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Evaluasi Invers

Rumus

(adj⁡A)ij=(−1)i+jMjiA−1=1det⁡Aadj⁡A.\begin{align*} (\operatorname{adj} A)_{ij} &= (-1)^{i+j} M_{ji}\\ A^{-1} &= \frac{1}{\det A} \operatorname{adj} A. \end{align*}
(1)

Soal 1

A=[−7−514]A = \begin{bmatrix} -7 & -5 \\ 1 & 4 \end{bmatrix}
(2)

Determinan A

det⁡(A)=a.d−b.cdet⁡(A)=(−7∗4)−(−5∗1)det⁡(A)=−28+5=−23\begin{align*} \det(A) = a.d - b.c\\ \det(A) = (-7 * 4) - (-5 * 1)\\ \det(A) = -28 + 5 = -23 \end{align*}
(3)

Cari adjoin

adj⁡(A)=[d−b−ca]adj⁡(A)=[45−1−7]\begin{align*} \operatorname{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}\\ \operatorname{adj}(A) = \begin{bmatrix} 4 & 5 \\ -1 & -7 \end{bmatrix} \end{align*}
(4)
A−1=1det⁡Aadj⁡A−1=1−23[45−1−7]A−1=[−4/23−5/231/237/23] \begin{align*} A^{-1} &= \frac{1}{\det A} \operatorname{adj}\\ A^{-1} &= \frac{1}{-23} \begin{bmatrix} 4 & 5 \\ -1 & -7 \end{bmatrix}\\ A^{-1} &= \begin{bmatrix} -4/23 & -5/23 \\ 1/23 & 7/23 \end{bmatrix}\\ \end{align*}
(5)

Soal 2

A=[02−31−2−1001]A = \begin{bmatrix} 0 & 2 & -3 \\ 1 & -2 & -1 \\ 0 & 0 & 1 \end{bmatrix}
(6)

hitung determinan A dengan mengekspansi sepanjang baris ketiga:

det⁡A=0∣2−3−2−1∣−0∣0−31−1∣+1∣021−2∣=1((0)(−2)−(1)(2))=1(0−2)=−2\begin{align*} \det A &= 0 \begin{vmatrix} 2 & -3 \\ -2 & -1 \end{vmatrix} - 0 \begin{vmatrix} 0 & -3 \\ 1 & -1 \end{vmatrix} + 1 \begin{vmatrix} 0 & 2 \\ 1 & -2 \end{vmatrix} \\ &= 1((0)(-2) - (1)(2)) \\ &= 1(0 - 2) = -2 \end{align*}
(7)

hitung adjoin dari A menggunakan rumus (adj⁡A)ij=(−1)i+jMji(\operatorname{adj} A)_{ij} = (-1)^{i+j} M_{ji}:

