SOLUTION: Find the inverse of each matrix, if it exists. 1.[ -2 -1] [10 7] 2.[-4 2] [-5 1] 3.[9 -3] [-6 2] 4. [3 4] [2 3] x=[6/5] 5. [2 -3] x+[-1 7] [9 12]

 



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take terms and  and switch their positions. This is simply switching the elements on the main diagonal 

 

take terms and  and change those numbers to their opposites keeping there positions.





:now we find the determinant of this matrix

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product of the main diagonal-product of the other diagonal

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we take that result and divide every element in the matrix and the result is our inverse matrix

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the way you can check is to multiply the two together and see if your result is the identity matrix....if it is you can rest easy you have the correct answer.

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2)and 3)

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now I will let you do the steps to 2 and 3

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answers to 2 is 

...........3 is no solution as the determinant = 0

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4)X or=

3a+4b=6

2a+3b=5

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3R2-2R1(Row 2)---->R1-4R2(row 1)---->1/3R1---> so we end with 

a=-2

b=3

so X=

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please re write equation 5  making it clear what their after

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6)6.[1 -6 0]      [1}

    [0 1 -7] [X} =[4]

    [3 0 2]       [11]



In order to go from a 3x3 to a 3x1 X also has to be a 3x1

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call X=

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a-6b=1.....eq 1

b-7c=4.....eq 2

3a-2c=11...eq 3

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lets solve by substution method

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revise eq 1 to a=1+6b and plug that value into eq 3

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3(1+6b)-2c=11

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3+18b-2c=11

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18b-2c=8....eq 3 revised

  b-7c=4....eq 2

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multiply eq 2 by -18 and add the results to revised eq 3

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18b-2c=8.......eq 3 revised

-18b+126c=-72..eq 2 revised

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as you can see the b terms are eliminated. We are left with -2c+126c=8-72

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124c=-64

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c=-64/124=-32/62=-16/31

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b-7(-16/31)=4

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b+112/31=4...multiply by 31

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31b+112=124

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31b=12

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b=12/31

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a=1+6(12/31)

a=31/31+72/31

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a=103/31

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X=