线性代数 示例

求出反函数 [[4,-10,29],[1,-2,5],[-3,7,-19]]
[4-10291-25-37-19]410291253719
解题步骤 1
Find the determinant.
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解题步骤 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 11 by its cofactor and add.
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解题步骤 1.1.1
Consider the corresponding sign chart.
|+-+-+-+-+|∣ ∣+++++∣ ∣
解题步骤 1.1.2
The cofactor is the minor with the sign changed if the indices match a - position on the sign chart.
解题步骤 1.1.3
The minor for a11a11 is the determinant with row 11 and column 11 deleted.
|-257-19|25719
解题步骤 1.1.4
Multiply element a11a11 by its cofactor.
4|-257-19|425719
解题步骤 1.1.5
The minor for a12a12 is the determinant with row 11 and column 22 deleted.
|15-3-19|15319
解题步骤 1.1.6
Multiply element a12a12 by its cofactor.
10|15-3-19|1015319
解题步骤 1.1.7
The minor for a13a13 is the determinant with row 11 and column 33 deleted.
|1-2-37|1237
解题步骤 1.1.8
Multiply element a13a13 by its cofactor.
29|1-2-37|291237
解题步骤 1.1.9
Add the terms together.
4|-257-19|+10|15-3-19|+29|1-2-37|425719+1015319+291237
4|-257-19|+10|15-3-19|+29|1-2-37|425719+1015319+291237
解题步骤 1.2
计算 |-257-19|25719
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解题步骤 1.2.1
可以使用公式 |abcd|=ad-cbabcd=adcb2×22×2 矩阵的行列式。
4(-2-19-75)+10|15-3-19|+29|1-2-37|4(21975)+1015319+291237
解题步骤 1.2.2
化简行列式。
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解题步骤 1.2.2.1
化简每一项。
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解题步骤 1.2.2.1.1
-22 乘以 -1919
4(38-75)+10|15-3-19|+29|1-2-37|4(3875)+1015319+291237
解题步骤 1.2.2.1.2
-77 乘以 55
4(38-35)+10|15-3-19|+29|1-2-37|4(3835)+1015319+291237
4(38-35)+10|15-3-19|+29|1-2-37|4(3835)+1015319+291237
解题步骤 1.2.2.2
3838 中减去 3535
43+10|15-3-19|+29|1-2-37|43+1015319+291237
43+10|15-3-19|+29|1-2-37|43+1015319+291237
43+10|15-3-19|+29|1-2-37|43+1015319+291237
解题步骤 1.3
计算 |15-3-19|15319
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解题步骤 1.3.1
可以使用公式 |abcd|=ad-cbabcd=adcb2×22×2 矩阵的行列式。
43+10(1-19-(-35))+29|1-2-37|43+10(119(35))+291237
解题步骤 1.3.2
化简行列式。
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解题步骤 1.3.2.1
化简每一项。
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解题步骤 1.3.2.1.1
-1919 乘以 11
43+10(-19-(-35))+29|1-2-37|43+10(19(35))+291237
解题步骤 1.3.2.1.2
乘以 -(-35)(35)
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解题步骤 1.3.2.1.2.1
-33 乘以 55
43+10(-19--15)+29|1-2-37|43+10(1915)+291237
解题步骤 1.3.2.1.2.2
-11 乘以 -1515
43+10(-19+15)+29|1-2-37|43+10(19+15)+291237
43+10(-19+15)+29|1-2-37|43+10(19+15)+291237
43+10(-19+15)+29|1-2-37|43+10(19+15)+291237
解题步骤 1.3.2.2
-19191515 相加。
43+10-4+29|1-2-37|43+104+291237
43+10-4+29|1-2-37|43+104+291237
43+10-4+29|1-2-37|43+104+291237
解题步骤 1.4
计算 |1-2-37|1237
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解题步骤 1.4.1
