إدخال مسألة...
الرياضيات المتناهية الأمثلة
[1012-2-1300]⎡⎢⎣1012−2−1300⎤⎥⎦
خطوة 1
خطوة 1.1
Choose the row or column with the most 00 elements. If there are no 0 elements choose any row or column. Multiply every element in column 2 by its cofactor and add.
خطوة 1.1.1
Consider the corresponding sign chart.
|+-+-+-+-+|
خطوة 1.1.2
The cofactor is the minor with the sign changed if the indices match a - position on the sign chart.
خطوة 1.1.3
The minor for a12 is the determinant with row 1 and column 2 deleted.
|2-130|
خطوة 1.1.4
Multiply element a12 by its cofactor.
0|2-130|
خطوة 1.1.5
The minor for a22 is the determinant with row 2 and column 2 deleted.
|1130|
خطوة 1.1.6
Multiply element a22 by its cofactor.
-2|1130|
خطوة 1.1.7
The minor for a32 is the determinant with row 3 and column 2 deleted.
|112-1|
خطوة 1.1.8
Multiply element a32 by its cofactor.
0|112-1|
خطوة 1.1.9
Add the terms together.
0|2-130|-2|1130|+0|112-1|
0|2-130|-2|1130|+0|112-1|
خطوة 1.2
اضرب 0 في |2-130|.
0-2|1130|+0|112-1|
خطوة 1.3
اضرب 0 في |112-1|.
0-2|1130|+0
خطوة 1.4
احسِب قيمة |1130|.
خطوة 1.4.1
يمكن إيجاد محدد المصفوفة 2×2 باستخدام القاعدة |abcd|=ad-cb.
0-2(1⋅0-3⋅1)+0
خطوة 1.4.2
بسّط المحدد.
خطوة 1.4.2.1
بسّط كل حد.
خطوة 1.4.2.1.1
اضرب 0 في 1.
0-2(0-3⋅1)+0
خطوة 1.4.2.1.2
اضرب -3 في 1.
0-2(0-3)+0
0-2(0-3)+0
خطوة 1.4.2.2
اطرح 3 من 0.
0-2⋅-3+0
0-2⋅-3+0
0-2⋅-3+0
خطوة 1.5
بسّط المحدد.
خطوة 1.5.1
اضرب -2 في -3.
0+6+0
خطوة 1.5.2
أضف 0 و6.
6+0
خطوة 1.5.3
أضف 6 و0.
6
6
6
خطوة 2
Since the determinant is non-zero, the inverse exists.
خطوة 3
Set up a 3×6 matrix where the left half is the original matrix and the right half is its identity matrix.
[1011002-2-1010300001]
خطوة 4
خطوة 4.1
Perform the row operation R2=R2-2R1 to make the entry at 2,1 a 0.
خطوة 4.1.1
Perform the row operation R2=R2-2R1 to make the entry at 2,1 a 0.
[1011002-2⋅1-2-2⋅0-1-2⋅10-2⋅11-2⋅00-2⋅0300001]
خطوة 4.1.2
بسّط R2.
[1011000-2-3-210300001]
[1011000-2-3-210300001]
خطوة 4.2
Perform the row operation R3=R3-3R1 to make the entry at 3,1 a 0.
خطوة 4.2.1
Perform the row operation R3=R3-3R1 to make the entry at 3,1 a 0.
[1011000-2-3-2103-3⋅10-3⋅00-3⋅10-3⋅10-3⋅01-3⋅0]
خطوة 4.2.2
بسّط R3.
[1011000-2-3-21000-3-301]
[1011000-2-3-21000-3-301]
خطوة 4.3
Multiply each element of R2 by -12 to make the entry at 2,2 a 1.
خطوة 4.3.1
Multiply each element of R2 by -12 to make the entry at 2,2 a 1.
[101100-12⋅0-12⋅-2-12⋅-3-12⋅-2-12⋅1-12⋅000-3-301]
خطوة 4.3.2
بسّط R2.
[10110001321-12000-3-301]
[10110001321-12000-3-301]
خطوة 4.4
Multiply each element of R3 by -13 to make the entry at 3,3 a 1.
خطوة 4.4.1
Multiply each element of R3 by -13 to make the entry at 3,3 a 1.
[10110001321-120-13⋅0-13⋅0-13⋅-3-13⋅-3-13⋅0-13⋅1]
خطوة 4.4.2
بسّط R3.
[10110001321-12000110-13]
[10110001321-12000110-13]
خطوة 4.5
Perform the row operation R2=R2-32R3 to make the entry at 2,3 a 0.
خطوة 4.5.1
Perform the row operation R2=R2-32R3 to make the entry at 2,3 a 0.
[1011000-32⋅01-32⋅032-32⋅11-32⋅1-12-32⋅00-32(-13)00110-13]
خطوة 4.5.2
بسّط R2.
[101100010-12-121200110-13]
[101100010-12-121200110-13]
خطوة 4.6
Perform the row operation R1=R1-R3 to make the entry at 1,3 a 0.
خطوة 4.6.1
Perform the row operation R1=R1-R3 to make the entry at 1,3 a 0.
[1-00-01-11-10-00+13010-12-121200110-13]
خطوة 4.6.2
بسّط R1.
[1000013010-12-121200110-13]
[1000013010-12-121200110-13]
[1000013010-12-121200110-13]
خطوة 5
The right half of the reduced row echelon form is the inverse.
[0013-12-121210-13]