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Idham Muqoddas
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Berikut ini adalah kumpulan soal mengenai matriks. Jika ingin bertanya soal, silahkan gabung ke grup Telegram, Signal, Discord, atau WhatsApp.

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STANDARSNBTHOTS


No.

Jika matriks \(A=\begin{pmatrix}a&-3\\1&1\end{pmatrix}\) merupakan matriks yang mempunyai invers, maka hasil kali semua nilai a yang mungkin sehingga 3det (A) = det (A−1) − 2 adalah
  1. \(-\dfrac{29}3\)
  2. \(-\dfrac{20}3\)
  3. \(\dfrac{20}3\)
  1. \(\dfrac{29}3\)
  2. \(\dfrac{32}3\)
ALTERNATIF PENYELESAIAN
\(\begin{aligned} \det(A)&=a\cdot1-(-3)\cdot1\\ &=a+3 \end{aligned}\)

CARA 1

\(\begin{aligned} A^{-1}&=\dfrac1{\det(A)}\begin{pmatrix}1&3\\-1&a\end{pmatrix}\\ &=\dfrac1{a+3}\begin{pmatrix}1&3\\-1&a\end{pmatrix}\\ &=\begin{pmatrix}\dfrac1{a+3}&\dfrac3{a+3}\\-\dfrac1{a+3}&\dfrac{a}{a+3}\end{pmatrix}\\ \det\left(A^{-1}\right)&=\dfrac1{a+3}\cdot\dfrac{a}{a+3}-\dfrac3{a+3}\cdot\left(-\dfrac1{a+3}\right)\\ &=\dfrac{a}{(a+3)^2}+\dfrac3{(a+3)^2}\\ &=\dfrac{a+3}{(a+3)^2}\\ &=\dfrac1{a+3} \end{aligned}\)

\(\begin{aligned} 3\det(A)&=\det\left(A^{-1}\right)-2\\ 3(a+3)&=\dfrac1{a+3}-2\qquad\color{red}{\times (a+3)}\\ 3(a+3)^2&=1-2(a+3)\\ 3\left(a^2+6a+9\right)&=1-2a-6\\ 3a^2+18a+27&=-2a-5\\ 3a^2+18a+27+2a+5&=0\\ 3a^2+20a+32&=0 \end{aligned}\)

\(a_1a_2=\dfrac{32}3\)

CARA 2

\(\begin{aligned} 3\det(A)&=\det\left(A^{-1}\right)-2\\ 3\det(A)&=\dfrac1{\det(A)}-2&\color{red}{\times\det(A)}\\ 3\left(\det(A)\right)^2&=1-2\det(A)\\ 3\left(\det(A)\right)^2+2\det(A)-1&=0\\ \left(3\det(A)-1\right)\left(\det(A)+1\right)&=0\\ (3(a+3)-1)(a+3+1)&=0\\ (3a+8)(a+4)&=0 \end{aligned}\)

\(\begin{aligned} a_1a_2&=\left(-\dfrac83\right)(-4)\\ &=\dfrac{32}3 \end{aligned}\)
Jadi, hasil kali semua nilai a yang mungkin adalah \(\dfrac{32}3\).
JAWAB: E

No.

Diketahui matriks \(A=\begin{pmatrix}3&a\\b&2\end{pmatrix}\) dan \(B=\begin{pmatrix}a&b\\3&2\end{pmatrix}\). Jika C adalah matriks berukuran 2×2 yang memiliki invers dan matriks AC maupun matriks BC tidak memiliki invers, maka nilai 4a2 + 9b2 =
  1. 72
  2. 74
  3. 76
  1. 78
  2. 80
ALTERNATIF PENYELESAIAN
|A| = 6 − ab
|B| = 2a − 3b

\(\begin{aligned} AC&=0\\ |A||C|&=0\\ (6-ab)|C|&=0\\ 6-ab&=0\\ ab&=6 \end{aligned}\)

\(\begin{aligned} BC&=0\\ |B||C|&=0\\ (2a-3b)|C|&=0\\ 2a-3b&=0 \end{aligned}\)

\(\begin{aligned} 4a^2+9b^2&=(2a)^2+(-3b)^2\\ &=(2a+(-3b))^2-2(2a)(-3b)\\ &=(2a-3b)^2+12ab\\ &=0^2+12(6)\\ &=\boxed{\boxed{72}} \end{aligned}\)
Jadi, 4a2 + 9b2 = 72.
JAWAB: A

No.

