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Evaluasi Determinan

A. Hitunglah determinan matrik berikut dengan menggunakan rumus expansi baris

SOAL 1.

A=(−7−514)A = \begin{pmatrix} -7 & -5 \\ 1 & 4 \end{pmatrix}
(1)
det⁡(A)=ad−bc\det(A) = ad - bc
(2)
det⁡(A)=(−7∗4)−(−5∗1)\det(A) = ( -7 * 4 ) - ( -5 * 1 )
(3)
det(A)=−28+5=−23det(A) = -28 + 5 = -23
(4)

SOAL 2.

A=[02−31−2−1001]A = \begin{bmatrix} 0 & 2 & -3 \\ 1 & -2 & -1 \\ 0 & 0 & 1 \end{bmatrix}
(5)
(−1) 0+1 det⁡(11)∗M11(−1) 2+1 det⁡(12)∗M12(−1) −3+1 det⁡(13)∗M13(-1)\ 0+1 \ \det(11) * M_{11}\\ (-1)\ 2+1 \ \det(12) * M_{12}\\ (-1)\ -3+1 \ \det(13) * M_{13}\\
(6)
(+)a11∗M11=(+)0∗((−2∗1)−(−1∗0))(−)a11∗M11=(−)2∗((1∗1)−(−1∗0))(+)a11∗M11=(+)−3∗((1∗0)−(−2∗0))(+) a_{11} * M_{11} = (+) 0 * ((-2 * 1) - (-1 * 0))\\ (-) a_{11} * M_{11} = (-) 2 * ((1 * 1) - (-1 * 0))\\ (+) a_{11} * M_{11} = (+) -3 * ((1 * 0) - (-2 * 0))\\
(7)
+0∗(−2−0)−2∗(1−0)+(−3∗(0−0))=−2+ 0 * (-2 - 0 ) \\ - 2 * (1 - 0) \\ + (-3 * (0 - 0))\\ = -2
(8)

SOAL 3.

A=[1−311−311111−31111−3].A = \begin{bmatrix} 1 & -3 & 1 & 1 \\ -3 & 1 & 1 & 1 \\ 1 & 1 & -3 & 1 \\ 1 & 1 & 1 & -3 \end{bmatrix}.
(9)

Ekspansi Baris 1 matriks A (4x4):

(−1) 1+1 det⁡(11)∗M11(−1) 1+2 det⁡(12)∗M12(−1) 1+3 det⁡(13)∗M13(−1) 1+4 det⁡(14)∗M14(-1)\ 1+1 \ \det(11) * M_{11}\\ (-1)\ 1+2 \ \det(12) * M_{12}\\ (-1)\ 1+3 \ \det(13) * M_{13}\\ (-1)\ 1+4 \ \det(14) * M_{14}\\
(10)
(+)a11∗M11=(+)1∗det⁡(M11)(−)a12∗M12=(−)−3∗det⁡(M12)(+)a13∗M13=(+)1∗det⁡(M13)(−)a14∗M14=(−)1∗det⁡(M14)(+) a_{11} * M_{11} = (+) 1 * \det(M_{11})\\ (-) a_{12} * M_{12} = (-) -3 * \det(M_{12})\\ (+) a_{13} * M_{13} = (+) 1 * \det(M_{13})\\ (-) a_{14} * M_{14} = (-) 1 * \det(M_{14})\\
(11)

Mencari nilai det⁡(M11)\det(M_{11}):

M11=[1111−3111−3]M_{11} = \begin{bmatrix} 1 & 1 & 1 \\ 1 & -3 & 1 \\ 1 & 1 & -3 \end{bmatrix}
(12)
(−1) 1+1 det⁡(11)∗M11(−1) 1+2 det⁡(12)∗M12(−1) 1+3 det⁡(13)∗M13(-1)\ 1+1 \ \det(11) * M_{11}\\ (-1)\ 1+2 \ \det(12) * M_{12}\\ (-1)\ 1+3 \ \det(13) * M_{13}\\
(13)
(+)a11∗M11=(+)1∗((−3∗−3)−(1∗1))(−)a12∗M12=(−)1∗((1∗−3)−(1∗1))(+)a13∗M13=(+)1∗((1∗1)−(−3∗1))(+) a_{11} * M_{11} = (+) 1 * ((-3 * -3) - (1 * 1))\\ (-) a_{12} * M_{12} = (-) 1 * ((1 * -3) - (1 * 1))\\ (+) a_{13} * M_{13} = (+) 1 * ((1 * 1) - (-3 * 1))\\
(14)
+1∗(9−1)−1∗(−3−1)+(1∗(1−−3))=16+ 1 * (9 - 1) \\ - 1 * (-3 - 1) \\ + (1 * (1 - -3))\\ = 16
(15)

