Consider the system of linear equations

-x | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$\Delta=\left|\begin{array}{ccc}-1 & 1 & 2 \ 3 & -a & 5 \ 2 & -2 & -a\end{array}ight|$

$=-1\left(a^{2}+10ight)-1(-3 a

Vedclass · Accepted Answer

$\mathrm{n}\left(\mathrm{S}_{1}\right)=2, \mathrm{n}\left(\mathrm{S}_{2}\right)=0$

Vedclass · Answer

$\Delta=\left|\begin{array}{ccc}-1 & 1 & 2 \ 3 & -a & 5 \ 2 & -2 & -a\end{array}ight|$

$=-1\left(a^{2}+10ight)-1(-3 a-10)+2(-6+2 a)$

$=-a^{2}-10+3 a+10-12+4 a$

$\Delta=-a^{2}+7 a-12$

$\Delta=-\left[a^{2}-7 a+12ight]$

$\Delta=-[(a-3)(a-4)]$

$\Delta_{1}=\left|\begin{array}{ccc}0 & 1 & 2 \ 1 & -a & 5 \ 7 & -2 & -a\end{array}ight|$

$=0-1(-\mathrm{a}-35)+2(-2+7 \mathrm{a})$

$\Rightarrow \mathrm{a}+35-4+14 \mathrm{a}$

$15 \mathrm{a}+31$

Now $\quad \Delta_{1}=15 \mathrm{a}+31$

For inconsistent $\Delta=0 	herefore \mathrm{a}=3, \mathrm{a}=4$

and for $\mathrm{a}=3$ and $4 \quad \Delta_{1} 
eq 0$

$\mathrm{n}\left(\mathrm{S}_{1}ight)=2$

For infinite solution : $\Delta=0$

and $\Delta_{1}=\Delta_{2}=\Delta_{3}=0$

Not possible

$	herefore \mathrm{n}\left(\mathrm{S}_{2}ight)=0$

Similar Questions

Set of equations $a + b - 2c = 0,$ $2a - 3b + c = 0$ and $a - 5b + 4c = \alpha $ is consistent for $\alpha$ equal to

if $\left| \begin{gathered}
- 6\ \ \,\,1\ \ \,\,\lambda \ \ \hfill \\
\,0\ \ \,\,\,\,3\ \ \,\,7\ \ \hfill \\
- 1\ \ \,\,0\ \ \,\,5\ \ \hfill \\
\end{gathered} \right| = 5948 $, then $\lambda $ is

Three digit numbers $x17, 3y6$ and $12z$ where $x, y, z$ are integers from $0$ to $9$, are divisible by a fixed constant $k$. Then the determinant $\left| {\,\begin{array}{*{20}{c}}x&3&1\\7&6&z\\1&y&2\end{array}\,} \right|$ + $48$ must be divisible by

The roots of the equation $\left| {\,\begin{array}{*{20}{c}}0&x&{16}\\x&5&7\\0&9&x\end{array}\,} \right| = 0$ are

How many values of $k $ , systeam of linear equations $\left( {k + 1} \right)x + 8y = 4k\;,\;kx + \left( {k + 3} \right)y$$ = 3k - 1$ has no solutions.