Jika \({\begin{pmatrix}1&2\\1&3\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}a\\b\end{pmatrix}}\) dan \({\begin{pmatrix}5&2\\3&1\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix}=\begin{pmatrix}1\\4\end{pmatrix}}\), maka nilai x + y =
  1. 30
  2. 31
  3. 32
  1. 33
  2. 34
ALTERNATIF PENYELESAIAN
\(\begin{aligned} \begin{pmatrix}5&2\\3&1\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix}&=\begin{pmatrix}1\\4\end{pmatrix}\\ \begin{pmatrix}5&2\\3&1\end{pmatrix}\begin{pmatrix}1&2\\1&3\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}&=\begin{pmatrix}1\\4\end{pmatrix}\\ \begin{pmatrix}7&16\\4&9\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}&=\begin{pmatrix}1\\4\end{pmatrix}\\ \begin{pmatrix}x\\y\end{pmatrix}&=\begin{pmatrix}7&16\\4&9\end{pmatrix}^{-1}\begin{pmatrix}1\\4\end{pmatrix}\\ &=\dfrac1{(7)(9)-(16)(4)}\begin{pmatrix}9&-16\\-4&7\end{pmatrix}\begin{pmatrix}1\\4\end{pmatrix}\\ &=\dfrac1{(-1}\begin{pmatrix}-55\\24\end{pmatrix}\\ &=\begin{pmatrix}55\\-24\end{pmatrix} \end{aligned}\)
x = 55
y = −24

\(\begin{aligned} x+y&=55+(-24)\\ &=\boxed{\boxed{31}} \end{aligned}\)
Jadi, x + y = 31.
JAWAB: B

No.

Diketahui matriks A berukuran 3×3 dan memenuhi \({A\begin{pmatrix}3\\2\\1\end{pmatrix}=\begin{pmatrix}2\\4\\2\end{pmatrix}}\) dan \({A\begin{pmatrix}3\\1\\2\end{pmatrix}=\begin{pmatrix}2\\2\\2\end{pmatrix}}\), matriks \({A\begin{pmatrix}6\\4\\2\end{pmatrix}=}\)
  1. \(\begin{pmatrix}4\\8\\4\end{pmatrix}\)
  2. \(\begin{pmatrix}8\\4\\8\end{pmatrix}\)
  3. \(\begin{pmatrix}12\\8\\4\end{pmatrix}\)
  1. \(\begin{pmatrix}4\\8\\12\end{pmatrix}\)
  2. \(\begin{pmatrix}12\\8\\12\end{pmatrix}\)
ALTERNATIF PENYELESAIAN

CARA BIASA

Misal \(A=\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}\)

\(\begin{aligned} A\begin{pmatrix}3\\2\\1\end{pmatrix}&=\begin{pmatrix}2\\4\\2\end{pmatrix}\\ \begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}\begin{pmatrix}3\\2\\1\end{pmatrix}&=\begin{pmatrix}2\\4\\2\end{pmatrix}\\ \begin{pmatrix}3a+2b+c\\3d+2e+f\\3g+2h+i\end{pmatrix}&=\begin{pmatrix}2\\4\\2\end{pmatrix} \end{aligned}\)

3a + 2b + c = 2
3d + 2e + f = 4
3g + 2h + i = 2

\(\begin{aligned} A\begin{pmatrix}6\\4\\2\end{pmatrix}&=\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}\begin{pmatrix}6\\4\\2\end{pmatrix}\\ &=\begin{pmatrix}6a+4b+2c\\6d+4e+2f\\6g+4h+2i\end{pmatrix}\\ &=\begin{pmatrix}2(3a+2b+c)\\2(3d+2e+f)\\2(3g+2h+i)\end{pmatrix}\\ &=\begin{pmatrix}2(2)\\2(4)\\2(2)\end{pmatrix}\\ &=\boxed{\boxed{\begin{pmatrix}4\\8\\4\end{pmatrix}}} \end{aligned}\)

CARA CEPAT

\(\begin{aligned} A\begin{pmatrix}6\\4\\2\end{pmatrix}&=A\cdot2\begin{pmatrix}3\\2\\1\end{pmatrix}\\ &=2A\begin{pmatrix}3\\2\\1\end{pmatrix}\\ &=2\begin{pmatrix}2\\4\\2\end{pmatrix}\\ &=\boxed{\boxed{\begin{pmatrix}4\\8\\4\end{pmatrix}}} \end{aligned}\)
Jadi, \(A\begin{pmatrix}6\\4\\2\end{pmatrix}=\begin{pmatrix}4\\8\\4\end{pmatrix}\).
JAWAB: A

No.