Mencari nilai det⁡(M12)\det(M_{12}):

M12=[−3111−3111−3]M_{12} = \begin{bmatrix} -3 & 1 & 1 \\ 1 & -3 & 1 \\ 1 & 1 & -3 \end{bmatrix}
(16)
(−1) 1+1 det⁡(11)∗M11(−1) 1+2 det⁡(12)∗M12(−1) 1+3 det⁡(13)∗M13(-1)\ 1+1 \ \det(11) * M_{11}\\ (-1)\ 1+2 \ \det(12) * M_{12}\\ (-1)\ 1+3 \ \det(13) * M_{13}\\
(17)
(+)a11∗M11=(+)−3∗((−3∗−3)−(1∗1))(−)a12∗M12=(−)1∗((1∗−3)−(1∗1))(+)a13∗M13=(+)1∗((1∗1)−(−3∗1))(+) a_{11} * M_{11} = (+) -3 * ((-3 * -3) - (1 * 1))\\ (-) a_{12} * M_{12} = (-) 1 * ((1 * -3) - (1 * 1))\\ (+) a_{13} * M_{13} = (+) 1 * ((1 * 1) - (-3 * 1))\\
(18)
+−3∗(9−1)−1∗(−3−1)+(1∗(1−−3))=−16+ -3 * (9 - 1) \\ - 1 * (-3 - 1) \\ + (1 * (1 - -3))\\ = -16
(19)

Mencari nilai det⁡(M13)\det(M_{13}):

M13=[−31111111−3]M_{13} = \begin{bmatrix} -3 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & -3 \end{bmatrix}
(20)
(−1) 1+1 det⁡(11)∗M11(−1) 1+2 det⁡(12)∗M12(−1) 1+3 det⁡(13)∗M13(-1)\ 1+1 \ \det(11) * M_{11}\\ (-1)\ 1+2 \ \det(12) * M_{12}\\ (-1)\ 1+3 \ \det(13) * M_{13}\\
(21)
(+)a11∗M11=(+)−3∗((1∗−3)−(1∗1))(−)a12∗M12=(−)1∗((1∗−3)−(1∗1))(+)a13∗M13=(+)1∗((1∗1)−(1∗1))(+) a_{11} * M_{11} = (+) -3 * ((1 * -3) - (1 * 1))\\ (-) a_{12} * M_{12} = (-) 1 * ((1 * -3) - (1 * 1))\\ (+) a_{13} * M_{13} = (+) 1 * ((1 * 1) - (1 * 1))\\
(22)
+−3∗(−3−1)−1∗(−3−1)+(1∗(1−1))=16+ -3 * (-3 - 1) \\ - 1 * (-3 - 1) \\ + (1 * (1 - 1))\\ = 16
(23)

Mencari nilai det⁡(M14)\det(M_{14}):

M14=[−31111−3111]M_{14} = \begin{bmatrix} -3 & 1 & 1 \\ 1 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}
(24)
(−1) 1+1 det⁡(11)∗M11(−1) 1+2 det⁡(12)∗M12(−1) 1+3 det⁡(13)∗M13(-1)\ 1+1 \ \det(11) * M_{11}\\ (-1)\ 1+2 \ \det(12) * M_{12}\\ (-1)\ 1+3 \ \det(13) * M_{13}\\
(25)
(+)a11∗M11=(+)−3∗((1∗1)−(−3∗1))(−)a12∗M12=(−)1∗((1∗1)−(−3∗1))(+)a13∗M13=(+)1∗((1∗1)−(1∗1))(+) a_{11} * M_{11} = (+) -3 * ((1 * 1) - (-3 * 1))\\ (-) a_{12} * M_{12} = (-) 1 * ((1 * 1) - (-3 * 1))\\ (+) a_{13} * M_{13} = (+) 1 * ((1 * 1) - (1 * 1))\\
(26)
+−3∗(1−−3)−1∗(1−−3)+(1∗(1−1))=−16+ -3 * (1 - -3) \\ - 1 * (1 - -3) \\ + (1 * (1 - 1))\\ = -16
(27)

Hasil Akhir Determinan AA:

+1∗(16)−−3∗(−16)+1∗(16)−1∗(−16)=16−48+16+16=0+ 1 * (16) \\ - -3 * (-16) \\ + 1 * (16)\\ - 1 * (-16)\\ = 16 - 48 + 16 + 16 = 0
(28)
KOMPUTASI ALJABAR LINEAR
Evaluasi Determinan dan Invers
KOMPUTASI ALJABAR LINEAR
Evaluasi Invers