可以使用公式 |abcd|=ad-cbabcd=adcb2×22×2 矩阵的行列式。
43+10-4+29(17-(-3-2))43+104+29(17(32))
解题步骤 1.4.2
化简行列式。
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解题步骤 1.4.2.1
化简每一项。
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解题步骤 1.4.2.1.1
77 乘以 11
43+10-4+29(7-(-3-2))43+104+29(7(32))
解题步骤 1.4.2.1.2
乘以 -(-3-2)(32)
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解题步骤 1.4.2.1.2.1
-33 乘以 -22
43+10-4+29(7-16)43+104+29(716)
解题步骤 1.4.2.1.2.2
-11 乘以 66
43+10-4+29(7-6)43+104+29(76)
43+10-4+29(7-6)43+104+29(76)
43+10-4+29(7-6)43+104+29(76)
解题步骤 1.4.2.2
77 中减去 66
43+10-4+29143+104+291
43+10-4+29143+104+291
43+10-4+29143+104+291
解题步骤 1.5
化简行列式。
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解题步骤 1.5.1
化简每一项。
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解题步骤 1.5.1.1
44 乘以 33
12+10-4+29112+104+291
解题步骤 1.5.1.2
1010 乘以 -44
12-40+2911240+291
解题步骤 1.5.1.3
2929 乘以 11
12-40+291240+29
12-40+291240+29
解题步骤 1.5.2
1212 中减去 4040
-28+2928+29
解题步骤 1.5.3
-28282929 相加。
11
11
11
解题步骤 2
Since the determinant is non-zero, the inverse exists.
解题步骤 3
Set up a 3×63×6 matrix where the left half is the original matrix and the right half is its identity matrix.
[4-10291001-25010-37-19001]410291001250103719001
解题步骤 4
求行简化阶梯形矩阵。
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解题步骤 4.1
Multiply each element of R1R1 by 1414 to make the entry at 1,11,1 a 11.
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解题步骤 4.1.1
Multiply each element of R1R1 by 1414 to make the entry at 1,11,1 a 11.
[44-1042941404041-25010-37-19001]⎢ ⎢441042941404041250103719001⎥ ⎥
解题步骤 4.1.2
化简 R1R1
[1-5229414001-25010-37-19001]⎢ ⎢15229414001250103719001⎥ ⎥
[1-5229414001-25010-37-19001]⎢ ⎢15229414001250103719001⎥ ⎥
解题步骤 4.2
Perform the row operation R2=R2-R1R2=R2R1 to make the entry at 2,12,1 a 00.
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解题步骤 4.2.1
Perform the row operation R2=R2-R1R2=R2R1 to make the entry at 2,12,1 a 00.
[1-5229414001-1-2+525-2940-141-00-0-37-19001]⎢ ⎢1522941400112+52529401410003719001⎥ ⎥
解题步骤 4.2.2
化简 R2R2
[1-522941400012-94-1410-37-19001]⎢ ⎢15229414000129414103719001⎥ ⎥
[1-522941400012-94-1410-37-19001]⎢ ⎢15229414000129414103719001⎥ ⎥
解题步骤 4.3
Perform the row operation R3=R3+3R1R3=R3+3R1 to make the entry at 3,13,1 a 00.
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解题步骤 4.3.1
Perform the row operation R3=R3+3R1R3=R3+3R1 to make the entry at 3,13,1 a 00.
[1-522941400012-94-1410-3+317+3(-52)-19+3(294)0+3(14)0+301+30]⎢ ⎢ ⎢ ⎢15229414000129414103+317+3(52)19+3(294)0+3(14)0+301+30⎥ ⎥ ⎥ ⎥
解题步骤 4.3.2
化简 R3R3
[1-522941400012-94-14100-121143401]⎢ ⎢ ⎢15229414000129414100121143401⎥ ⎥ ⎥
[1-522941400012-94-14100-121143401]⎢ ⎢ ⎢15229414000129414100121143401⎥ ⎥ ⎥
解题步骤 4.4
Multiply each element of R2R2 by 22 to make the entry at 2,22,2 a 11.
