If 3A + 4B' = {array}{ccc} 7 & -10 & 17 \\ 0 & 6 & 31 {array} and 2B - 3A' = {array}{cc} -1 & 18 \\ 4 & 0 \\ -5 & -7 {array} then B = ?

Correct Answer is : 13−1124{array}{cc} 1 & 3 \\ -1 & 1 \\ 2 & 4 {array}1−12314

{array}{cc} 1 & 3 \\ -1 & 1 \\ 2 & 4 {array}

{array}{cc} 1 & -3 \\ -1 & 1 \\ 2 & 4 {array}

{array}{cc} -1 & -18 \\ 4 & -16 \\ -5 & -7 {array}

{array}{cc} 1 & 3 \\ -1 & 1 \\ 2 & -4 {array}

Correct Answer is : 13−1124{array}{cc} 1 & 3 \\ -1 & 1 \\ 2 & 4 {array}1−12314

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UPSC NDA Math Model Paper 4

Question : 64 of 120

Marks: +1, -0

3A + 4B' = \begin{array}{ccc} 7 & -10 & 17 \\ 0 & 6 & 31 \end{array}

2B - 3A' = \begin{array}{cc} -1 & 18 \\ 4 & 0 \\ -5 & -7 \end{array}

B = ?

Solution:

Using the Properties of Transpose of a Matrix:

3A + 4B' = \begin{array}{ccc} 7 & -10 & 17 \\ 0 & 6 & 31 \end{array}

\Rightarrow (3A+4B')' = \left[\begin{array}{ccc} 7 & -10 & 17 \\ 0 & 6 & 31 \end{array}\right]'

\Rightarrow 3A' + 4B = \begin{array}{ccc} 7 & -10 & 17 \\ 0 & 6 & 31 \end{array}'

\Rightarrow 3A' + 4B = \begin{array}{cc} 7 & 0 \\ -10 & 6 \\ 17 & 31 \end{array}

. . .(1)

2B - 3A' = \begin{array}{cc} -1 & 18 \\ 4 & 0 \\ -5 & -7 \end{array}

. . .(2)

Adding equations (1) and (2), we get,

(3A' + 4B) + (2B - 3A')

= \begin{array}{cc} 7 & 0 \\ -10 & 6 \\ 17 & 31 \end{array} + \begin{array}{cc} -1 & 18 \\ 4 & 0 \\ -5 & -7 \end{array}

\Rightarrow 6B + 0 = \begin{array}{cc} 7 - 1 & -10 + 4 \\ 17 - 5 & 0 + 18 \\ 6 + 0 & 31 - 7 \end{array}

\Rightarrow 6B = \begin{array}{cc} 6 & 18 \\ -6 & 6 \\ 12 & 24 \end{array}

\Rightarrow B = \begin{array}{cc} 1 & 3 \\ -1 & 1 \\ 2 & 4 \end{array}

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