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AP EAMCET Engineering 2013 Solved Paper

Section: Mathematics

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Question : 95 of 160

Marks: +1, -0

If

A = \begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix}

satisfies the equation

x^{2} + 4x - p = 0

p

is equal to

64
42
36
24

Solution:

Given,

A = \begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix}

A^{2} = \begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix}

= \begin{bmatrix} 64+10 & -40+20 \\ -16+8 & 10+16 \end{bmatrix}

= \begin{bmatrix} 74 & -20 \\ -8 & 26 \end{bmatrix}

4A = \begin{bmatrix} -32 & 20 \\ 8 & 16 \end{bmatrix}

-pI = \begin{bmatrix} -p & 0 \\ 0 & -p \end{bmatrix}

Since, the matrix A satisfies the equation

x^{2} + 4x - p = 0,

then

A^{2} + 4A - pI = 0

\begin{bmatrix} 74 & -20 \\ -8 & 26 \end{bmatrix} + \begin{bmatrix} -32 & 20 \\ 8 & 16 \end{bmatrix}

+ \begin{bmatrix} -p & 0 \\ 0 & -p \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}

\begin{bmatrix} 74-32-p & -20+20+0 \\ -8+8+0 & 26+16-p \end{bmatrix} =

\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}

⇒

\begin{bmatrix} 42-p & 0 \\ 0 & 42-p \end{bmatrix}

\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}

On comparing, we get

42 - p = 0 \Rightarrow p = 42

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