 20 – 3 = 17[/tex]
8. 1 soal ini aja ( matriks kelas 11 )
semoga membantu yaaaa
9. Soal matriks kelas 11
maksudnya soal matriks itu apa ya?
10. Soal matriks kelas 11, tolong dijawab yaa
8.
[tex](\text{AB\:}+\:\text{C})=\left[\begin{array}{ccc}2&3\\1&0\end{array}\right]\left[\begin{array}{ccc}2&0\\1&3\end{array}\right]+\left[\begin{array}{ccc}-5&-6\\2&8\end{array}\right][/tex]
[tex](\text{AB\:}+\:\text{C})=\left[\begin{array}{ccc}(2).(2)+(3).(1)&(2).(0)+(3).(3)\\(1).(2)+(0).(1)&(1).(0)+(0).(3)\end{array}\right]+\left[\begin{array}{ccc}-5&-6\\2&8\end{array}\right][/tex]
[tex](\text{AB\:}+\:\text{C})=\left[\begin{array}{ccc}7&9\\2&0\end{array}\right]+\left[\begin{array}{ccc}-5&-6\\2&8\end{array}\right][/tex]
[tex](\text{AB\:}+\:\text{C})=\left[\begin{array}{ccc}2&3\\4&8\end{array}\right][/tex]
[tex](\text{AB\:}+\:\text{C})^{-1}=\frac{1}{(2).(8)-(3).(4)}\left[\begin{array}{ccc}8&-3\\-4&2\end{array}\right][/tex]
[tex]\boxed{\boxed{(\text{AB\:}+\:\text{C})^{-1}=\frac{1}{4}\left[\begin{array}{ccc}8&-3\\-4&2\end{array}\right]}}[/tex]
[tex]\to(\bold{C})[/tex]
9.
[tex]\text{C}^{-1}=\frac{1}{(2).(5)-(3).(3)}[\begin{array}{ccc}5&-3\\-3&2\end{array}][/tex]
[tex]\text{C}^{-1}=\frac{1}{10-9}[\begin{array}{ccc}5&-3\\-3&2\end{array}][/tex]
[tex]\text{C}^{-1}=[\begin{array}{ccc}5&-3\\-3&2\end{array}][/tex]
[tex]\text{A\:}+\text{\:B\:}=\text{\:C}^{-1}[/tex]
[tex][\begin{array}{ccc}6&2\\-3&-2\end{array}]+[\begin{array}{ccc}-1&-5\\0&3n\:+\:1\end{array}]=[\begin{array}{ccc}5&-3\\-3&2\end{array}][/tex]
[tex][\begin{array}{ccc}5&-3\\-3&3n\:+\:1\end{array}]=[\begin{array}{ccc}5&-3\\-3&2\end{array}][/tex]
Maka, didapatkan persamaan :
3n + 1 = 2
3n = 2 – 1
3n = 1
[tex]\boxed{\boxed{n=\frac{1}{3}}}\to\:(\bold{B})[/tex]
10.
