Matematica discreta Esempi
[330103020]⎡⎢⎣330103020⎤⎥⎦
Passaggio 1
Passaggio 1.1
Choose the row or column with the most 00 elements. If there are no 00 elements choose any row or column. Multiply every element in row 33 by its cofactor and add.
Passaggio 1.1.1
Consider the corresponding sign chart.
|+-+-+-+-+|∣∣
∣∣+−+−+−+−+∣∣
∣∣
Passaggio 1.1.2
The cofactor is the minor with the sign changed if the indices match a -− position on the sign chart.
Passaggio 1.1.3
The minor for a31a31 is the determinant with row 33 and column 11 deleted.
|3003|∣∣∣3003∣∣∣
Passaggio 1.1.4
Multiply element a31a31 by its cofactor.
0|3003|0∣∣∣3003∣∣∣
Passaggio 1.1.5
The minor for a32a32 is the determinant with row 33 and column 22 deleted.
|3013|∣∣∣3013∣∣∣
Passaggio 1.1.6
Multiply element a32a32 by its cofactor.
-2|3013|−2∣∣∣3013∣∣∣
Passaggio 1.1.7
The minor for a33a33 is the determinant with row 33 and column 33 deleted.
|3310|∣∣∣3310∣∣∣
Passaggio 1.1.8
Multiply element a33a33 by its cofactor.
0|3310|0∣∣∣3310∣∣∣
Passaggio 1.1.9
Add the terms together.
0|3003|-2|3013|+0|3310|0∣∣∣3003∣∣∣−2∣∣∣3013∣∣∣+0∣∣∣3310∣∣∣
0|3003|-2|3013|+0|3310|0∣∣∣3003∣∣∣−2∣∣∣3013∣∣∣+0∣∣∣3310∣∣∣
Passaggio 1.2
Moltiplica 00 per |3003|∣∣∣3003∣∣∣.
0-2|3013|+0|3310|0−2∣∣∣3013∣∣∣+0∣∣∣3310∣∣∣
Passaggio 1.3
Moltiplica 00 per |3310|∣∣∣3310∣∣∣.
0-2|3013|+00−2∣∣∣3013∣∣∣+0
Passaggio 1.4
Calcola |3013|∣∣∣3013∣∣∣.
Passaggio 1.4.1
È possibile trovare il determinante di una matrice 2×22×2 usando la formula |abcd|=ad-cb∣∣∣abcd∣∣∣=ad−cb.
0-2(3⋅3-1⋅0)+00−2(3⋅3−1⋅0)+0
Passaggio 1.4.2
Semplifica il determinante.
Passaggio 1.4.2.1
Moltiplica 33 per 33.
0-2(9-1⋅0)+00−2(9−1⋅0)+0
Passaggio 1.4.2.2
Sottrai 00 da 99.
0-2⋅9+00−2⋅9+0
0-2⋅9+00−2⋅9+0
0-2⋅9+00−2⋅9+0
Passaggio 1.5
Semplifica il determinante.
Passaggio 1.5.1
Moltiplica -2−2 per 99.
0-18+00−18+0
Passaggio 1.5.2
Sottrai 1818 da 00.
-18+0−18+0
Passaggio 1.5.3
Somma -18−18 e 00.
-18−18
-18−18
-18−18
Passaggio 2
Since the determinant is non-zero, the inverse exists.
Passaggio 3
Set up a 3×63×6 matrix where the left half is the original matrix and the right half is its identity matrix.
[330100103010020001]⎡⎢⎣330100103010020001⎤⎥⎦
Passaggio 4
Passaggio 4.1
Multiply each element of R1R1 by 1313 to make the entry at 1,11,1 a 11.
Passaggio 4.1.1
Multiply each element of R1R1 by 1313 to make the entry at 1,11,1 a 11.
[333303130303103010020001]⎡⎢
⎢⎣333303130303103010020001⎤⎥
⎥⎦
Passaggio 4.1.2
Semplifica R1R1.
[1101300103010020001]⎡⎢
⎢⎣1101300103010020001⎤⎥
⎥⎦
[1101300103010020001]⎡⎢
⎢⎣1101300103010020001⎤⎥
⎥⎦
Passaggio 4.2
Perform the row operation R2=R2-R1R2=R2−R1 to make the entry at 2,12,1 a 00.
Passaggio 4.2.1
Perform the row operation R2=R2-R1R2=R2−R1 to make the entry at 2,12,1 a 00.
