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AP EAMCET Engineering 2018 Apr 22 Shift 1 Paper

Section: Mathematics

Question : 5 of 160

Marks: +1, -0

A = \begin{bmatrix} \frac{k}{2} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{m}{4} \end{bmatrix}

A^{-1} = \begin{bmatrix} \frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{4} \end{bmatrix}

k+l+m=

Solution:

The given matrices are.

A = \begin{bmatrix} \frac{k}{2} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{m}{4} \end{bmatrix}

and

A^{-1} = \begin{bmatrix} \frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{4} \end{bmatrix}

From property of matrices

A A^{-1}=1

Here, the matrices l is identity matrices.

∴

\begin{bmatrix} \frac{k}{2} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{m}{4} \end{bmatrix} \begin{bmatrix} \frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{4} \end{bmatrix}

= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

On solving we get.

k=4 \times 1=4

l=9 \times 1=9

m=16 \times 1=16

\therefore k+1+m=4+9+16=29

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