[tex]\text{AB}=[\begin{array}{ccc}a&1&a\\-1&a&2\end{array}][\begin{array}{ccc}-2&1\\1&0\\1&-1\end{array}][/tex]
[tex]\text{AB}=[\begin{array}{ccc}(a).(-2)+(1).(1)+(a).(1)&(a).(1)+(1).(0)+(a).(-1)\\(-1).(-2)+(a).(1)+(2).(1)&(-1).(1)+(a).(0)+(2).(-1)\end{array}][/tex]
[tex]\text{AB}=[\begin{array}{ccc}-2a+1+a&a+0-a\\2+a+2&-1+0-2\end{array}][/tex]
[tex]\text{AB}=[\begin{array}{ccc}-a+1&0\\a+4&-3\end{array}][/tex]
[tex]det\:(\text{AB})\:=\:6[/tex]
[tex](-a+1)(-3)\:-\:(0).(a+4)\:=\:6[/tex]
[tex]3a\:-\:3\:-\:0\:=\:6[/tex]
[tex]3a\:=\:6\:+\:3[/tex]
[tex]3a\:=\:9\:\:\to\:a\:=\:3[/tex]
Sehingga:
[tex]\text{AB}=[\begin{array}{ccc}-a+1&0\\a+4&-3\end{array}][/tex]
[tex]\to\:\text{AB}=[\begin{array}{ccc}-3+1&0\\3+4&-3\end{array}][/tex]
[tex]\to\:\text{AB}=[\begin{array}{ccc}-2&0\\7&-3\end{array}][/tex]
\to\:
[tex](\text{AB})^{-1}=\frac{1}{6}[\begin{array}{ccc}-3&0\\-7&-2\end{array}][/tex]
[tex]\text{B}^{-1}.\text{A}^{-1}=(\text{AB})^{-1}[/tex]
[tex]\boxed{\boxed{\text{B}^{-1}.\text{A}^{-1}=\frac{1}{6}[\begin{array}{ccc}-3&0\\-7&-2\end{array}]}}[/tex]
[tex]\to(\bold{D})[/tex]
11. tolong yg bisa soal matriks soal kelas 11
Maaf klo slh. Semoga membantu
Penjelasan dengan langkah-langkah:
[tex]{\pink{\fcolorbox{blue}{black}{\boxed{\bold{\blue{by }}}\boxed{\mathfrak{}\red{ pushrank21}}}}}[/tex]
12. SOAL MATRIKS KELAS 11 SMK
Jawaban:
Jawabannya adalah B.
Penjelasan ada pada gambar⬆️⬆️⬆️
13. 1 soal ini aja(matriks kelas 11)
Jawab :
P = [tex] \left[\begin{array}{ccc}2&-1\\4&8\end{array}\right] [/tex]
Q = [tex] \left[\begin{array}{ccc}2&2s\\s+t&8\end{array}\right] [/tex]
Sehingga
Pᵀ = [tex] \left[\begin{array}{ccc}2&4\\-1&8\end{array}\right] [/tex]
Pᵀ = Q
[[tex] \left[\begin{array}{ccc}2&4\\-1&8\end{array}\right] = \left[\begin{array}{ccc}2&2s\\s+t&8\end{array}\right] [/tex]
Lalu didapatkan
-1 = s+t … (1)
4 = 2s … (2)
Sederhanakan persamaan (2)
4 = 2s
4/2 = s
s = 2
Substitusikan ke persamaan (1)
-1 = s+t
-1 = 2+t
-1-2 = t
t = -3
Jadi, nilai s dan t adalah 2 dan -3
vin
ㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡ
Mapel : Matematika
Kelas : X SMA
Materi : Matriks
Kata Kunci : Persamaan Matriks
Kode Soal : 2 (Matematika)
Kode Kategorisasi : 10.2.8
14. jawab dong kak soal mtk wajib kelas 11 tentang matriks
maaf bila tulis
an acak acakan
15. SOAL UN TENTANG MATRIKS KELAS 11
Jawab:
Penjelasan dengan langkah-langkah:
matriks
kesamaan
elemen seletak bernilai sama
___
soal
[tex]\sf A = \left[\begin{array}{ccc}-4&2x- 2\\-2x+11z &6\end{array}\right] ~ dan~ B = \left[\begin{array}{ccc}-2&-5\\26 &z-y\end{array}\right][/tex]
Jika A = 2B , maka
2x-2 = 2(-5)
2x – 2 = -10
2x = -8
x= – 4
-2x + 11 z = 2(26)
-2(-4) + 11 z = 52
8 + 11 z = 52
11z = 44
z = 4
6= 2 (z-y)
3 = z – y
3 = 4 – y
y = 1
nilai (x – 2y + z) =
= (-4) – 2(1) + (4)
= – 2
16. soal matriks kelas 11, ada yang bisa gaa?
SEMOGA BISA MEMBANTU NYA*_*
17. soal matriks kelas 11
Jawaban:
A.