adj⁡A=[+M11−M21+M31−M12+M22−M32+M13−M23+M33]=[+∣−2−101∣−∣2−301∣+∣2−3−2−1∣−∣1−101∣+∣0−301∣−∣0−31−1∣+∣1−200∣−∣0200∣+∣021−2∣]=[+((−2)(1)−(−1)(0))−((2)(1)−(−3)(0))+((2)(−1)−(−3)(−2))−((1)(1)−(−1)(0))+((0)(1)−(−3)(0))−((0)(−1)−(−3)(1))+((1)(0)−(−2)(0))−((0)(0)−(2)(0))+((0)(−2)−(2)(1))]=[+(−2−0)−(2−0)+(−2−6)−(1−0)+(0−0)−(0−(−3))+(0−0)−(0−0)+(0−2)]=[−2−2−8−10−300−2]\begin{align*} \operatorname{adj} A &= \begin{bmatrix} +M_{11} & -M_{21} & +M_{31} \\ -M_{12} & +M_{22} & -M_{32} \\ +M_{13} & -M_{23} & +M_{33} \end{bmatrix} \\ &= \begin{bmatrix} +\begin{vmatrix} -2 & -1 \\ 0 & 1 \end{vmatrix} & -\begin{vmatrix} 2 & -3 \\ 0 & 1 \end{vmatrix} & +\begin{vmatrix} 2 & -3 \\ -2 & -1 \end{vmatrix} \\[10pt] -\begin{vmatrix} 1 & -1 \\ 0 & 1 \end{vmatrix} & +\begin{vmatrix} 0 & -3 \\ 0 & 1 \end{vmatrix} & -\begin{vmatrix} 0 & -3 \\ 1 & -1 \end{vmatrix} \\[10pt] +\begin{vmatrix} 1 & -2 \\ 0 & 0 \end{vmatrix} & -\begin{vmatrix} 0 & 2 \\ 0 & 0 \end{vmatrix} & +\begin{vmatrix} 0 & 2 \\ 1 & -2 \end{vmatrix} \end{bmatrix} \\ &= \begin{bmatrix} +((-2)(1) - (-1)(0)) & -((2)(1) - (-3)(0)) & +((2)(-1) - (-3)(-2)) \\ -((1)(1) - (-1)(0)) & +((0)(1) - (-3)(0)) & -((0)(-1) - (-3)(1)) \\ +((1)(0) - (-2)(0)) & -((0)(0) - (2)(0)) & +((0)(-2) - (2)(1)) \end{bmatrix} \\ &= \begin{bmatrix} +(-2 - 0) & -(2 - 0) & +(-2 - 6) \\ -(1 - 0) & +(0 - 0) & -(0 - (-3)) \\ +(0 - 0) & -(0 - 0) & +(0 - 2) \end{bmatrix} = \begin{bmatrix} -2 & -2 & -8 \\ -1 & 0 & -3 \\ 0 & 0 & -2 \end{bmatrix} \end{align*}
(8)

Sehingga invers matriksnya adalah:

A−1=1−2adj⁡A=1−2[−2−2−8−10−300−2]A^{-1} = \frac{1}{-2} \operatorname{adj} A = \frac{1}{-2} \begin{bmatrix} -2 & -2 & -8 \\ -1 & 0 & -3 \\ 0 & 0 & -2 \end{bmatrix}
(9)

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}
(10)

hitung determinan A dengan mengekspansi sepanjang baris pertama:

det⁡A=1∣1111−3111−3∣−(−3)∣−3111−3111−3∣+1∣−31111111−3∣−1∣−31111−3111∣=1(1(−3⋅−3−1⋅1)−1(1⋅−3−1⋅1)+1(1⋅1−−3⋅1))+3(−3(−3⋅−3−1⋅1)−1(1⋅−3−1⋅1)+1(1⋅1−−3⋅1))+1(−3(1⋅−3−1⋅1)−1(1⋅−3−1⋅1)+1(1⋅1−1⋅1))−1(−3(1⋅1−−3⋅1)−1(1⋅1−−3⋅1)+1(1⋅1−1⋅1))=1(1(8)−1(−4)+1(4))+3(−3(8)−1(−4)+1(4))+1(−3(−4)−1(−4)+1(0))−1(−3(4)−1(4)+1(0))=1(8+4+4)+3(−24+4+4)+1(12+4+0)−1(−12−4+0)=1(16)+3(−16)+1(16)−1(−16)=16−48+16+16=0\begin{align*} \det A &= 1 \begin{vmatrix} 1 & 1 & 1 \\ 1 & -3 & 1 \\ 1 & 1 & -3 \end{vmatrix} - (-3) \begin{vmatrix} -3 & 1 & 1 \\ 1 & -3 & 1 \\ 1 & 1 & -3 \end{vmatrix} + 1 \begin{vmatrix} -3 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & -3 \end{vmatrix} - 1 \begin{vmatrix} -3 & 1 & 1 \\ 1 & 1 & -3 \\ 1 & 1 & 1 \end{vmatrix} \\ &= 1( 1(-3 \cdot -3 - 1 \cdot 1) - 1(1 \cdot -3 - 1 \cdot 1) + 1(1 \cdot 1 - -3 \cdot 1) ) \\ &\quad + 3( -3(-3 \cdot -3 - 1 \cdot 1) - 1(1 \cdot -3 - 1 \cdot 1) + 1(1 \cdot 1 - -3 \cdot 1) ) \\ &\quad + 1( -3(1 \cdot -3 - 1 \cdot 1) - 1(1 \cdot -3 - 1 \cdot 1) + 1(1 \cdot 1 - 1 \cdot 1) ) \\ &\quad - 1( -3(1 \cdot 1 - -3 \cdot 1) - 1(1 \cdot 1 - -3 \cdot 1) + 1(1 \cdot 1 - 1 \cdot 1) ) \\ &= 1( 1(8) - 1(-4) + 1(4) ) + 3( -3(8) - 1(-4) + 1(4) ) \\ &\quad + 1( -3(-4) - 1(-4) + 1(0) ) - 1( -3(4) - 1(4) + 1(0) ) \\ &= 1(8 + 4 + 4) + 3(-24 + 4 + 4) + 1(12 + 4 + 0) - 1(-12 - 4 + 0) \\ &= 1(16) + 3(-16) + 1(16) - 1(-16) \\ &= 16 - 48 + 16 + 16 = 0 \end{align*}
(11)

