A={bmatrix} x+y & y \\ 2x & x-y {bmatrix},B={bmatrix} 2 \\ -1 {bmatrix} and C={bmatrix} 3 \\ 2 {bmatrix} If AB = C, then what is A² equal to?

Correct Answer is : [6−10426]{bmatrix} 6 & -10 \\ 4 & 26 {bmatrix}[64−1026]

{bmatrix} 6 & -10 \\ 4 & 26 {bmatrix}

{bmatrix} -10 & 5 \\ 4 & 24 {bmatrix}

{bmatrix} -5 & -6 \\ -4 & -20 {bmatrix}

{bmatrix} -5 & -7 \\ -5 & 20 {bmatrix}

Correct Answer is : [6−10426]{bmatrix} 6 & -10 \\ 4 & 26 {bmatrix}[64−1026]

Test Index

UPSC NDA I 2015 Mathematics Paper

Question : 23 of 120

Marks: +1, -0

A=\begin{bmatrix} x+y & y \\ 2x & x-y \end{bmatrix},B=\begin{bmatrix} 2 \\ -1 \end{bmatrix}

C=\begin{bmatrix} 3 \\ 2 \end{bmatrix}

AB = C

A^{2}

equal to?

Solution:

Concept:

Use the matrix equation

AB = C

to find

x

and

y

, then compute

A^2

by matrix multiplication.

Explanation:

Given

A = \begin{bmatrix} x+y & y \\ 2x & x-y \end{bmatrix}

B = \begin{bmatrix} 2 \\ -1 \end{bmatrix}

C = \begin{bmatrix} 3 \\ 2 \end{bmatrix}

, and

AB = C

Compute

AB

AB = \begin{bmatrix} x+y & y \\ 2x & x-y \end{bmatrix} \begin{bmatrix} 2 \\ -1 \end{bmatrix} = \begin{bmatrix} 2(x+y) + y(-1) \\ 2x(2) + (x-y)(-1) \end{bmatrix} = \begin{bmatrix} 2x + 2y - y \\ 4x - x + y \end{bmatrix} = \begin{bmatrix} 2x + y \\ 3x + y \end{bmatrix}

Set equal to

C

\begin{bmatrix} 2x+y \\ 3x+y \end{bmatrix} = \begin{bmatrix} 3 \\ 2 \end{bmatrix}

This gives

2x+y=3

and

3x+y=2

Subtract first from second:

(3x+y)-(2x+y)=2-3 \Rightarrow x = -1

Substitute

x=-1

into

2(-1)+y=3 \Rightarrow -2+y=3 \Rightarrow y=5

Thus

A = \begin{bmatrix} -1+5 & 5 \\ 2(-1) & -1-5 \end{bmatrix} = \begin{bmatrix} 4 & 5 \\ -2 & -6 \end{bmatrix}

Now compute

A^2 = A \cdot A

A^2 = \begin{bmatrix} 4 & 5 \\ -2 & -6 \end{bmatrix} \begin{bmatrix} 4 & 5 \\ -2 & -6 \end{bmatrix} = \begin{bmatrix} 4\cdot4 + 5\cdot(-2) & 4\cdot5 + 5\cdot(-6) \\ (-2)\cdot4 + (-6)\cdot(-2) & (-2)\cdot5 + (-6)\cdot(-6) \end{bmatrix}

Calculate each entry:

Top-left:

16 + (-10) = 6

Top-right:

20 + (-30) = -10

Bottom-left:

(-8) + 12 = 4

Bottom-right:

(-10) + 36 = 26

Thus

A^2 = \begin{bmatrix} 6 & -10 \\ 4 & 26 \end{bmatrix}

Answer:

\begin{bmatrix} 6 & -10 \\ 4 & 26 \end{bmatrix}

which is option A.

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