3 and 4 .Determinants and Matrices
hard

If $A\, = \,\left[ {\begin{array}{*{20}{c}}
1&2&x\\
3&{ - 1}&2
\end{array}} \right]$ and $B\, = \,\left[ {\begin{array}{*{20}{c}}
y\\
x\\
1
\end{array}} \right]$ be such that $AB\, = \,\left[ {\begin{array}{*{20}{c}}
6\\
8
\end{array}} \right],$ then

A

$y = 2x$

B

$y = -2x$

C

$y = x$

D

$y = -x$

(JEE MAIN-2014)

Solution

Let $A = \left[ {\begin{array}{*{20}{c}}
1&2&x\\
3&{ – 1}&2
\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}
y\\
x\\
1
\end{array}} \right]$

$AB = \left[ {\begin{array}{*{20}{c}}
1&2&x\\
3&{ – 1}&2
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
y\\
x\\
1
\end{array}} \right]$

$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
6\\
8
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{y + 2x + x}\\
{3y – x + 2}
\end{array}} \right]$

$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
6\\
8
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{y + 3x}\\
{3y – x + 2}
\end{array}} \right]$

$ \Rightarrow y + 3x = 6$ and $3y – x = 6$

On solving, we gat 

$x = \frac{6}{5}$ and $y = \frac{{12}}{5}$

$ \Rightarrow y = 2x$

Standard 12
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.