hitung adjoin dari A menggunakan rumus (adj⁡A)ij=(−1)i+jMji(\operatorname{adj} A)_{ij} = (-1)^{i+j} M_{ji}:

adj⁡A=[+M11−M21+M31−M41−M12+M22−M32+M42+M13−M23+M33−M43−M14+M24−M34+M44]=[+∣1111−3111−3∣−∣−3111−3111−3∣+∣−31111111−3∣−∣−31111−3111∣−∣−3111−3111−3∣+∣1111−3111−3∣−∣1−3111111−3∣+∣1−3111−3111∣+∣−31111111−3∣−∣111−31111−3∣+∣1−31−31111−3∣−∣1−31−311111∣−∣−31111−3111∣+∣111−31111−3∣−∣1−31−311111∣+∣1−31−31111−3∣]=[+(1((−3)(−3)−(1)(1))−1((1)(−3)−(1)(1))+1((1)(1)−(−3)(1)))−(−3((−3)(−3)−(1)(1))−1((1)(−3)−(1)(1))+1((1)(1)−(−3)(1)))+(−3((1)(−3)−(1)(1))−1((1)(−3)−(1)(1))+1((1)(1)−(1)(1)))−(−3((1)(1)−(−3)(1))−1((1)(1)−(−3)(1))+1((1)(1)−(1)(1)))−(−3((−3)(−3)−(1)(1))−1((1)(−3)−(1)(1))+1((1)(1)−(−3)(1)))+(1((−3)(−3)−(1)(1))−1((1)(−3)−(1)(1))+1((1)(1)−(−3)(1)))−(1((1)(−3)−(1)(1))−(−3)((1)(−3)−(1)(1))+1((1)(1)−(1)(1)))+(1((1)(1)−(−3)(1))−(−3)((1)(1)−(−3)(1))+1((1)(1)−(1)(1)))+(−3((1)(−3)−(1)(1))−1((1)(−3)−(1)(1))+1((1)(1)−(1)(1)))−(1((1)(−3)−(1)(1))−1((−3)(−3)−(1)(1))+1((−3)(1)−(1)(1)))+(1((1)(−3)−(1)(1))−(−3)((−3)(−3)−(1)(1))+1((−3)(1)−(1)(1)))−(1((1)(1)−(1)(1))−(−3)((−3)(1)−(1)(1))+1((−3)(1)−(1)(1)))−(−3((1)(1)−(−3)(1))−1((1)(1)−(−3)(1))+1((1)(1)−(1)(1)))+(1((1)(−3)−(1)(1))−1((−3)(−3)−(1)(1))+1((−3)(1)−(1)(1)))−(1((1)(1)−(1)(1))−(−3)((−3)(1)−(1)(1))+1((−3)(1)−(1)(1)))+(1((1)(−3)−(1)(1))−(−3)((−3)(−3)−(1)(1))+1((−3)(1)−(1)(1)))]=[+(1(8)−1(−4)+1(4))−(−3(8)−1(−4)+1(4))+(−3(−4)−1(−4)+1(0))−(−3(4)−1(4)+1(0))−(−3(8)−1(−4)+1(4))+(1(8)−1(−4)+1(4))−(1(−4)−(−3)(−4)+1(0))+(1(4)−(−3)(4)+1(0))+(−3(−4)−1(−4)+1(0))−(1(−4)−1(8)+1(−4))+(1(−4)−(−3)(8)+1(−4))−(1(0)−(−3)(−4)+1(−4))−(−3(4)−1(4)+1(0))+(1(−4)−1(8)+1(−4))−(1(0)−(−3)(−4)+1(−4))+(1(−4)−(−3)(8)+1(−4))]=[+(16)−(−16)+(16)−(−16)−(−16)+(16)−(−16)+(16)+(16)−(−16)+(16)−(−16)−(−16)+(16)−(−16)+(16)]=[16161616161616161616161616161616]\begin{align*} \operatorname{adj} A &= \begin{bmatrix} +M_{11} & -M_{21} & +M_{31} & -M_{41} \\ -M_{12} & +M_{22} & -M_{32} & +M_{42} \\ +M_{13} & -M_{23} & +M_{33} & -M_{43} \\ -M_{14} & +M_{24} & -M_{34} & +M_{44} \end{bmatrix} \\[10pt] &= \begin{bmatrix} +\begin{vmatrix} 1 & 1 & 1 \\ 1 & -3 & 1 \\ 1 & 1 & -3 \end{vmatrix} & -\begin{vmatrix} -3 & 1 & 1 \\ 1 & -3 & 1 \\ 1 & 1 & -3 \end{vmatrix} & +\begin{vmatrix} -3 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & -3 \end{vmatrix} & -\begin{vmatrix} -3 & 1 & 1 \\ 1 & 1 & -3 \\ 1 & 1 & 1 \end{vmatrix} \\[15pt] -\begin{vmatrix} -3 & 1 & 1 \\ 1 & -3 & 1 \\ 1 & 1 & -3 \end{vmatrix} & +\begin{vmatrix} 1 & 1 & 1 \\ 1 & -3 & 1 \\ 1 & 1 & -3 \end{vmatrix} & -\begin{vmatrix} 1 & -3 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & -3 \end{vmatrix} & +\begin{vmatrix} 1 & -3 & 1 \\ 1 & 1 & -3 \\ 1 & 1 & 1 \end{vmatrix} \\[15pt] +\begin{vmatrix} -3 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & -3 \end{vmatrix} & -\begin{vmatrix} 1 & 1 & 1 \\ -3 & 1 & 1 \\ 1 & 1 & -3 \end{vmatrix} & +\begin{vmatrix} 1 & -3 & 1 \\ -3 & 1 & 1 \\ 1 & 1 & -3 \end{vmatrix} & -\begin{vmatrix} 1 & -3 & 1 \\ -3 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix} \\[15pt] -\begin{vmatrix} -3 & 1 & 1 \\ 1 & 1 & -3 \\ 1 & 1 & 1 \end{vmatrix} & +\begin{vmatrix} 1 & 1 & 1 \\ -3 & 1 & 1 \\ 1 & 1 & -3 \end{vmatrix} & -\begin{vmatrix} 1 & -3 & 1 \\ -3 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix} & +\begin{vmatrix} 1 & -3 & 1 \\ -3 & 1 & 1 \\ 1 & 1 & -3 \end{vmatrix} \end{bmatrix} \\[10pt] &= \begin{bmatrix} +( 1((-3)(-3) - (1)(1)) - 1((1)(-3) - (1)(1)) + 1((1)(1) - (-3)(1)) ) & -( -3((-3)(-3) - (1)(1)) - 1((1)(-3) - (1)(1)) + 1((1)(1) - (-3)(1)) ) & +( -3((1)(-3) - (1)(1)) - 1((1)(-3) - (1)(1)) + 1((1)(1) - (1)(1)) ) & -( -3((1)(1) - (-3)(1)) - 1((1)(1) - (-3)(1)) + 1((1)(1) - (1)(1)) ) \\[15pt] -( -3((-3)(-3) - (1)(1)) - 1((1)(-3) - (1)(1)) + 1((1)(1) - (-3)(1)) ) & +( 1((-3)(-3) - (1)(1)) - 1((1)(-3) - (1)(1)) + 1((1)(1) - (-3)(1)) ) & -( 1((1)(-3) - (1)(1)) - (-3)((1)(-3) - (1)(1)) + 1((1)(1) - (1)(1)) ) & +( 