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

Hence the answer is 0.

Question 4 :

If A =

Solution :

|A|  =  1/4

det (Ak)  =  (1/4)k

if k = 1

det (A1) = (1/4)1

if k = 2

det (A2) = (1/4)2

if k = 3

det (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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