If left| {,begin{array}{*{20}{c}}a&b&00&a&bb&0&aend{array},} right| = 0 , th

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3 and 4 .Determinants and Matrices

easy

If $\left| {\,\begin{array}{*{20}{c}}a&b&0\\0&a&b\\b&0&a\end{array}\,} \right| = 0$, then

$a $ is one of the cube roots of unity

$b$ is one of the cube roots of unity

$\left( {\frac{a}{b}} \right)$is one of the cube roots of unity

$\left( {\frac{a}{b}} \right)$is one of the cube roots of $ -1$

Solution

(d) Given, $\Delta = \left| {\,\begin{array}{*{20}{c}}a&b&0\\0&a&b\\b&0&a\end{array}\,} \right|\, = \,0.$
Expanding the given determinant, we get $a({a^2} – 0) – b(0 – {b^2}) = 0$ or ${a^3} + {b^3} = 0.$
This equation may be written as ${\left( {\frac{a}{b}} \right)^3} = – 1.$
Therefore, $\left( {\frac{a}{b}} \right)$ is one of the cube roots of $ -1$.

Standard 12

Mathematics

If $D = \left| {\,\begin{array}{*{20}{c}}1&1&1\\1&{1 + x}&1\\1&1&{1 + y}\end{array}\,} \right|$ for $x \ne 0,y \ne 0$ then $D$ is

medium

(AIEEE-2007)

The system of equations $x+y+z=6$, $x+2 y+5 z=9$, $x+5 y+\lambda z=\mu$ has no solution if

medium

(JEE MAIN-2025)

The roots of the determinant equation (in $x$) $\left| {\,\begin{array}{*{20}{c}}a&a&x\\m&m&m\\b&x&b\end{array}\,} \right| = 0$

easy

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

easy

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

hard

If $\left| {\,\begin{array}{*{20}{c}}a&b&0\\0&a&b\\b&0&a\end{array}\,} \right| = 0$, then

Solution

Similar Questions

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