Ordered pair (a, b) , for which system of linear equations 3 x-2 y+z=b ; 5 x-8 y+

English

Gujarati

Hindi

3 and 4 .Determinants and Matrices

medium

The ordered pair $(a, b)$, for which the system of linear equations $3 x-2 y+z=b$ ; $5 x-8 y+9 z=3$ ; $2 x+y+a z=-1$ has no solution, is

A

$\left(3, \frac{1}{3}\right)$

B

$\left(-3, \frac{1}{3}\right)$

C

$\left(-3,-\frac{1}{3}\right)$

D

$\left(3,-\frac{1}{3}\right)$

(JEE MAIN-2022)

Solution

$\left|\begin{array}{ccc} 3 & -2 & 1 \\ 5 & -8 & 9 \\ 2 & 1 & a \end{array}\right|=0$

$3(-8 a-9)+2(5 a-18)+1(21)=0$

$\Rightarrow a=-3$

Also $\Delta_{2}=\left|\begin{array}{ccc}3 & -2 & b^{\frac{1}{3}} \\ 5 & 8 & 3 \\ 2 & 1 & -1\end{array}\right|$

If $b =\frac{1}{3}$

$\Delta_{2}=0$

So $b$ must be equal to $-\frac{1}{3}$

Standard 12

Mathematics

Share

Similar Questions

If the system of equations $x + ay = 0,$ $az + y = 0$ and $ax + z = 0$ has infinite solutions, then the value of $a$ is

easy

(IIT-2003)

Let $\alpha, \beta(\alpha \neq \beta)$ be the values of $m$, for which the equations $x+y+z=1 ; x+2 y+4 z=m$ and $x+4 y+10 z=m^2$ have infinitely many solutions. Then the value of $\sum_{n=1}^{10}\left(n^\alpha+n^\beta\right)$ is equal to :

hard

(JEE MAIN-2025)

Consider system of equations in $x$ , $y$ and $z$

$12x + by + cz = 0$ ; $ax + 24y + cz = 0$ ; $ax + by + 36z = 0$ .

(where $a$ , $b$ , $c$ are real numbers, $a \ne 12$ , $b \ne 24$ , $c \ne 36$ ).

If system of equation has solution and $z \ne 0$, then value of $\frac{1}{{a – 12}} + \frac{2}{{b – 24}} + \frac{3}{{c – 36}}$ is

normal

If $\left| {\,\begin{array}{*{20}{c}}{x + 1}&1&1\\2&{x + 2}&2\\3&3&{x + 3}\end{array}\,} \right| = 0,$ then $x$ is

easy

The solutions of the equation $\left| {\,\begin{array}{*{20}{c}}x&2&{ – 1}\\2&5&x\\{ – 1}&2&x\end{array}\,} \right| = 0$ are

medium