SOLVING DETERMINANTS WITHOUT EXPANDING

Question 1 :

Prove that

Solution :

First let us subtract row 2 from row 1

The value of sec2A - tan2A is 1 and the value of tan2A - sec2A  is -1.So, we get

Here column 1 and 3 are identical, so the answer is 0.

Hence proved.

Question 2 :

Show that 

Solution :

First let us subtract row 2 from row 1

After factor out "a" from 1st row, row 1 and row 3 will be equal.

So the answer is 0.

Hence proved.

Question 3 :

Write the general form of a 3 × 3 skew-symmetric matrix and prove that its determinant is 0.

Solution :

By expanding this, we get

  =  -a(0 + bc) + b(ac - 0)

  =  -a(bc) + b(ac)

  =  -abc + abc

  =  0

Hence determinant of 3 x 3 skew matrix is 0.

Question 4 :

If

prove that a, b, c are in G.P. or a is a root of ax2 + 2bx + c = 0.

Solution :

2 + 2bα + c  =  0

α is the root of + 2bα + c  = 0

ac - b2  =  0

ac  =  b2

ac  =  bb

a/b  =  b/c

Hence a,b and c are in G.P

Question 5 :

prove that

Solution :

By using the property let us write the given determinant as sum of two determinants. 

Now let us factor out a,b and c from 1st, 2nd and 3rd row respectively.

By interchanging R1 and R3, we get

  =   0

Hence proved.

Question 6 :

If a, b, c are pth, qth and rth terms of an A.P, find the value of

Solution :

General term of A.P

an  =  a + (n - 1)d

pth term  =  a

qth term  =  b

rth term  =  c

ap  =  a + (p - 1)d   -----(1)

aq  =  a + (q - 1)d   -----(1)

ar  =  a + (r - 1)d   -----(1)

Now we have to replace a, b and c by (1), (2) and (3) respectively.

In the first determinant, 1st and 3rd rows are identical.

In the second determinant, 1st and 2nd rows are identical.

In the third determinant, 1st and 3rd rows are identical.

  =  a(0) + d(0) - 0

  =  0

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