Linear Algebra Examples

[203300024]203300024
Step 1
Find the determinant.
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Step 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 22 by its cofactor and add.
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Step 1.1.1
Consider the corresponding sign chart.
|+-+-+-+-+|∣ ∣+++++∣ ∣
Step 1.1.2
The cofactor is the minor with the sign changed if the indices match a - position on the sign chart.
Step 1.1.3
The minor for a21a21 is the determinant with row 22 and column 11 deleted.
|0324|0324
Step 1.1.4
Multiply element a21a21 by its cofactor.
-3|0324|30324
Step 1.1.5
The minor for a22a22 is the determinant with row 22 and column 22 deleted.
|2304|2304
Step 1.1.6
Multiply element a22a22 by its cofactor.
0|2304|02304
Step 1.1.7
The minor for a23a23 is the determinant with row 22 and column 33 deleted.
|2002|2002
Step 1.1.8
Multiply element a23a23 by its cofactor.
0|2002|02002
Step 1.1.9
Add the terms together.
-3|0324|+0|2304|+0|2002|30324+02304+02002
-3|0324|+0|2304|+0|2002|30324+02304+02002
Step 1.2
Multiply 00 by |2304|2304.
-3|0324|+0+0|2002|30324+0+02002
Step 1.3
Multiply 00 by |2002|2002.
-3|0324|+0+030324+0+0
Step 1.4
Evaluate |0324|0324.
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Step 1.4.1
The determinant of a 2×22×2 matrix can be found using the formula |abcd|=ad-cbabcd=adcb.
-3(04-23)+0+03(0423)+0+0
Step 1.4.2
Simplify the determinant.
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Step 1.4.2.1
Simplify each term.
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Step 1.4.2.1.1
Multiply 00 by 44.
-3(0-23)+0+03(023)+0+0
Step 1.4.2.1.2
Multiply -22 by 33.
-3(0-6)+0+03(06)+0+0
-3(0-6)+0+03(06)+0+0
Step 1.4.2.2
Subtract 66 from 00.
-3-6+0+036+0+0
-3-6+0+036+0+0
-3-6+0+036+0+0
Step 1.5
Simplify the determinant.
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Step 1.5.1
Multiply -33 by -66.
18+0+018+0+0
Step 1.5.2
Add 1818 and 00.
18+018+0
Step 1.5.3
Add 1818 and 00.
1818
1818
1818
Step 2
Since the determinant is non-zero, the inverse exists.
Step 3
Set up a 3×63×6 matrix where the left half is the original matrix and the right half is its identity matrix.
[203100300010024001]203100300010024001
Step 4
Find the reduced row echelon form.
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Step 4.1
Multiply each element of R1R1 by 1212 to make the entry at 1,11,1 a 11.
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Step 4.1.1
Multiply each element of R1R1 by 1212 to make the entry at 1,11,1 a 11.
[220232120202300010024001]⎢ ⎢220232120202300010024001⎥ ⎥
Step 4.1.2
Simplify R1R1.
[10321200300010024001]⎢ ⎢10321200300010024001⎥ ⎥
[10321200300010024001]⎢ ⎢10321200300010024001⎥ ⎥
Step 4.2
Perform the row operation R2=R2-3R1 to make the entry at 2,1 a 0.
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Step 4.2.1
Perform the row operation R2=R2-3R1 to make the entry at 2,1 a 0.
[103212003-310-300-3(32)0-3(12)1-300-30024001]
Step 4.2.2
Simplify R2.
[1032120000-92-3210024001]
[1032120000-92-3210024001]
Step 4.3
Swap R3 with R2 to put a nonzero entry at 2,2.
[1032120002400100-92-3210]
Step 4.4
Multiply each element of R2 by 12 to make the entry at 2,2 a 1.
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Step 4.4.1
Multiply each element of R2 by 12 to make the entry at 2,2 a 1.
[1032120002224202021200-92-3210]
Step 4.4.2
Simplify R2.
[10321200012001200-92-3210]
[10321200012001200-92-3210]
Step 4.5
Multiply each element of R3 by -29 to make the entry at 3,3 a 1.
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Step 4.5.1
Multiply each element of R3 by -29 to make the entry at 3,3 a 1.
[103212000120012-290-290-29(-92)-29(-32)-291-290]
Step 4.5.2
Simplify R3.
[10321200012001200113-290]
[10321200012001200113-290]
Step 4.6
Perform the row operation R2=R2-2R3 to make the entry at 2,3 a 0.
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Step 4.6.1
Perform the row operation R2=R2-2R3 to make the entry at 2,3 a 0.
[103212000-201-202-210-2(13)0-2(-29)12-2000113-290]
Step 4.6.2
Simplify R2.
[10321200010-23491200113-290]
[10321200010-23491200113-290]
Step 4.7
Perform the row operation R1=R1-32R3 to make the entry at 1,3 a 0.
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Step 4.7.1
Perform the row operation R1=R1-32R3 to make the entry at 1,3 a 0.
[1-3200-32032-32112-32130-32(-29)0-320010-23491200113-290]
Step 4.7.2
Simplify R1.
[1000130010-23491200113-290]
[1000130010-23491200113-290]
[1000130010-23491200113-290]
Step 5
The right half of the reduced row echelon form is the inverse.
[0130-23491213-290]
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 [x2  12  π  xdx ] 
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