Subjects algebra

Determinant Singular

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1. Tentukan determinan dari matriks-matriks berikut. **a.** Matriks $\begin{bmatrix} 3 & -2 \\ -4 & 5 \end{bmatrix}$ Determinannya adalah $$\det = (3)(5) - (-2)(-4) = 15 - 8 = 7$$ **b.** Matriks $\begin{bmatrix} x & 3 \\ -x & 5 \end{bmatrix}$ Determinannya adalah $$\det = x \cdot 5 - 3 \cdot (-x) = 5x + 3x = 8x$$ **c.** Matriks $\begin{bmatrix} x^2 & 2x \\ x^2 & 2x + 3 \end{bmatrix}$ Determinannya adalah $$\det = x^2(2x + 3) - 2x(x^2) = 2x^3 + 3x^2 - 2x^3 = 3x^2$$ **d.** Matriks $\begin{bmatrix} 5 & 2 & 3 \\ 1 & 2 & 6 \\ 2 & -3 & 4 \end{bmatrix}$ Gunakan ekspansi kofaktor baris pertama: $$\det = 5 \begin{vmatrix} 2 & 6 \\ -3 & 4 \end{vmatrix} - 2 \begin{vmatrix} 1 & 6 \\ 2 & 4 \end{vmatrix} + 3 \begin{vmatrix} 1 & 2 \\ 2 & -3 \end{vmatrix}$$ Hitung minor: $$= 5(2 \cdot 4 - (-3) \cdot 6) - 2(1 \cdot 4 - 2 \cdot 6) + 3(1 \cdot (-3) - 2 \cdot 2)$$ $$= 5(8 + 18) - 2(4 - 12) + 3(-3 - 4)$$ $$= 5(26) - 2(-8) + 3(-7) = 130 + 16 - 21 = 125$$ **e.** Matriks $\begin{bmatrix} 1 & 4 & 1 \\ 6 & 24 & 3 \\ 2 & 8 & 7 \end{bmatrix}$ Ekspansi kofaktor baris pertama: $$\det = 1 \begin{vmatrix} 24 & 3 \\ 8 & 7 \end{vmatrix} - 4 \begin{vmatrix} 6 & 3 \\ 2 & 7 \end{vmatrix} + 1 \begin{vmatrix} 6 & 24 \\ 2 & 8 \end{vmatrix}$$ Hitung minor: $$= 1(24 \cdot 7 - 3 \cdot 8) - 4(6 \cdot 7 - 3 \cdot 2) + 1(6 \cdot 8 - 24 \cdot 2)$$ $$= 1(168 - 24) - 4(42 - 6) + 1(48 - 48)$$ $$= 144 - 4(36) + 0 = 144 - 144 = 0$$ **f.** Matriks $3 \times \begin{bmatrix} 4 & 5 & 4 \\ 1 & 1 & 4 \\ 2 & 6 & 7 \end{bmatrix}$ Hitung determinan matriks dalam terlebih dahulu: $$\det = 4 \begin{vmatrix} 1 & 4 \\ 6 & 7 \end{vmatrix} - 5 \begin{vmatrix} 1 & 4 \\ 2 & 7 \end{vmatrix} + 4 \begin{vmatrix} 1 & 1 \\ 2 & 6 \end{vmatrix}$$ Hitung minor: $$= 4(1 \cdot 7 - 4 \cdot 6) - 5(1 \cdot 7 - 4 \cdot 2) + 4(1 \cdot 6 - 1 \cdot 2)$$ $$= 4(7 - 24) - 5(7 - 8) + 4(6 - 2)$$ $$= 4(-17) - 5(-1) + 4(4) = -68 + 5 + 16 = -47$$ Kalikan dengan 3: $$3 \times (-47) = -141$$ 2. Manakah di antara matriks-matriks berikut yang merupakan singular? Sebuah matriks singular jika determinannya $0$. **a.** $\begin{bmatrix} 2 & 3 \\ 5 & 4 \end{bmatrix}$ $$\det = 2 \cdot 4 - 3 \cdot 5 = 8 - 15 = -7 \neq 0$$ Tidak singular. **b.** $\begin{bmatrix} -4 & 3 \\ 2 & 1 \end{bmatrix}$ $$\det = (-4)(1) - 3(2) = -4 - 6 = -10 \neq 0$$ Tidak singular. **c.** $\begin{bmatrix} 1 & 2 & 4 \\ 2 & -2 & 5 \\ -1 & 2 & 3 \end{bmatrix}$ Hitung determinan: $$= 1 \begin{vmatrix} -2 & 5 \\ 2 & 3 \end{vmatrix} - 2 \begin{vmatrix} 2 & 5 \\ -1 & 3 \end{vmatrix} + 4 \begin{vmatrix} 2 & -2 \\ -1 & 2 \end{vmatrix}$$ $$= 1((-2)(3) - 5(2)) - 2(2 \cdot 3 - 5(-1)) + 4(2 \cdot 2 - (-2)(-1))$$ $$= 1(-6 - 10) - 2(6 + 5) + 4(4 - 2) = -16 - 22 + 8 = -30 \neq 0$$ Tidak singular. **d.** $\begin{bmatrix} 6 & 3 \\ 2 & 1 \end{bmatrix}$ $$\det = 6 \cdot 1 - 3 \cdot 2 = 6 - 6 = 0$$ Singular. **e.** $\begin{bmatrix} -3 & 5 \\ 3 & -5 \end{bmatrix}$ $$\det = (-3)(-5) - 5(3) = 15 - 15 = 0$$ Singular. **f.** $\begin{bmatrix} 1 & -1 & 2 \\ 4 & 1 & 3 \\ 7 & 2 & 5 \end{bmatrix}$ Hitung determinan: $$= 1 \begin{vmatrix} 1 & 3 \\ 2 & 5 \end{vmatrix} - (-1) \begin{vmatrix} 4 & 3 \\ 7 & 5 \end{vmatrix} + 2 \begin{vmatrix} 4 & 1 \\ 7 & 2 \end{vmatrix}$$ $$= 1(1 \cdot 5 - 3 \cdot 2) + 1(4 \cdot 5 - 3 \cdot 7) + 2(4 \cdot 2 - 1 \cdot 7)$$ $$= 1(5 - 6) + 1(20 - 21) + 2(8 - 7) = -1 - 1 + 2 = 0$$ Singular. **Kesimpulan:** Matriks singular adalah d, e, dan f pada soal nomor 2.