# MATRIX AND DETERMINANTS EXAMPLE PROBLEMS

## About "Matrix Determinant Example Problems"

Matrix Determinant Example Problems :

Here we are going to see some example problems to understand solving determinants using properties.

To know properties of determinants, please visit the page "Properties of determinants".

## Matrix Determinant Example Problems - Questions

Question 1 :

Prove that

Solution :

First let us factor "a" from the 1st row, "b" from the 2nd row and c from the 3rd row.

Now we have to multiply column 1, 2 and  by a, b and c respectively.

Let us subtract 2nd row from 1st row and subtract 3rd row from the 2nd row.

=  x2(c2x2 + x4 + b2x2) + x2(0 + a2 x2

=  x2(c2x2 + x4 + b2x2) + x2(0 + a2 x2

=  x4 c2 + x6 + b2x4 + a2x4

=  x4(c2 + x2 + b2+ a2)

Hence it is divisible by x4.

Question 2 :

If a, b, c are all positive, and are pth, qth and rth terms of a G.P., show that

Solution :

nth term of G.P

an  =  arn-1

a = pth term of G.P  =  arp-1   --(1)

b = qth term of G.P  =  arq-1   --(2)

c = rth term of G.P  =  arr-1   --(3)

By using properties of determinants, let us write them as sum of two determinants.

In the second determinant, let us add 1st and 3rd column.

In the first determinant column 1 and  are identical. In the second determinant column 1 and 2 are identical.

=   log a (0) + log r (0)

=  0

Hence it is proved.

Question 3 :

Find the value of

if x, y and z ≠ 1

Solution :

By expanding the above determinant, we get

=  1[1 - logzy logyz] - logxy[logyx - logzx logyz] + logxz[logyxlogzy - logzx]

By using the properties of logarithms

=  [1 - logyy] - logxylogyx + logxylogzx logyz + logxzlogyxlogzy - logxzlogzx

=  [1 - logyy] - logyy + logzylogyz + logyzlogzy - logzz

=  [1 - 1] - 1 + logyy + logzz - 1

=  - 1 + 1 + 1 - 1

=  0

Question 4 :

If A =

Solution :

|A|  =  1/4

det (Ak)  =  (1/4)k

 if k = 1det (A1) = (1/4)1 if k = 2det (A2) = (1/4)2 if k = 3det (A3)=(1/4)3

By finding the sum, we get

=  (1/4) + (1/4)2 + (1/4)3 + ..................n terms

Sum of geometric series

Sn  =   a(rn - 1) / (r - 1)

a  =  1/4,  r  =  1/4

Sn  =   (1/4)((1/4)n - 1) / ((1/4) - 1)

=   (1/4)((1/4)n - 1) / (-3/4)

=  (-1/3) ((1/4)n - 1)

=  (1/3)(1 - (1/4)n)

Hence it is proved.

After having gone through the stuff given above, we hope that the students would have understood, "Matrix Determinant Example Problems".

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