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4 Maths -- System of Linear Equations

Solve (Gauss Elimination):3x1 – x2 + x3 = 2,- 15x1 + 6x2 – 5x3 = 5,x1 – 2x2 + 2x3 = 1

Solve (Gauss Elimination):

3x1 – x2 + x3 = 2,

- 15x1 + 6x2 – 5x3 = 5,

x1 – 2x2 + 2x3 = 1

1 Answers

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Soln:


Or, 3x1 – x2 + x3 = 2 …(i)


Or, - 15x1 + 6x2 – 5x3 = 5 …(ii)


Or, 5x1 – 2x2 + 2x3 = 1 …(iii),


Multiplying equation (i) by 5 and adding with (ii),


Or, x2 = 15 ….(iv)


Multiplying equation (i) by 5 and (iii) by 3 and subtracting (iii) from (i),


Or, x2 + x3 = 7….(v)


Now, we have three equations,


Or, 3x1 – x2 + x3 = 2 …(i)


Or, x2 = 15 …(iv)


Or, x2 – x3 = 7 …(v)


From (iv), we have, x2 = 15.


Using x2 = 15 in (v), we have, x3 = 8.


Using x2 = 15, x3 = 8 in equation (i), we have, x1 = 3.


So, x1 = 3, x2 = 15, x3 = 8.

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Examine consistency. Solve if possible.x1 + x2 + x3 = -3,3x1 + x2 – 2x3 = -2,2x1 + 4x2 + 7x3 = 7

Soln:


We have,


Or, x1 + x2 + x3 = -3 …(i)


Or, 3x1 + x2 – 2x3 = -2 …(ii)


Or, 2x1 + 4x2 + 7x3 = 7 …(iii)


Multiplying equation (I) by 3 and subtracting (ii) from (i),


Or, 2x2 + 5x3 = 7 ….(iv)


Multiplying equation (i) by 2 and subtracting (iii) from (i),


Or, -2x2 – 5x3 = 13 …(v)


Adding equation (iv) and (v),


0 = 20.


Now, we have the following three equations.


Or, x1 + x2 + x3 = -3….(i)


Or, 2x2 + 5x3 = 7 ….(iv)


Or, 0.x3 = 20 ….(vi)


Hence, no value of x3 satisfies the equation (vi),


So, the system has no solution....

Examine consistency. Solve if possible.x1 + 2x2 + 3x3 = 4,4x1 + 5x2 + 6x3 = -7,7x1 + 8x2 + 9x3 = 10

Soln:


Or, x1 + 2x2 + 3x3 = 4 …(i)


Or, 4x1 + 5x2 + 6x3 = -7 ….(ii)


Or, 7x1 + 8x2 + 9x3 = 10 …… (iii)


Multiplying equation (i) by 4 and then subtracting from (ii),


Or, -3x1 – 6x3 = -23 ….(iv).


Again, multiplying equation (i) by 7 and then subtracting from (iii),


Or, - 6x2 – 12x3 = -18….(v)


Multiplying equation (iv) by 2 and then subtracting from (v),


0 = 28 …(vi)


Now, we have the following equations,


Or, x1 + 2x2 + 3x3 = 4….(i)


Or, - 3x2 – 6x3 = -23 …(iv)


Or, 0.x3 = 28 ….(vi).


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