Determinant ,left| {,begin{array}{*{20}{c}}1&1&11&2&31&3&6end{array},} right

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3 and 4 .Determinants and Matrices

medium

The determinant $\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&2&3\\1&3&6\end{array}\,} \right|$ is not equal to

$\left| {\,\begin{array}{*{20}{c}}2&1&1\\2&2&3\\2&3&6\end{array}\,} \right|$

$\left| {\,\begin{array}{*{20}{c}}2&1&1\\3&2&3\\4&3&6\end{array}\,} \right|$

$\left| {\begin{array}{*{20}{c}}1&2&1\\1&5&3\\1&9&6\end{array}} \right|$

$\left| {\,\begin{array}{*{20}{c}}3&1&1\\6&2&3\\{10}&3&6\end{array}} \right|\,$

Solution

(a) $\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&2&3\\1&3&6\end{array}\,} \right|\, = \,\left| {\,\begin{array}{*{20}{c}}2&1&1\\3&2&3\\4&3&6\end{array}\,} \right|$ by ${C_1} \to {C_1} + {C_2}$

= $\left| {\,\begin{array}{*{20}{c}}
1&2&1\\
1&5&3\\
1&9&6
\end{array}\,} \right|\,$ by ${C_2} \to {C_2} + {C_3}$

= $\left| {\,\begin{array}{*{20}{c}}3&1&1\\6&2&3\\{10}&3&6\end{array}\,} \right|$, by ${C_1} \to {C_1} + {C_2} + {C_3}$.

But $ \ne \left| {\,\begin{array}{*{20}{c}}2&1&1\\2&2&3\\2&3&6\end{array}\,} \right|$.

Standard 12

Mathematics

In a third order determinant, each element of the first column consists of sum of two terms, each element of the second column consists of sum of three terms and each element of the third column consists of sum of four terms. Then it can be decomposed into $n $determinants, where $ n$ has the value

medium

The roots of the equation $\left| {\,\begin{array}{*{20}{c}}{1 + x}&1&1\\1&{1 + x}&1\\1&1&{1 + x}\end{array}\,} \right| = 0$ are

easy

If the system of linear equations $3 x+y+\beta z=3$, $2 x+\alpha y-z=-3$, $x+2 y+z=4$ has infinitely many solutions, then the value of $22 \beta-9 \alpha$ is :

medium

(JEE MAIN-2025)

If $\omega $ is a cube root of unity and $\Delta = \left| {\begin{array}{*{20}{c}}1&{2\omega }\\\omega &{{\omega ^2}}\end{array}} \right|$, then ${\Delta ^2}$ is equal to

easy

$S$ denote the set of all real values of $\lambda$ such that the system of equations $\lambda x + y + z =1$ ; $x +\lambda y + z =1$ ; $x + y +\lambda z =1$ is inconsistent, then $\sum_{\lambda \in S}\left(|\lambda|^2+|\lambda|\right)$ is equal to

hard

(JEE MAIN-2023)

The determinant $\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&2&3\\1&3&6\end{array}\,} \right|$ is not equal to

Solution

Similar Questions

In a third order determinant, each element of the first column consists of sum of two terms, each element of the second column consists of sum of three terms and each element of the third column consists of sum of four terms. Then it can be decomposed into $n $determinants, where $ n$ has the value

The roots of the equation $\left| {\,\begin{array}{*{20}{c}}{1 + x}&1&1\\1&{1 + x}&1\\1&1&{1 + x}\end{array}\,} \right| = 0$ are

If the system of linear equations $3 x+y+\beta z=3$, $2 x+\alpha y-z=-3$, $x+2 y+z=4$ has infinitely many solutions, then the value of $22 \beta-9 \alpha$ is :

If $\omega $ is a cube root of unity and $\Delta = \left| {\begin{array}{*{20}{c}}1&{2\omega }\\\omega &{{\omega ^2}}\end{array}} \right|$, then ${\Delta ^2}$ is equal to

$S$ denote the set of all real values of $\lambda$ such that the system of equations $\lambda x + y + z =1$ ; $x +\lambda y + z =1$ ; $x + y +\lambda z =1$ is inconsistent, then $\sum_{\lambda \in S}\left(|\lambda|^2+|\lambda|\right)$ is equal to

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