Let k_1 and k_2 be the maximum and minimum va | vedclass

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

Question

(D) For the system of equations to be consistent,the determinant of the augmented matrix must be zero:$\Delta = \begin{vmatrix} 1 & k & 1 \ k & 1 & 2

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(D) For the system of equations to be consistent,the determinant of the augmented matrix must be zero:$\Delta = \begin{vmatrix} 1 & k & 1 \ k & 1 & 2 \ 1 & 1 & k \end{vmatrix} = 0$Expanding the determinant:$1(k - 2) - k(k^2 - 2) + 1(k - 1) = 0$$k - 2 - k^3 + 2k + k - 1 = 0$$-k^3 + 4k - 3 = 0$$k^3 - 4k + 3 = 0$By inspection,$k = 1$ is a root. Dividing by $(k - 1)$:$(k - 1)(k^2 + k - 3) = 0$If $k = 1$,the equations become $x + y = 1$,$x + y = 2$,and $x + y = 1$,which is inconsistent.Thus,$k^2 + k - 3 = 0$. The roots are $k = \frac{-1 \pm \sqrt{1 - 4(1)(-3)}}{2} = \frac{-1 \pm \sqrt{13}}{2}$.Let $k_1 = \frac{-1 + \sqrt{13}}{2}$ and $k_2 = \frac{-1 - \sqrt{13}}{2}$.Then $k_1^2 + k_2^2 = (k_1 + k_2)^2 - 2k_1k_2 = (-1)^2 - 2(-3) = 1 + 6 = 7$.

Let $k_1$ and $k_2$ be the maximum and minimum values of $k$ for which the system of equations $x + ky = 1$,$kx + y = 2$,and $x + y = k$ are consistent. Then $k_1^2 + k_2^2$ is equal to:

Similar Questions

If the augmented matrix corresponding to the system of equations $x+y-z=1$, $2x+4y-z=0$ and $3x+4y+5z=18$ is transformed to $\left[\begin{array}{cccc}1 & a & 0 & -1 \\ 0 & 2 & 1 & b \\ 0 & 0 & c & 32\end{array}\right]$, then $\sqrt{a+b+c}=$

The system of equations $2x + 6y = -11$,$6x + 20y - 6z = -3$ and $6y - 18z = -1$ are

Examine the consistency of the system of equations: $x+y+z=1$,$2x+3y+2z=2$,and $ax+ay+2az=4$.

Number of triplets of $(a, b, c)$ for which the system of equations $ax - by = 2a - b$ and $(c + 1)x + cy = 10 - a + 3b$ has infinitely many solutions and $(x = 1, y = 3)$ is one of the solutions,is:

If the system of equations $x+2y-3z=2$,$2x+\lambda y+5z=5$,$14x+3y+\mu z=33$ has infinitely many solutions,then $\lambda+\mu$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module