If left| {,begin{array}{*{20}{c}}{{a₁}}&{{b₁}}&{{c₁}}{{a₂}}&{{b₂}}&{{c₂}

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3 and 4 .Determinants and Matrices

medium

If $\left| {\,\begin{array}{{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\\{{a_3}}&{{b_3}}&{{c_3}}\end{array}\,} \right| = 5$; then the value of $\left| {\,\begin{array}{{20}{c}}{{b_2}{c_3} - {b_3}{c_2}}&{{c_2}{a_3} - {c_3}{a_2}}&{{a_2}{b_3} - {a_3}{b_2}}\\{{b_3}{c_1} - {b_1}{c_3}}&{{c_3}{a_1} - {c_1}{a_3}}&{{a_3}{b_1} - {a_1}{b_3}}\\{{b_1}{c_2} - {b_2}{c_1}}&{{c_1}{a_2} - {c_2}{a_1}}&{{a_1}{b_2} - {a_2}{b_1}}\end{array}\,} \right|$is

$5$

$25$

$125$

$0$

Solution

(b) Required determinant
$|adj\,A|$=$|A{|^{3 – 1}}$,$\,{\rm{where}}A = \left| {\,\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\\{{a_3}}&{{b_3}}&{{c_3}}\end{array}\,} \right|$
$ = {5^2} = 25,$ $(\because \,\,|adj\,A| = |a{|^{n – 1}})$

Standard 12

Mathematics

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

normal

How many values of $k $ , systeam of linear equations $\left( {k + 1} \right)x + 8y = 4k\;,\;kx + \left( {k + 3} \right)y$$ = 3k – 1$ has no solutions.

medium

(IIT-2002)

If $A=\left[\begin{array}{lll}1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9\end{array}\right],$ find $|A|$.

easy

If $\left| {\,\begin{array}{*{20}{c}}a&b&{a\alpha – b}\\b&c&{b\alpha – c}\\2&1&0\end{array}\,} \right| = 0$ and $\alpha \ne \frac{1}{2},$ then

medium

Let $\alpha $ and $\beta $ be the roots of the equation $x^2 + x + 1 = 0.$ Then for $y \ne 0$ in $R,$ $\left| {\begin{array}{*{20}{c}}
{y\, + \,1}&\alpha &\beta \\
\alpha &{y\, + \,\beta }&1\\
\beta &1&{y\, + \,\alpha }
\end{array}} \right|$ is equal to

hard

(JEE MAIN-2019)

Solution

Similar Questions

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

How many values of $k $ , systeam of linear equations $\left( {k + 1} \right)x + 8y = 4k\;,\;kx + \left( {k + 3} \right)y$$ = 3k – 1$ has no solutions.

If $A=\left[\begin{array}{lll}1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9\end{array}\right],$ find $|A|$.

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Let $\alpha $ and $\beta $ be the roots of the equation $x^2 + x + 1 = 0.$ Then for $y \ne 0$ in $R,$ $\left| {\begin{array}{*{20}{c}} {y\, + \,1}&\alpha &\beta \\ \alpha &{y\, + \,\beta }&1\\ \beta &1&{y\, + \,\alpha } \end{array}} \right|$ is equal to

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Let $\alpha $ and $\beta $ be the roots of the equation $x^2 + x + 1 = 0.$ Then for $y \ne 0$ in $R,$ $\left| {\begin{array}{*{20}{c}}
{y\, + \,1}&\alpha &\beta \\
\alpha &{y\, + \,\beta }&1\\
\beta &1&{y\, + \,\alpha }
\end{array}} \right|$ is equal to