FINDING DETERMINANT OF A SQUARE MATRIX

Evaluate the following determinants.

Problem 1 :

02-41

Solution:

=02-41=0+8=8

Problem 2 :

-40-55

Solution:

=-40-55=-20+0=-20

Problem 3 :

5-5-5-3-35345

Solution:

=5-5-5-3-35345=5-3545+5-3535-5-3-334=5(-15-20)+5(-15-15)-5(-12+9)=5(-35)+5(-30)-5(-3)=-175-150+15=-310

Problem 4 :

3-142-254-10

Solution:

=3-142-254-10=3-25-10+12540+42-24-1=3(0+5)+1(0-20)+4(-2+8)=3(5)+1(-20)+4(6)=15-20+24=19

Evaluate each determinant.

Problem 5 :

1-2-1-233-1-34

Solution:

=1-2-1-233-1-34=133-34+2-23-14-1-23-1-3=1(12+9)+2(-8+3)-1(6+3)=1(21)+2(-5)-1(9)=21-10-9=2

Problem 6 :

32-51-2-455-1

Solution:

=32-51-2-455-1=3-2-45-1-21-45-1-51-255=3(2+20)-2(-1+20)-5(5+10)=3(22)-2(19)-5(15)=66-38-75=-47

Find the area of a triangle with the given vertices.

Problem 7 :

(-2, 5), (6, 3) and (0, -8)

Solution:

To find the area of the triangle by joining the point given in the formula,

K=12x1y11x2y21x3y31

Substitute the given values in the above formula, we get

k=12-2516310-81k=12[-2(3+8)-5(6-0)+1(-48-0)]=12[-2(11)-5(6)+1(-48)]=12[-22-30-48]=12[-100]k=-50

Since the area of the triangle cannot be negative, the value of k = 50 square units.

Problem 8 :

(2, 5.5), (-3.5, 12) and (-5, -1.5)

Solution:

To find the area of the triangle by joining the point given in the formula,

K=12x1y11x2y21x3y31

Substitute the given values in the above formula, we get

k=1225.51-3.5121-5-1.51k=12[2(12+1.5)-5.5(-3.5+5)+1(5.25+60)]=12[2(13.5)-5.5(1.5)+1(65.25)]=12[27-8.25+65.25]=12[84]k=42 square units

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