If A = ≤ft[ {{array}{cc} {4i - 6}&{10i}\\ {14i}&{6 + 4i} {array}} ] and k = {1}{{2i}}, where i = { - 1}, then kA is equal to

Correct Answer is : [2+3i572−3i]{bmatrix} 2+3i & 5 \\ 7 & 2-3i {bmatrix}[2+3i752−3i]

{bmatrix} 2+3i & 5 \\ 7 & 2-3i {bmatrix}

{bmatrix} 2-3i & 5 \\ 7 & 2+3i {bmatrix}

{bmatrix} 2-3i & 7 \\ 5 & 2+3i {bmatrix}

{bmatrix} 2+3i & 5\\ 7 & 2+3i {bmatrix}

Correct Answer is : [2+3i572−3i]{bmatrix} 2+3i & 5 \\ 7 & 2-3i {bmatrix}[2+3i752−3i]

Test Index

UPSC NDA 2 2017 Mathematics Solved Paper

Question : 11 of 120

Marks: +1, -0

A = \left[ {\begin{array}{cc} {4i - 6}&{10i}\\ {14i}&{6 + 4i} \end{array}} \right]

k = \frac{1}{{2i}}

i = \sqrt { - 1}

, then kA is equal to

Solution:

Concept:

Scalar multiplication: each element of matrix

A

is multiplied by the scalar

k

Explanation:

Given

k = \frac{1}{2i}

and

A = \begin{bmatrix} 4i-6 & 10i \\ 14i & 6+4i \end{bmatrix}

Multiply every element by

k

kA = \frac{1}{2i} \begin{bmatrix} 4i-6 & 10i \\ 14i & 6+4i \end{bmatrix} = \begin{bmatrix} \frac{4i-6}{2i} & \frac{10i}{2i} \\ \frac{14i}{2i} & \frac{6+4i}{2i} \end{bmatrix}

Simplify each entry.

Entry (1,1):

\frac{4i-6}{2i} = 2 - \frac{3}{i}

. Multiply numerator and denominator of

\frac{3}{i}

i

\frac{3}{i} = \frac{3i}{i^2} = \frac{3i}{-1} = -3i

. So

2 - (-3i) = 2+3i

Entry (1,2):

\frac{10i}{2i} = 5

Entry (2,1):

\frac{14i}{2i} = 7

Entry (2,2):

\frac{6+4i}{2i} = \frac{6}{2i} + \frac{4i}{2i} = \frac{3}{i} + 2

. Rationalize

\frac{3}{i} = -3i

. Hence

2 + (-3i) = 2-3i

Thus

kA = \begin{bmatrix} 2+3i & 5 \\ 7 & 2-3i \end{bmatrix}

Answer:

\begin{bmatrix} 2+3i & 5 \\ 7 & 2-3i \end{bmatrix}

(Option A).

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