If Delta = left| {,begin{array}{*{20}{c}}a&b&cx&y&zp&q&rend{array},} right|

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3 and 4 .Determinants and Matrices

easy

If $\Delta = \left| {\,\begin{array}{{20}{c}}a&b&c\\x&y&z\\p&q&r\end{array}\,} \right|$, then $\left| {\,\begin{array}{{20}{c}}{ka}&{kb}&{kc}\\{kx}&{ky}&{kz}\\{kp}&{kq}&{kr}\end{array}\,} \right|$=

$\Delta $

$k\Delta $

$3k\Delta $

${k^3}\Delta $

Solution

(d)$\left| {\,\begin{array}{*{20}{c}}{ka}&{kb}&{kc}\\{kx}&{ky}&{kz}\\{kp}&{kq}&{kr}\end{array}\,} \right|\, = \,{k^3}\,\left| {\,\begin{array}{*{20}{c}}a&b&c\\x&y&z\\p&q&r\end{array}\,} \right| = {k^3}\Delta $.

Standard 12

Mathematics

If $A =$ $\left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]$ (where $bc \ne 0$) satisfies the equations $x^2 + k = 0$, then

normal

Let integers $a , b \in[-3,3]$ be such that $a + b \neq 0$. Then the number of all possible ordered pairs $(a, b)$, for which $\left|\frac{z-a}{z+b}\right|=1$ and $\left|\begin{array}{ccc}z+1 & \omega & \omega^2 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega\end{array}\right|$ $=1, z \in C$, where $\omega$ and $\omega^2$ are the roots of $x^2+x+$ $1=0$, is equal to________

hard

(JEE MAIN-2025)

Without expanding the determinant, prove that

$\left|\begin{array}{lll}a & a^{2} & b c \\ b & b^{2} & c a \\ c & c^{2} & a b\end{array}\right|=\left|\begin{array}{lll}1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3}\end{array}\right|$

medium

By using properties of determinants, show that:

$\left|\begin{array}{ccc}x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|=(5 x+4)(4-x)^{2}$

medium

The value of the determinant $\left| {\,\begin{array}{*{20}{c}}a&{a\, + \,b}&{a\, + \,2b}\\{a\, + \,2b}&a&{a\, + \,b}\\{a\, + \,b}&{a\, + \,2b}&a\end{array}\,} \right|$ is

medium

If $\Delta = \left| {\,\begin{array}{*{20}{c}}a&b&c\\x&y&z\\p&q&r\end{array}\,} \right|$, then $\left| {\,\begin{array}{*{20}{c}}{ka}&{kb}&{kc}\\{kx}&{ky}&{kz}\\{kp}&{kq}&{kr}\end{array}\,} \right|$=

Solution

Similar Questions

If $A =$ $\left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]$ (where $bc \ne 0$) satisfies the equations $x^2 + k = 0$, then

Without expanding the determinant, prove that $\left|\begin{array}{lll}a & a^{2} & b c \\ b & b^{2} & c a \\ c & c^{2} & a b\end{array}\right|=\left|\begin{array}{lll}1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3}\end{array}\right|$

By using properties of determinants, show that: $\left|\begin{array}{ccc}x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|=(5 x+4)(4-x)^{2}$

The value of the determinant $\left| {\,\begin{array}{*{20}{c}}a&{a\, + \,b}&{a\, + \,2b}\\{a\, + \,2b}&a&{a\, + \,b}\\{a\, + \,b}&{a\, + \,2b}&a\end{array}\,} \right|$ is

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If $\Delta = \left| {\,\begin{array}{{20}{c}}a&b&c\\x&y&z\\p&q&r\end{array}\,} \right|$, then $\left| {\,\begin{array}{{20}{c}}{ka}&{kb}&{kc}\\{kx}&{ky}&{kz}\\{kp}&{kq}&{kr}\end{array}\,} \right|$=

Without expanding the determinant, prove that

$\left|\begin{array}{lll}a & a^{2} & b c \\ b & b^{2} & c a \\ c & c^{2} & a b\end{array}\right|=\left|\begin{array}{lll}1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3}\end{array}\right|$

By using properties of determinants, show that:

$\left|\begin{array}{ccc}x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|=(5 x+4)(4-x)^{2}$