Question 1 :
Prove that
Solution :
First let us factor "a" from the 1^{st} row, "b" from the 2^{nd} row and c from the 3^{rd} row.
Now we have to multiply column 1, 2 and by a, b and c respectively.
Let us subtract 2^{nd} row from 1^{st} row and subtract 3^{rd} row from the 2^{nd} row.
= x^{2}(c^{2}x^{2} + x^{4} + b^{2}x^{2}) + x^{2}(0 + a^{2} x^{2})
= x^{2}(c^{2}x^{2} + x^{4} + b^{2}x^{2}) + x^{2}(0 + a^{2} x^{2})
= x^{4} c^{2} + x^{6} + b^{2}x^{4} + a^{2}x^{4}
= x^{4}(c^{2} + x^{2} + b^{2}+ a^{2})
Hence it is divisible by x^{4}.
Question 2 :
If a, b, c are all positive, and are p^{th}, q^{th} and r^{th} terms of a G.P., show that
Solution :
n^{th} term of G.P
a_{n} = ar^{n-1}
a = p^{th} term of G.P = ar^{p-1 } --(1)
b = q^{th} term of G.P = ar^{q-1 } --(2)
c = r^{th} term of G.P = ar^{r-1 } --(3)
By using properties of determinants, let us write them as sum of two determinants.
In the second determinant, let us add 1^{st} and 3^{rd} 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 - log_{z}y log_{y}z] - log_{x}y[log_{y}x - log_{z}x log_{y}z] + log_{x}z[log_{y}xlog_{z}y - log_{z}x]
By using the properties of logarithms
= [1 - log_{y}y] - log_{x}ylog_{y}x + log_{x}ylog_{z}x log_{y}z + log_{x}zlog_{y}xlog_{z}y - log_{x}zlog_{z}x
= [1 - log_{y}y] - log_{y}y + log_{z}ylog_{y}z + log_{y}zlog_{z}y - log_{z}z
= [1 - 1] - 1 + log_{y}y + log_{z}z - 1
= - 1 + 1 + 1 - 1
= 0
Hence the answer is 0.
Question 4 :
If A =
Solution :
|A| = 1/4
det (A^{k}) = (1/4)^{k}
if k = 1 det (A^{1}) = (1/4)^{1} |
if k = 2 det (A^{2}) = (1/4)^{2} |
if k = 3 det (A^{3})=(1/4)^{3} |
By finding the sum, we get
= (1/4) + (1/4)^{2} + (1/4)^{3} + ..................n terms
Sum of geometric series
S_{n} = a(r^{n} - 1) / (r - 1)
a = 1/4, r = 1/4
S_{n} = (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.
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