2,,left| {,begin{array}{*{20}{c}}1&1&1a&b&c{{a^2} - bc}&{{b^2}

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3 and 4 .Determinants and Matrices

hard

$2\,\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\a&b&c\\{{a^2} - bc}&{{b^2} - ac}&{{c^2} - ab}\end{array}\,} \right| = $

$0$

$1$

$2$

$3abc$

Solution

(a) We have $2\,\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\a&b&c\\{{a^2} – bc}&{{b^2} – ac}&{{c^2} – ab}\end{array}\,} \right|$

= $2\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\a&b&c\\{{a^2}}&{{b^2}}&{{c^2}}\end{array}\,} \right| – 2\left| {\,\begin{array}{*{20}{c}}1&1&1\\a&b&c\\{bc}&{ac}&{ab}\end{array}\,} \right|$

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

{ Applying ${C_1}(a),{C_2}(b),{C_3}(c)$}

$ = 2\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\a&b&c\\{{a^2}}&{{b^2}}&{{c^2}}\end{array}\,} \right| – \frac{2}{{abc}}(abc)\,\left| {\,\begin{array}{*{20}{c}}a&b&c\\{{a^2}}&{{b^2}}&{{c^2}}\\1&1&1\end{array}\,} \right| = 0$.

Standard 12

Mathematics

Evaluate $\left|\begin{array}{ccc}102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6\end{array}\right|$

easy

Let the numbers $2, b, c$ be in an $A.P$ and $A = \left[ {\begin{array}{*{20}{c}}
1&1&1 \\
2&b&c \\
4&{{b^2}}&{{c^2}}
\end{array}} \right]$. If $det(A) \in [2,16]$ then $c$ lies in the interval

hard

(JEE MAIN-2019)

The value of $x$ obtained from the equation $\left| {\,\begin{array}{*{20}{c}}{x + \alpha }&\beta &\gamma \\\gamma &{x + \beta }&\alpha \\\alpha &\beta &{x + \gamma }\end{array}\,} \right| = 0$ will be

medium

Suppose $a_1, a_2, …….$ real numbers, with $a_1 \ne 0$. If $a_1, a_2, a_3, ……….$ are in $A.P$. then

normal

The parameter on which the value of the determinant $\left| {\,\begin{array}{*{20}{c}}1&a&{{a^2}}\\{\cos (p – d)x}&{\cos px}&{\cos (p + d)x}\\{\sin (p – d)x}&{\sin px}&{\sin (p + d)x}\end{array}\,} \right|$ does not depend upon

medium

(IIT-1997)