If A , B and C are angles of a triangle then value of determinant $left| {begin{array}{*{20}{c}} &nb

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3 and 4 .Determinants and Matrices

normal

If $A$, $B$ and $C$ are the angles of a triangle then the value of the determinant

$\left| {\begin{array}{*{20}{c}}
{ - 1 + \cos B}&{\cos C + \cos B}&{\cos B} \\
{\cos C + \cos A}&{ - 1 + \cos A}&{\cos A} \\
{ - 1 + \cos B}&{ - 1 + \cos A}&{ - 1}
\end{array}} \right|$

$1$

$2$

$-1$

$0$

Solution

Given $\left|\begin{array}{ccc}-1+\cos B & \cos C+\cos B & \cos B \\ \cos C+\cos A & -1+\cos A & \cos A \\ -1+\cos B & -1+\cos A & -1\end{array}\right|$

$=\left|\begin{array}{ccc}-1 & \cos C & \cos B \\ \cos C & -1 & \cos A \\ \cos B & \cos A & -1\end{array}\right|\left(\begin{array}{cc}C_{1} \rightarrow C_{1}-C_{3} \\ C_{2} \rightarrow C_{2}-C_{3}\end{array}\right)$

$=\frac{1}{a}\left|\begin{array}{ccc}-a & \cos C & \cos B \\ a \cos C & -1 & \cos A \\ a \cos B & \cos A & -1\end{array}\right|$

$=\frac{1}{a}\left|\begin{array}{ccc}0 & \cos C & \cos B \\ 0 & -1 & \cos A \\ 0 & \cos A & -1\end{array}\right|$

$=0$

Standard 12

Mathematics

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

If ${D_r} = \left| {\begin{array}{*{20}{c}}{{2^{r – 1}}}&{{{2.3}^{r – 1}}}&{{{4.5}^{r – 1}}}\\x&y&z\\{{2^n} – 1}&{{3^n} – 1}&{{5^n} – 1}\end{array}} \right|$, then the value of $\sum\limits_{r = 1}^n {{D_r} = } $

hard

The value of $\left| {\,\begin{array}{*{20}{c}}{265}&{240}&{219}\\{240}&{225}&{198}\\{219}&{198}&{181}\end{array}\,} \right|$ is equal to

easy

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

hard

(IIT-1980)

If $a, b, c,$ are non zero complex numbers satisfying $a^2 + b^2 + c^2 = 0$ and $\left| {\begin{array}{*{20}{c}}
{{b^2} + {c^2}}&{ab}&{ac}\\
{ab}&{{c^2} + {a^2}}&{bc}\\
{ac}&{bc}&{{a^2} + {b^2}}
\end{array}} \right| = k{a^2}{b^2}{c^2},$ then $k$ is equal to

hard

(AIEEE-2012)

If $A$, $B$ and $C$ are the angles of a triangle then the value of the determinant $\left| {\begin{array}{*{20}{c}} { - 1 + \cos B}&{\cos C + \cos B}&{\cos B} \\ {\cos C + \cos A}&{ - 1 + \cos A}&{\cos A} \\ { - 1 + \cos B}&{ - 1 + \cos A}&{ - 1} \end{array}} \right|$

Solution

Similar Questions

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

If ${D_r} = \left| {\begin{array}{*{20}{c}}{{2^{r – 1}}}&{{{2.3}^{r – 1}}}&{{{4.5}^{r – 1}}}\\x&y&z\\{{2^n} – 1}&{{3^n} – 1}&{{5^n} – 1}\end{array}} \right|$, then the value of $\sum\limits_{r = 1}^n {{D_r} = } $

The value of $\left| {\,\begin{array}{*{20}{c}}{265}&{240}&{219}\\{240}&{225}&{198}\\{219}&{198}&{181}\end{array}\,} \right|$ is equal to

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

If $a, b, c,$ are non zero complex numbers satisfying $a^2 + b^2 + c^2 = 0$ and $\left| {\begin{array}{*{20}{c}} {{b^2} + {c^2}}&{ab}&{ac}\\ {ab}&{{c^2} + {a^2}}&{bc}\\ {ac}&{bc}&{{a^2} + {b^2}} \end{array}} \right| = k{a^2}{b^2}{c^2},$ then $k$ is equal to

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If $A$, $B$ and $C$ are the angles of a triangle then the value of the determinant

$\left| {\begin{array}{*{20}{c}}
{ - 1 + \cos B}&{\cos C + \cos B}&{\cos B} \\
{\cos C + \cos A}&{ - 1 + \cos A}&{\cos A} \\
{ - 1 + \cos B}&{ - 1 + \cos A}&{ - 1}
\end{array}} \right|$

If $a, b, c,$ are non zero complex numbers satisfying $a^2 + b^2 + c^2 = 0$ and $\left| {\begin{array}{*{20}{c}}
{{b^2} + {c^2}}&{ab}&{ac}\\
{ab}&{{c^2} + {a^2}}&{bc}\\
{ac}&{bc}&{{a^2} + {b^2}}
\end{array}} \right| = k{a^2}{b^2}{c^2},$ then $k$ is equal to