The value of a for which the system of equati | vedclass

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

Question

(C) For a system of homogeneous linear equations to have a non-zero solution,the determinant of the coefficient matrix must be zero. $\left|\begin{arr

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(C) For a system of homogeneous linear equations to have a non-zero solution,the determinant of the coefficient matrix must be zero. $\left|\begin{array}{ccc} a^3 & (a+1)^3 & (a+2)^3 \ a & (a+1) & (a+2) \ 1 & 1 & 1 \end{array}ight| = 0$ Interchanging rows to simplify: $-\left|\begin{array}{ccc} 1 & 1 & 1 \ a & (a+1) & (a+2) \ a^3 & (a+1)^3 & (a+2)^3 \end{array}ight| = 0$ Using the property of the Vandermonde-like determinant $\left|\begin{array}{ccc} 1 & 1 & 1 \ x & y & z \ x^3 & y^3 & z^3 \end{array}ight| = (x-y)(y-z)(z-x)(x+y+z)$,where $x=a, y=a+1, z=a+2$: $-(a-(a+1))((a+1)-(a+2))((a+2)-a)(a+(a+1)+(a+2)) = 0$ $-(-1)(-1)(2)(3a+3) = 0$ $-2(3a+3) = 0$ $3a+3 = 0 \Rightarrow a = -1$.

The value of $a$ for which the system of equations has a non-zero solution is
$a^{3} x+(a+1)^{3} y+(a+2)^{3} z=0$
$a x+(a+1) y+(a+2) z=0$
$x+y+z=0$

Similar Questions

The roots of the equation $\left| \begin{array}{ccc} x-1 & 1 & 1 \\ 1 & x-1 & 1 \\ 1 & 1 & x-1 \end{array} \right| = 0$ are

If $A = \begin{bmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix}$,then $\det A$ =

If $x^2+y^2+z^2 \neq 0, \quad x=cy+bz, \quad y=az+cx$ and $z=bx+ay$,then $a^2+b^2+c^2+2abc$ is equal to

Solve the equation $\left|\begin{array}{ccc}x+a & x & x \\ x & x+a & x \\ x & x & x+a\end{array}\right|=0$,where $a \neq 0$.

If $\left|\begin{array}{ll}3 & x \\ x & 1\end{array}\right|=\left|\begin{array}{ll}3 & 2 \\ 4 & 1\end{array}\right|$,then $x$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

The value of $a$ for which the system of equations has a non-zero solution is $a^{3} x+(a+1)^{3} y+(a+2)^{3} z=0$ $a x+(a+1) y+(a+2) z=0$ $x+y+z=0$