logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Matrices
0 votes

If $A=\begin{vmatrix}\alpha &0\\1 &1\end{vmatrix}$ and $B=\begin{vmatrix}1 &0\\5 &1\end{vmatrix}$then the value of $\alpha$ for which $A^2=B$ is

$(a)\;1\qquad(b)\;-1\qquad(c)\;4\qquad(d)\;No\;real\;value$

Can you answer this question?
 
 

1 Answer

0 votes
Given that :
$A=\begin{vmatrix}\alpha &0\\1 &1\end{vmatrix}$ and $B=\begin{vmatrix}1 &0\\5 &1\end{vmatrix}$
$A^2=B$
$\Rightarrow \begin{vmatrix}\alpha &0\\1 &1\end{vmatrix}\begin{vmatrix}\alpha &0\\1 &1\end{vmatrix}=\begin{vmatrix}1 &0\\5 &1\end{vmatrix}$
$\Rightarrow \begin{vmatrix}\alpha^2&0\\\alpha+1&1\end{vmatrix}=\begin{vmatrix}1 &0\\5&1\end{vmatrix}$
$\Rightarrow \alpha^2=1$ and $\alpha+1=5$
$\Rightarrow \alpha=\pm 1$ and $\alpha=4$
There is no common value.
There is no real value of $\alpha$ for which $A^2=B$
Hence (d) is the correct answer.
answered Nov 20, 2013 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...