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Mahitesh Rawat 1 year, 4 months ago

Continuity ?
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Tanushri Bhayani 1 year, 8 months ago

Kindly provide complete question !!
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Preeti Dabral 1 year, 8 months ago

The effective rate is the actual rate compounded annually. Therefore, the required sum S is given by
S = 12000 (1 + {tex}\frac{3}{100}{/tex})10 (1 + {tex}\frac{4}{100}{/tex})4 (1 + {tex}\frac{5}{100}{/tex})2
{tex}\Rightarrow{/tex} S = 12000 (1.03)10 (1.04)4 (1.05)2
{tex}\Rightarrow{/tex} S = 12000 (1.34391638) (1.16985856) (1.1025) = 20800.10
Hence, the amount is ₹20,800.10

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Tanushri Bhayani 1 year, 8 months ago

Ohh' alright

Nishant Dogra 1 year, 8 months ago

Using the properties of matrix algebra, we can expand the expression (I + A)3 as follows: (I + A)3 = (I + A)(I + A)(I + A) = (I + A)(I2 + 2IA + A2) = (I + A)(I + 2A + A) = I2 + 3IA + 3A2 + A3 = I + 3A + 3A2 + A3 (since A2 = A) Substituting this expression into the original equation, we get: (I + A)3 - 7A = (I + 3A + 3A2 + A3) - 7A = I + 2A + 3A2 - 7A = I - 5A + 3A2 Therefore, (I + A)3 - 7A is equal to I - 5A + 3A2.
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Tanushri Bhayani 1 year, 8 months ago

Hmm' got it 👍 thanks

Nishant Dogra 1 year, 8 months ago

Solution We can calculate the product AB as follows: AB = [[-i, 0], [0, i]] [[1, 0], [0, -1]] = [[-i, 0], [0, -i]] Next, we can calculate the determinant of AB as follows: det(AB) = (-i) * (-i) - 0 * 0 (determinant of a 2x2 matrix) = 1 Now, we can use the formula for the product of determinants to find the determinant of A multiplied by the determinant of (i^2-1)B: det(A) * det((i^2-1)B) = det(AB) det(A) * (i^2-1)^2 * det(B) = 1 Substituting the known values, we get: det(A) * (i^2-1)^2 * (-1/1) = 1 det(A) * 4 = 1 det(A) = 1/4 Since A is a 2x2 matrix, its determinant can be calculated as follows: det(A) = (-i) * i - 0 * 0 (determinant of a 2x2 matrix) = -1 Therefore, we have a contradiction: det(A) = -1 and det(A) = 1/4. Hence, there is no value of a that satisfies the given conditions.

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