If $A = \left| {\,\begin{array}{*{20}{c}}1&1&1\\a&b&c\\{{a^3}}&{{b^3}}&{{c^3}}\end{array}\,} \right|,B = \left| {\,\begin{array}{*{20}{c}}1&1&1\\{{a^2}}&{{b^2}}&{{c^2}}\\{{a^3}}&{{b^3}}&{{c^3}}\end{array}\,} \right|,C = \left| {\,\begin{array}{*{20}{c}}a&b&c\\{{a^2}}&{{b^2}}&{{c^2}}\\{{a^3}}&{{b^3}}&{{c^3}}\end{array}\,} \right|,$ then which relation is correct
$A = B$
$A = C$
$B = C$
None of these
If ${x^a}{y^b} = {e^m},{x^c}{y^d} = {e^n},{\Delta _1} = \left| {\,\begin{array}{*{20}{c}}m&b\\n&d\end{array}\,} \right|\,\,{\Delta _2} = \left| {\,\begin{array}{*{20}{c}}a&m\\c&n\end{array}\,} \right|$ and ${\Delta _3} = \left| {\,\begin{array}{*{20}{c}}a&b\\c&d\end{array}\,} \right|$, then the values of $x$ and $y$ are respectively
If ${\Delta _1} = \left| {\,\begin{array}{*{20}{c}}x&b&b\\a&x&b\\a&a&x\end{array}\,} \right|$ and ${\Delta _2} = \left| {\,\begin{array}{*{20}{c}}x&b\\a&x\end{array}\,} \right|$ are the given determinants, then
If the system of linear equation $x - 4y + 7z = g,\,3y - 5z = h, \,-\,2x + 5y - 9z = k$ is
consistent, then
Statement $-1 :$Determinant of a skew-symmetric matrix of order $3$ is zero
Statement $-2 :$ For any matrix $A,$ $\det \left( {{A^T}} \right) = {\rm{det}}\left( A \right)$ and $\det \left( { - A} \right) = - {\rm{det}}\left( A \right)$ Where $\det \left( A \right) = A$. Then :
Let $\alpha, \beta$ and $\gamma$ be real numbers. consider the following system of linear equations
$x+2 y+z=7$
$x+\alpha z=11$
$2 x-3 y+\beta z=\gamma$
Match each entry in List - $I$ to the correct entries in List-$II$
List - $I$ | List - $II$ |
($P$) If $\beta=\frac{1}{2}(7 \alpha-3)$ and $\gamma=28$, then the system has | ($1$) a unique solution |
($Q$) If $\beta=\frac{1}{2}(7 \alpha-3)$ and $\gamma \neq 28$, then the system has | ($2$) no solution |
($R$) If $\beta \neq \frac{1}{2}(7 \alpha-3)$ where $\alpha=1$ and $\gamma \neq 28$, then the system has |
($3$) infinitely many solutions |
($S$) If $\beta \neq \frac{1}{2}(7 \alpha-3)$ where $\alpha=1$ and $\gamma=28$, then the system has | ($4$) $x=11, y=-2$ and $z=0$ as a solution |
($5$) $x=-15, y=4$ and $z=0$ as a solution |
Then the system has