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解题步骤 4.4.1
Multiply each element of R2R2 by 22 to make the entry at 2,22,2 a 11.
[1-522941400202(12)2(-94)2(-14)21200-121143401]⎢ ⎢ ⎢ ⎢1522941400202(12)2(94)2(14)21200121143401⎥ ⎥ ⎥ ⎥
解题步骤 4.4.2
化简 R2R2
[1-52294140001-92-12200-121143401]⎢ ⎢ ⎢1522941400019212200121143401⎥ ⎥ ⎥
[1-52294140001-92-12200-121143401]⎢ ⎢ ⎢1522941400019212200121143401⎥ ⎥ ⎥
解题步骤 4.5
Perform the row operation R3=R3+12R2R3=R3+12R2 to make the entry at 3,23,2 a 00.
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解题步骤 4.5.1
Perform the row operation R3=R3+12R2R3=R3+12R2 to make the entry at 3,23,2 a 00.
[1-52294140001-92-12200+120-12+121114+12(-92)34+12(-12)0+1221+120]⎢ ⎢ ⎢ ⎢1522941400019212200+12012+121114+12(92)34+12(12)0+1221+120⎥ ⎥ ⎥ ⎥
解题步骤 4.5.2
化简 R3R3
[1-52294140001-92-122000121211]⎢ ⎢ ⎢15229414000192122000121211⎥ ⎥ ⎥
[1-52294140001-92-122000121211]⎢ ⎢ ⎢15229414000192122000121211⎥ ⎥ ⎥
解题步骤 4.6
Multiply each element of R3R3 by 22 to make the entry at 3,33,3 a 11.
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解题步骤 4.6.1
Multiply each element of R3R3 by 22 to make the entry at 3,33,3 a 11.
[1-52294140001-92-122020202(12)2(12)2121]⎢ ⎢ ⎢15229414000192122020202(12)2(12)2121⎥ ⎥ ⎥
解题步骤 4.6.2
化简 R3R3
[1-52294140001-92-1220001122]⎢ ⎢152294140001921220001122⎥ ⎥
[1-52294140001-92-1220001122]⎢ ⎢152294140001921220001122⎥ ⎥
解题步骤 4.7
Perform the row operation R2=R2+92R3R2=R2+92R3 to make the entry at 2,32,3 a 00.
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解题步骤 4.7.1
Perform the row operation R2=R2+92R3R2=R2+92R3 to make the entry at 2,32,3 a 00.
[1-5229414000+9201+920-92+921-12+9212+9220+922001122]⎢ ⎢15229414000+9201+92092+92112+9212+9220+922001122⎥ ⎥
解题步骤 4.7.2
化简 R2
[1-5229414000104119001122]
[1-5229414000104119001122]
解题步骤 4.8
Perform the row operation R1=R1-294R3 to make the entry at 1,3 a 0.
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解题步骤 4.8.1
Perform the row operation R1=R1-294R3 to make the entry at 1,3 a 0.
[1-2940-52-2940294-294114-29410-29420-29420104119001122]
解题步骤 4.8.2
化简 R1
[1-520-7-292-2920104119001122]
[1-520-7-292-2920104119001122]
解题步骤 4.9
Perform the row operation R1=R1+52R2 to make the entry at 1,2 a 0.
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解题步骤 4.9.1
Perform the row operation R1=R1+52R2 to make the entry at 1,2 a 0.
[1+520-52+5210+520-7+524-292+5211-292+5290104119001122]
解题步骤 4.9.2
化简 R1
[10031380104119001122]
[10031380104119001122]
[10031380104119001122]
解题步骤 5
The right half of the reduced row echelon form is the inverse.
[31384119122]
 [x2  12  π  xdx ]