[tex] \binom{12 \: \: \: 18}{17 \: \: – 8} [/tex]
Penjelasan dengan langkah-langkah:
Ada di lampiran ya….
Mapel : MatematikaKelas : XIMateri : MatriksKode soal : 2
[tex]{\boxed{\boxed{\color{Magenta} Semoga \ membantu \ dan \ semangat \ belajarnya :)}}}[/tex]
18. Soal matriks kelas 11
[tex]\text{AC}~=~\text{B}[/tex]
[tex][\begin{array}{ccc}2&-1\\3&-4\end{array}].\text{C}~=~[\begin{array}{ccc}3&2\\-13&-7\end{array}][/tex]
[tex]\text{C}~=~[\begin{array}{ccc}2&-1\\3&-4\end{array}]^{-1}[\begin{array}{ccc}3&2\\-13&-7\end{array}][/tex]
[tex]\text{C}~=~\frac{1}{(2).(-4)~-~ (-1).(3)}[\begin{array}{ccc}-4&1\\-3&2\end{array}][\begin{array}{ccc}3&2\\-13&-7\end{array}][/tex]
[tex]\text{C}~=~-\frac{1}{5}[\begin{array}{ccc}-4&1\\-3&2\end{array}][\begin{array}{ccc}3&2\\-13&-7\end{array}][/tex]
[tex]\text{C}~=~[\begin{array}{ccc}\frac{4}{5}&-\frac{1}{5}\\\frac{3}{5}&-\frac{2}{5}\end{array}][\begin{array}{ccc}3&2\\-13&-7\end{array}][/tex]
[tex]\text{C}~=~[\begin{array}{ccc}(\frac{4}{5}).(3)~+~(-\frac{1}{5}).(-13)&(\frac{4}{5}).(2)~+~(-\frac{1}{5}).(-7)\\(\frac{3}{5}).(3)~+~(-\frac{2}{5}).(-13)&(\frac{3}{5}).(2)~+~(-\frac{2}{5}).(-7)\end{array}][/tex]
[tex]\text{C}~=~[\begin{array}{ccc}\frac{12}{5}~+~\frac{13}{5}&\frac{8}{5}~+~\frac{7}{5}\\\frac{9}{5}~+~\frac{26}{5}&\frac{6}{5}~+~\frac{14}{5}\end{array}][/tex]
[tex]\text{C}~=~[\begin{array}{ccc}\frac{25}{5}&\frac{15}{5}\\\frac{35}{5}&\frac{20}{5}\end{array}][/tex]
[tex]\text{C}~=~[\begin{array}{ccc}5&3\\7&4\end{array}][/tex]
[tex]\text{C}^{-1}~=~\frac{1}{(5).(4)~-~(3).(7)}[\begin{array}{ccc}4&-3\\-7&5\end{array}][/tex]
[tex]\text{C}^{-1}~=~\frac{1}{-1}[\begin{array}{ccc}4&-3\\-7&5\end{array}][/tex]
[tex]\text{C}^{-1}~=~-[\begin{array}{ccc}4&-3\\-7&5\end{array}][/tex]
[tex]\text{C}^{-1}~=~[\begin{array}{ccc}-4&3\\7&-5\end{array}][/tex]
[tex]\to\boxed{\boxed{(\text{C}^{-1})^\text{T}~=~[\begin{array}{ccc}-4&7\\3&-5\end{array}]}}[/tex]
19. 1 soal ini aja ( matriks kelas 11 )
• 3p = p + 4
3p – p = 4
2p = 4
p = 2
• 3q = 6 + p + q
masukkan p = 2
3q – q = 6 + 2
2q = 8
q = 4
• 3s = 2s + 3
3s – 2s = 3
s = 3
• 3r = – 1 + r + s
3r – r = -1 + s
masukkan s = 3
2r = -1 + 3
2r = 2
r = 1
20. matriks kelas 11 soal dilampiran
Jawab:
Penjelasan dengan langkah-langkah: