3 and 4 .Determinants and Matrices
hard

જો $A = \left( {\begin{array}{*{20}{c}}
{\alpha  - 1}\\
0\\
0
\end{array}} \right),\,\,\,B = \left( {\begin{array}{*{20}{c}}
{\alpha  + 1}\\
0\\
0
\end{array}} \right)$ બે શ્રેણિક છે તો $AB^T$ એ શૂન્યતર શ્રેણિક થવા માટે $\left| \alpha  \right|$ ની કિમત  . . . શક્ય નથી.

A

$2$

B

$0$

C

$1$

D

$3$

(AIEEE-2012)

Solution

Let $A = \left( {\begin{array}{*{20}{c}}
{\alpha  – 1}\\
0\\
0
\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}
{\alpha  + 1}\\
0\\
0
\end{array}} \right)$

be two matrices.

$A{B^T} = \left( {\begin{array}{*{20}{c}}
{\alpha  – 1}\\
0\\
0
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{\alpha  + 1}&0&0
\end{array}} \right)$

           $ = \left( {\begin{array}{*{20}{c}}
{{\alpha ^2} – 1}&0&0\\
0&0&0\\
0&0&0
\end{array}} \right)$

Thus, $A{B^T}$ is non-zero matrix for $\left| \alpha  \right| \ne 1$

Standard 12
Mathematics

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