Diketahui matriks A berordo 2×2 dan \({A=\begin{pmatrix}-2&-5\\1&3\end{pmatrix},}\) \({C=\begin{pmatrix}4&6\\3&5\end{pmatrix}.}\) Jika B memenuhi A⋅B = C, maka det (2B−1) adalah
  1. 1
  2. 2
  3. −2
  1. −1
  2. −4
SKMP Februari 2021
ALTERNATIF PENYELESAIAN
det A = (−2)(3) − (−5)(1) = −1
det C = (4)(5) − (6)(3) = 2

\(\begin{aligned} A\cdot B&=C\\ B&=A^{-1}\cdot C \end{aligned}\)

\(\begin{aligned} \det\left(2B^{-1}\right)&=2^2\det\left(B^{-1}\right)\\ &=\dfrac4{\det B}\\ &=\dfrac4{\det \left(A^{-1}\cdot C\right)}\\ &=\dfrac4{\det A^{-1}\cdot\det C}\\ &=\dfrac{4\det A}{\det C}\\ &=\dfrac{4(-1)}2\\ &=\boxed{\boxed{-2}} \end{aligned}\)
Jadi, det (2B−1) = −2.
JAWAB: C

No.

Jika $A=\begin{pmatrix}-1&-1&0\\-1&1&2\end{pmatrix}$, $B=\begin{pmatrix}-1&x\\1&y\\0&z\end{pmatrix}$ dan $AB=\begin{pmatrix}0&2\\ 2&4\end{pmatrix}$, maka nilai $z-x$ adalah ....
SBMPTN 2014 kode 677
ALTERNATIF PENYELESAIAN
\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} (-1)(-1) + (-1)(1) + (0)(0) & (-1)(x) + (-1)(y) + (0)(z) \\ (-1)(-1) + (1)(1) + (2)(0)& (-1)(x) + (1)(y) + (2)(z) \end{pmatrix}&=\begin{pmatrix}0&2\\ 2&4\end{pmatrix}\\ \begin{pmatrix} 1 - 1 + 0 & -x - y \\ 1 + 1 + 0 & -x + y + 2z \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}

Berdasarkan kesamaan dua matriks, kita peroleh sistem persamaan:

$$-x - y = 2 \quad \text{... (Persamaan 1)}$$

$$-x + y + 2z = 4 \quad \text{... (Persamaan 2)}$$

Untuk mencari nilai $z - x$, kita dapat menjumlahkan Persamaan 1 dan Persamaan 2 secara langsung untuk mengeliminasi variabel $y$:

\begin{aligned}(-x - y) + (-x + y + 2z) &= 2 + 4\\ -2x + 2z &= 6\\ 2(z - x) &= 6\\ z - x &= \boxed{\boxed{3}}\end{aligned}
Jadi, nilai $z - x$ adalah $3$.

No.

Jika $A=\begin{pmatrix}-1&1&0\\-2&1&1\end{pmatrix}$, $B=\begin{pmatrix}1&x\\-1&y\\0&z\end{pmatrix}$ dan $AB=\begin{pmatrix}-2&-1\\ -3&-4\end{pmatrix}$, maka nilai $z-x$ adalah ....
Ganesha Operation
ALTERNATIF PENYELESAIAN
\begin{aligned}AB &=\begin{pmatrix}0&2\\ 2&4\end{pmatrix}\\ \begin{pmatrix}-1&1&0\\-2&1&1\end{pmatrix} \begin{pmatrix}1&x\\-1&y\\0&z\end{pmatrix}&=\begin{pmatrix}-2&-1\\ -3&-4\end{pmatrix}\\ \begin{pmatrix} (-1)(1) + (1)(-1) + (0)(0) & (-1)(x) + (1)(y) + (0)(z) \\ (-2)(1) + (1)(-1) + (1)(0) & (-2)(x) + (1)(y) + (1)(z) \end{pmatrix}&=\begin{pmatrix}-2&-1\\ -3&-4\end{pmatrix}\\ \begin{pmatrix} -1 - 1 + 0 & -x + y \\ -2 - 1 + 0 & -2x + y + z \end{pmatrix}&=\begin{pmatrix}-2&-1\\ -3&-4\end{pmatrix}\\ \begin{pmatrix} -2 & -x + y \\ -3 & -2x + y + z \end{pmatrix}&=\begin{pmatrix}-2&-1\\ -3&-4\end{pmatrix}\end{aligned}

Berdasarkan kesamaan dua matriks, kita peroleh sistem persamaan:

$$-x + y = -1 \quad \text{... (Persamaan 1)}$$

$$-2x + y + z = -4 \quad \text{... (Persamaan 2)}$$

Untuk mencari nilai $z - x$, kita dapat mengurangkan Persamaan 2 dengan Persamaan 1 untuk mengeliminasi variabel $y$:

\begin{aligned}(-2x + y + z) - (-x + y) &= -4 - (-1)\\ -2x + y + z + x - y &= -3\\ z - x &= \boxed{\boxed{-3}}\end{aligned}
Jadi, nilai $z - x$ adalah $-3$.


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