Test Index

TS EAMCET 10-Sep-2020 Shift 2 Solved Paper

Section: Mathematics

© examsnet.com

Question : 4 of 160

Marks: +1, -0

Let

A=\begin{bmatrix} 1 & 4 & 2 \\ 2 & -1 & 4 \\ -3 & 7 & -6 \end{bmatrix}

B=[b_{ij}]_{3\times 3}

b_{11}=2, b_{13}=-2, b_{12}=0

AB =\begin{bmatrix} 2 & 14 & -4 \\ 4 & 1 & -8 \\ -6 & 15 & 12 \end{bmatrix}

|B|+\operatorname{trace}(B)=

-2
10
-8
6

Solution:

We have,

A=\begin{bmatrix} 1 & 4 & 2 \\ 2 & -1 & 4 \\ -3 & 7 & -6 \end{bmatrix}

and

B=\begin{bmatrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{bmatrix}=\begin{bmatrix} 2 & 0 & -2 \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{bmatrix}

\therefore AB=\begin{bmatrix} 1 & 4 & 2 \\ 2 & -1 & 4 \\ -3 & 7 & -6 \end{bmatrix}\begin{bmatrix} 2 & 0 & -2 \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{bmatrix}

\begin{bmatrix} 2+4b_{21}+2b_{31} & 0+4b_{22}+2b_{32} & -2+4b_{23}+2b_{33} \\ 4-b_{21}+4b_{31} & 0+b_{22}+4b_{32} & -4-b_{23}+4b_{33} \\ -6+7b_{21}-6b_{31} & 0+7b_{22}-6b_{32} & 6+7b_{23}-6b_{33} \end{bmatrix}

=\begin{bmatrix} 2 & 14 & -4 \\ 4 & 1 & -8 \\ -6 & 15 & 12 \end{bmatrix}

On solving above equal matrices with corresponding elements, we get

b_{21}=b_{31}=0, b_{22}=3, b_{32}=1, b_{23}=0

and

b_{33}=-1

\therefore \;\; B=\begin{bmatrix} 2 & 0 & -2 \\ 0 & 3 & 0 \\ 0 & 1 & -1 \end{bmatrix}

\therefore |B|=2(-3-0)+(-2)(0-0)=-6

and Trace

(B)=2+3-1=4

\therefore |B|+\operatorname{trace}(B)=-6+4=-2

© examsnet.com

Go to Question:

Prev Question

Next Question