[11013001-10-13-00-131-00-0020001]⎡⎢
⎢⎣11013001−10−13−00−131−00−0020001⎤⎥
⎥⎦
Passaggio 4.2.2
Semplifica R2R2.
[11013000-13-1310020001]⎡⎢
⎢⎣11013000−13−1310020001⎤⎥
⎥⎦
[11013000-13-1310020001]⎡⎢
⎢⎣11013000−13−1310020001⎤⎥
⎥⎦
Passaggio 4.3
Multiply each element of R2R2 by -1−1 to make the entry at 2,22,2 a 11.
Passaggio 4.3.1
Multiply each element of R2R2 by -1−1 to make the entry at 2,22,2 a 11.
[1101300-0--1-1⋅3--13-1⋅1-0020001]⎡⎢
⎢⎣1101300−0−−1−1⋅3−−13−1⋅1−0020001⎤⎥
⎥⎦
Passaggio 4.3.2
Semplifica R2R2.
[110130001-313-10020001]⎡⎢
⎢⎣110130001−313−10020001⎤⎥
⎥⎦
[110130001-313-10020001]⎡⎢
⎢⎣110130001−313−10020001⎤⎥
⎥⎦
Passaggio 4.4
Perform the row operation R3=R3-2R2R3=R3−2R2 to make the entry at 3,23,2 a 00.
Passaggio 4.4.1
Perform the row operation R3=R3-2R2R3=R3−2R2 to make the entry at 3,23,2 a 00.
[110130001-313-100-2⋅02-2⋅10-2⋅-30-2(13)0-2⋅-11-2⋅0]⎡⎢
⎢
⎢
⎢⎣110130001−313−100−2⋅02−2⋅10−2⋅−30−2(13)0−2⋅−11−2⋅0⎤⎥
⎥
⎥
⎥⎦
Passaggio 4.4.2
Semplifica R3R3.
[110130001-313-10006-2321]⎡⎢
⎢
⎢⎣110130001−313−10006−2321⎤⎥
⎥
⎥⎦
[110130001-313-10006-2321]⎡⎢
⎢
⎢⎣110130001−313−10006−2321⎤⎥
⎥
⎥⎦
Passaggio 4.5
Multiply each element of R3R3 by 1616 to make the entry at 3,33,3 a 11.
Passaggio 4.5.1
Multiply each element of R3R3 by 1616 to make the entry at 3,33,3 a 11.
[110130001-313-10060666-2362616]⎡⎢
⎢
⎢
⎢⎣110130001−313−10060666−2362616⎤⎥
⎥
⎥
⎥⎦
Passaggio 4.5.2
Semplifica R3R3.
[110130001-313-10001-191316]⎡⎢
⎢
⎢⎣110130001−313−10001−191316⎤⎥
⎥
⎥⎦
[110130001-313-10001-191316]⎡⎢
⎢
⎢⎣110130001−313−10001−191316⎤⎥
⎥
⎥⎦
Passaggio 4.6
Perform the row operation R2=R2+3R3R2=R2+3R3 to make the entry at 2,32,3 a 00.
Passaggio 4.6.1
Perform the row operation R2=R2+3R3R2=R2+3R3 to make the entry at 2,32,3 a 00.
[11013000+3⋅01+3⋅0-3+3⋅113+3(-19)-1+3(13)0+3(16)001-191316]⎡⎢
⎢
⎢
⎢⎣11013000+3⋅01+3⋅0−3+3⋅113+3(−19)−1+3(13)0+3(16)001−191316⎤⎥
⎥
⎥
⎥⎦
Passaggio 4.6.2
Semplifica R2R2.
[11013000100012001-191316]⎡⎢
⎢
⎢⎣11013000100012001−191316⎤⎥
⎥
⎥⎦
[11013000100012001-191316]⎡⎢
⎢
⎢⎣11013000100012001−191316⎤⎥
⎥
⎥⎦
Passaggio 4.7
Perform the row operation R1=R1-R2R1=R1−R2 to make the entry at 1,21,2 a 00.
Passaggio 4.7.1
Perform the row operation R1=R1-R2R1=R1−R2 to make the entry at 1,21,2 a 00.
[1-01-10-013-00-00-120100012001-191316]⎡⎢
⎢
⎢⎣1−01−10−013−00−00−120100012001−191316⎤⎥
⎥
⎥⎦
Passaggio 4.7.2
Semplifica R1R1.
[100130-120100012001-191316]
[100130-120100012001-191316]
[100130-120100012001-191316]
Passaggio 5
The right half of the reduced row echelon form is the inverse.
[130-120012-191316]