(B) Given that $|A| = 2$,$|B| = 3$,and $|C| = 5$.Using the properties of determinants,we know that $|A^k| = |A|^k$,$|AB| = |A||B|$,and $|A^{-1}| = \fr

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(B) Given that $|A| = 2$,$|B| = 3$,and $|C| = 5$.Using the properties of determinants,we know that $|A^k| = |A|^k$,$|AB| = |A||B|$,and $|A^{-1}| = \frac{1}{|A|}$.Therefore,$\det(A^2BC^{-1}) = |A^2| \cdot |B| \cdot |C^{-1}|$.This simplifies to $|A|^2 \cdot |B| \cdot \frac{1}{|C|}$.Substituting the given values: $(2)^2 \cdot 3 \cdot \frac{1}{5} = 4 \cdot 3 \cdot \frac{1}{5} = \frac{12}{5}$.

Class 12 Mathematics 3-4. Matrices and Determinants Properties of determinants

If $A, B$ and $C$ are $n \times n$ matrices and $\det(A) = 2$,$\det(B) = 3$ and $\det(C) = 5$,then the value of $\det(A^2BC^{-1})$ is equal to

A
$\frac{6}{5}$
B
$\frac{12}{5}$
C
$\frac{18}{5}$
D
$\frac{24}{5}$

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3-4. Matrices and Determinants

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Properties of determinants

If $\Delta=\left|\begin{array}{lll}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2\end{array}\right|$ and $\Delta_1=\left|\begin{array}{ccc}1 & 1 & 1 \\ b c & c a & a b \\ a & b & c\end{array}\right|$,then

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Show that $\left|\begin{array}{ccc}1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c\end{array}\right|=abc\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=abc+bc+ca+ab$.

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Using the property of determinants and without expanding,prove that:
$\left|\begin{array}{lll}2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86\end{array}\right|=0$

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If $a, b$ and $c$ are non-zero numbers,then $\Delta = \left| \begin{array}{ccc} b^2c^2 & bc & b+c \\ c^2a^2 & ca & c+a \\ a^2b^2 & ab & a+b \end{array} \right|$ is equal to

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The value of the determinant $\left| \begin{array}{ccc} 1 & 1 & 1 \\ b+c & c+a & a+b \\ b+c-a & c+a-b & a+b-c \end{array} \right|$ is

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If $A, B$ and $C$ are $n \times n$ matrices and $\det(A) = 2$,$\det(B) = 3$ and $\det(C) = 5$,then the value of $\det(A^2BC^{-1})$ is equal to

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If $\Delta=\left|\begin{array}{lll}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2\end{array}\right|$ and $\Delta_1=\left|\begin{array}{ccc}1 & 1 & 1 \\ b c & c a & a b \\ a & b & c\end{array}\right|$,then

Show that $\left|\begin{array}{ccc}1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c\end{array}\right|=abc\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=abc+bc+ca+ab$.

Using the property of determinants and without expanding,prove that:$\left|\begin{array}{lll}2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86\end{array}\right|=0$

If $a, b$ and $c$ are non-zero numbers,then $\Delta = \left| \begin{array}{ccc} b^2c^2 & bc & b+c \\ c^2a^2 & ca & c+a \\ a^2b^2 & ab & a+b \end{array} \right|$ is equal to

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Using the property of determinants and without expanding,prove that:
$\left|\begin{array}{lll}2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86\end{array}\right|=0$