If q₁ , q₂ , q₃ are roots of equation x^3 + 64 = 0 , then value of $left| {begin{array}{*{20}{

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3 and 4 .Determinants and Matrices

normal

If $q_1$ , $q_2$ , $q_3$ are roots of the equation $x^3 + 64$ = $0$ , then the value of $\left| {\begin{array}{*{20}{c}}
{{q_1}}&{{q_2}}&{{q_3}} \\
{{q_2}}&{{q_3}}&{{q_1}} \\
{{q_3}}&{{q_1}}&{{q_2}}
\end{array}} \right|$ is

$1$

$4$

$16$

$0$

Solution

$\left|\begin{array}{lll}{\mathrm{q}_{1}} & {\mathrm{q}_{2}} & {\mathrm{q}_{3}} \\ {\mathrm{q}_{2}} & {\mathrm{q}_{3}} & {\mathrm{q}_{1}} \\ {\mathrm{q}_{3}} & {\mathrm{q}_{1}} & {\mathrm{q}_{2}}\end{array}\right| \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}$

$=\left(\mathrm{q}_{1}+\mathrm{q}_{2}+\mathrm{q}_{3}\right)\left|\begin{array}{lll}{1} & {\mathrm{q}_{2}} & {\mathrm{q}_{3}} \\ {1} & {\mathrm{q}_{3}} & {\mathrm{q}_{1}} \\ {1} & {\mathrm{q}_{1}} & {\mathrm{q}_{2}}\end{array}\right|$

$=0$ ($\because$ sum of roots is zero)

Standard 12

Mathematics

In a square matrix $A$ of order $3, a_{i i}'s$ are the sum of the roots of the equation $x^2 – (a + b)x + ab= 0$; $a_{i , i + 1}'s$ are the product of the roots, $a_{i , i – 1}'s$ are all unity and the rest of the elements are all zero. The value of the det. $(A)$ is equal to

normal

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medium

If $\left| {\,\begin{array}{*{20}{c}}{x – 1}&3&0\\2&{x – 3}&4\\3&5&6\end{array}\,} \right| = 0$, then $x =$

easy

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medium

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medium

(JEE MAIN-2025)

If $q_1$ , $q_2$ , $q_3$ are roots of the equation $x^3 + 64$ = $0$ , then the value of $\left| {\begin{array}{*{20}{c}} {{q_1}}&{{q_2}}&{{q_3}} \\ {{q_2}}&{{q_3}}&{{q_1}} \\ {{q_3}}&{{q_1}}&{{q_2}} \end{array}} \right|$ is

Solution

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If $q_1$ , $q_2$ , $q_3$ are roots of the equation $x^3 + 64$ = $0$ , then the value of $\left| {\begin{array}{*{20}{c}}
{{q_1}}&{{q_2}}&{{q_3}} \\
{{q_2}}&{{q_3}}&{{q_1}} \\
{{q_3}}&{{q_1}}&{{q_2}}
\end{array}} \right|$ is