If left| {,begin{array}{*{20}{c}}{ - {a^2}}&{ab}&{ac}{ab}&{

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3 and 4 .Determinants and Matrices

medium

If $\left| {\,\begin{array}{*{20}{c}}{ - {a^2}}&{ab}&{ac}\\{ab}&{ - {b^2}}&{bc}\\{ac}&{bc}&{ - {c^2}}\end{array}\,} \right| = K{a^2}{b^2}{c^2},$ then $K = $

$-4$

$2$

$4$

$8$

Solution

(c) $\left| {\,\begin{array}{*{20}{c}}{ – {a^2}}&{ab}&{ac}\\{ab}&{ – {b^2}}&{bc}\\{ac}&{bc}&{ – {c^2}}\end{array}} \right| = abc\left| {\,\begin{array}{*{20}{c}}{ – a}&b&c\\a&{ – b}&c\\a&b&{ – c}\end{array}\,} \right|$

$ = (abc)(abc)\left| {\,\begin{array}{*{20}{c}}{ – 1}&1&1\\1&{ – 1}&1\\1&1&{ – 1}\end{array}} \right| = {a^2}{b^2}{c^2}( – 1)( – 4)$

$ = 4{a^2}{b^2}{c^2} = K{a^2}{b^2}{c^2}$,

$(given) ==> K = 4.$

Standard 12

Mathematics

The determinant $\left| {\begin{array}{*{20}{c}}{\cos \,\,(\theta \, + \,\phi )}&{ – \,\sin \,\,(\theta \, + \,\phi )}&{\cos \,2\phi }\\{\sin \,\theta }&{\cos \,\theta }&{\sin \,\phi }\\{ – \,\cos \,\theta }&{\sin \,\theta }&{\cos \,\phi }\end{array}} \right|$ is :

normal

Let $S$ be the set of all real values of $k$ for which the system oflinear equations $x +y + z = 2$ ; $2x +y – z = 3$ ; $3x + 2y + kz = 4$ has a unique solution. Then $S$ is

hard

(JEE MAIN-2018)

One of the roots of the given equation $\left| {\,\begin{array}{*{20}{c}}{x + a}&b&c\\b&{x + c}&a\\c&a&{x + b}\end{array}\,} \right| = 0$ is

easy

The values of $\alpha$, for which $\left|\begin{array}{ccc}1 & \frac{3}{2} & \alpha+\frac{3}{2} \\ 1 & \frac{1}{3} & \alpha+\frac{1}{3} \\ 2 \alpha+3 & 3 \alpha+1 & 0\end{array}\right|=0$, lie in the interval

hard

(JEE MAIN-2024)

The value of the determinant$\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&{1 – x}&1\\1&1&{1 + y}\end{array}\,} \right|$is