If left| {,begin{array}{*{20}{c}}{{x^2} + x}&{x + 1}&{x - 2}{2{x^2} + 3x

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3 and 4 .Determinants and Matrices

medium

If $\left| {\,\begin{array}{*{20}{c}}{{x^2} + x}&{x + 1}&{x - 2}\\{2{x^2} + 3x - 1}&{3x}&{3x - 3}\\{{x^2} + 2x + 3}&{2x - 1}&{2x - 1}\end{array}\,} \right| = Ax - 12$, then the value of $A $ is

$12$

$24$

$-12$

$-24$

(IIT-1982) (JEE MAIN-2015)

Solution

(b) Trick : Put $x = 1$, then we have

$\left| {\,\begin{array}{*{20}{c}}2&2&{ – 1}\\4&3&0\\6&1&1\end{array}\,} \right| = A – 12 \Rightarrow \left| {\,\begin{array}{*{20}{c}}0&2&{ – 1}\\1&3&0\\5&1&1\end{array}\,} \right| = A – 12$

{Apply ${C_1} \to {C_1} – {C_2}$}

$ \Rightarrow $ $ – 2 + ( – 1)\,( – 14) = A – 12 \Rightarrow A = 24$.

Standard 12

Mathematics

$\left| {\,\begin{array}{*{20}{c}}{bc}&{bc' + b'c}&{b'c'}\\{ca}&{ca' + c'a}&{c'a'}\\{ab}&{ab' + a'b}&{a'b'}\end{array}\,} \right|$ is equal to

hard

If the system of linear equations $2 x + y – z =7$ ; $x-3 y+2 z=1$ ; $x +4 y +\delta z = k$, where $\delta, k \in R$ has infinitely many solutions, then $\delta+ k$ is equal to

medium

(JEE MAIN-2022)

The remainder when the determinant $\left|\begin{array}{lll} 2014^{2014} & 2015^{2015} & 2016^{2016} \\ 2017^{2017} & 2018^{2018} & 2019^{2019} \\ 2020^{2020} & 2021^{2021} & 2022^{2022} \end{array}\right|$ is divided by $5$ is

normal

(KVPY-2015)

The determinant $\left| {\,\begin{array}{*{20}{c}}a&b&{a – b}\\b&c&{b – c}\\2&1&0\end{array}\,} \right|$ is equal to zero if $a,b,c$ are in

easy

If the system of linear equations $2x + 2y + 3z = a$ ; $3x – y + 5z = b$ ; $x – 3y + 2z = c$ Where $a, b, c$ are non zero real numbers, has more than one solution, then