Question 1 :
Prove that
Solution :
First let us subtract row 2 from row 1
The value of sec^{2}A - tan^{2}A is 1 and the value of tan^{2}A - sec^{2}A 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 1^{st} 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 ax^{2} + 2bx + c = 0.
Solution :
aα^{2 }+ 2bα + c = 0 α is the root of aα^{2 }+ 2bα + c = 0 |
ac - b^{2} = 0 ac = b^{2} 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 1^{st}, 2^{nd} and 3^{rd} row respectively.
By interchanging R_{1} and R_{3}, we get
= 0
Hence proved.
Question 6 :
If a, b, c are p^{th}, q^{th} and r^{th} terms of an A.P, find the value of
Solution :
General term of A.P
a_{n} = a + (n - 1)d
p^{th} term = a q^{th} term = b r^{th} term = c |
a_{p} = a + (p - 1)d -----(1) a_{q} = a + (q - 1)d -----(1) a_{r} = a + (r - 1)d -----(1) |
Now we have to replace a, b and c by (1), (2) and (3) respectively.
In the first determinant, 1^{st} and 3^{rd} rows are identical.
In the second determinant, 1^{st} and 2^{nd} rows are identical.
In the third determinant, 1^{st} and 3^{rd} rows are identical.
= a(0) + d(0) - 0
= 0
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