1((1)(1) - (-3)(1)) - (-3)((1)(1) - (-3)(1)) + 1((1)(1) - (1)(1)) ) \\[15pt] +( -3((1)(-3) - (1)(1)) - 1((1)(-3) - (1)(1)) + 1((1)(1) - (1)(1)) ) & -( 1((1)(-3) - (1)(1)) - 1((-3)(-3) - (1)(1)) + 1((-3)(1) - (1)(1)) ) & +( 1((1)(-3) - (1)(1)) - (-3)((-3)(-3) - (1)(1)) + 1((-3)(1) - (1)(1)) ) & -( 1((1)(1) - (1)(1)) - (-3)((-3)(1) - (1)(1)) + 1((-3)(1) - (1)(1)) ) \\[15pt] -( -3((1)(1) - (-3)(1)) - 1((1)(1) - (-3)(1)) + 1((1)(1) - (1)(1)) ) & +( 1((1)(-3) - (1)(1)) - 1((-3)(-3) - (1)(1)) + 1((-3)(1) - (1)(1)) ) & -( 1((1)(1) - (1)(1)) - (-3)((-3)(1) - (1)(1)) + 1((-3)(1) - (1)(1)) ) & +( 1((1)(-3) - (1)(1)) - (-3)((-3)(-3) - (1)(1)) + 1((-3)(1) - (1)(1)) ) \end{bmatrix} \\[10pt] &= \begin{bmatrix} +( 1(8) - 1(-4) + 1(4) ) & -( -3(8) - 1(-4) + 1(4) ) & +( -3(-4) - 1(-4) + 1(0) ) & -( -3(4) - 1(4) + 1(0) ) \\[10pt] -( -3(8) - 1(-4) + 1(4) ) & +( 1(8) - 1(-4) + 1(4) ) & -( 1(-4) - (-3)(-4) + 1(0) ) & +( 1(4) - (-3)(4) + 1(0) ) \\[10pt] +( -3(-4) - 1(-4) + 1(0) ) & -( 1(-4) - 1(8) + 1(-4) ) & +( 1(-4) - (-3)(8) + 1(-4) ) & -( 1(0) - (-3)(-4) + 1(-4) ) \\[10pt] -( -3(4) - 1(4) + 1(0) ) & +( 1(-4) - 1(8) + 1(-4) ) & -( 1(0) - (-3)(-4) + 1(-4) ) & +( 1(-4) - (-3)(8) + 1(-4) ) \end{bmatrix} \\[10pt] &= \begin{bmatrix} +(16) & -(-16) & +(16) & -(-16) \\ -(-16) & +(16) & -(-16) & +(16) \\ +(16) & -(-16) & +(16) & -(-16) \\ -(-16) & +(16) & -(-16) & +(16) \end{bmatrix} = \begin{bmatrix} 16 & 16 & 16 & 16 \\ 16 & 16 & 16 & 16 \\ 16 & 16 & 16 & 16 \\ 16 & 16 & 16 & 16 \end{bmatrix} \end{align*}
(12)

Sehingga invers matriksnya dirumuskan dengan:

A−1=1det⁡Aadj⁡A=10[16161616161616161616161616161616]A^{-1} = \frac{1}{\det{A}} \operatorname{adj} A = \frac{1}{0} \begin{bmatrix} 16 & 16 & 16 & 16 \\ 16 & 16 & 16 & 16 \\ 16 & 16 & 16 & 16 \\ 16 & 16 & 16 & 16 \end{bmatrix}
(13)

Karena det⁡(A)=0\det(A) = 0, maka operasi invers ini bernilai tak terdefinisi

matriks A adalah singular atau tidak memiliki invers

KOMPUTASI ALJABAR LINEAR
Evaluasi Determinan