Finite Math Examples

[203300024]
Step 1
Find the determinant.
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Step 1.1
Choose the row or column with the most 0 elements. If there are no 0 elements choose any row or column. Multiply every element in row 2 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 a21 is the determinant with row 2 and column 1 deleted.
|0324|
Step 1.1.4
Multiply element a21 by its cofactor.
-3|0324|
Step 1.1.5
The minor for a22 is the determinant with row 2 and column 2 deleted.
|2304|
Step 1.1.6
Multiply element a22 by its cofactor.
0|2304|
Step 1.1.7
The minor for a23 is the determinant with row 2 and column 3 deleted.
|2002|
Step 1.1.8
Multiply element a23 by its cofactor.
0|2002|
Step 1.1.9
Add the terms together.
-3|0324|+0|2304|+0|2002|
-3|0324|+0|2304|+0|2002|
Step 1.2
Multiply 0 by |2304|.
-3|0324|+0+0|2002|
Step 1.3
Multiply 0 by |2002|.
-3|0324|+0+0
Step 1.4
Evaluate |0324|.
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Step 1.4.1
The determinant of a 2×2 matrix can be found using the formula |abcd|=ad-cb.
-3(04-23)+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 0 by 4.
-3(0-23)+0+0
Step 1.4.2.1.2
Multiply -2 by 3.
-3(0-6)+0+0
-3(0-6)+0+0
Step 1.4.2.2
Subtract 6 from 0.
-3-6+0+0
-3-6+0+0
-3-6+0+0
Step 1.5
Simplify the determinant.
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Step 1.5.1
Multiply -3 by -6.
18+0+0
Step 1.5.2
Add 18 and 0.
18+0
Step 1.5.3
Add 18 and 0.
18
18
18
Step 2
Since the determinant is non-zero, the inverse exists.
Step 3
Set up a 3×6 matrix where the left half is the original matrix and the right half is its identity matrix.
[203100300010024001]
Step 4
Find the reduced row echelon form.
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Step 4.1
Multiply each element of R1 by 12 to make the entry at 1,1 a 1.
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Step 4.1.1
Multiply each element of R1 by 12 to make the entry at 1,1 a 1.
[220232120202300010024001]
Step 4.1.2
Simplify R1.
[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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