Let D₁ = left| {,begin{array}{*{20}{c}}a&b&{a + b}c&d&{c + d}a&b&{a

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3 and 4 .Determinants and Matrices

normal

Let $D_1 =$ $\left| {\,\begin{array}{{20}{c}}a&b&{a + b}\\c&d&{c + d}\\a&b&{a - b}\end{array}\,} \right|$ and $D_2 =$ $\left| {\,\begin{array}{{20}{c}}a&c&{a + c}\\b&d&{b + d}\\a&c&{a + b + c}\end{array}\,} \right|$ then the value of $\frac{{{D_1}}}{{{D_2}}}$ where $b \ne 0$ and $ad \ne bc$, is

$-2$

$0$

$- 2b$

$2b$

Solution

Using $\rightarrow C_3 \rightarrow C_3 – (C_1 + C_2),$ $D_1 =$ $\left|{\,\begin{array}{*{20}{c}}a&b&{a + b}\\c&d&{c + d}\\a&b&{a – b}\end{array}\,} \right|$ and $D_2 = $ $\left| {\,\begin{array}{*{20}{c}}a&c&{a + c}\\b&d&{b + d}\\ a&c&{a + b + c}\end{array}\,} \right|$

$\therefore$ $\frac{{{D_1}}}{{{D_2}}}$ $=$ $\frac{{ – 2b(ad – bc)}}{{b(ad – bc)}}$ $=$ $- 2$

Standard 12

Mathematics

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

medium

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

Using the property of determinants and without expanding, prove that:

$\left|\begin{array}{lll}b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y\end{array}\right|=2\left|\begin{array}{lll}a & p & x \\ b & q & y \\ c & r & z\end{array}\right|$

hard

Prove that $\left|\begin{array}{ccc}a^{2} & b c & a c+c^{2} \\ a^{2}+a b & b^{2} & a c \\ a b & b^{2}+b c & c^{2}\end{array}\right|=4 a^{2} b^{2} c^{2}$

hard

If $f(x) = \left| {\begin{array}{*{20}{c}}{x – 3}&{2{x^2} – 18}&{3{x^3} – 81}\\{x – 5}&{2{x^2} – 50}&{4{x^3} – 500}\\1&2&3\end{array}} \right|$ then $f(1).f(3) + f(3).f(5) + f(5).f(1)$=

hard

Let $D_1 =$ $\left| {\,\begin{array}{*{20}{c}}a&b&{a + b}\\c&d&{c + d}\\a&b&{a - b}\end{array}\,} \right|$ and $D_2 =$ $\left| {\,\begin{array}{*{20}{c}}a&c&{a + c}\\b&d&{b + d}\\a&c&{a + b + c}\end{array}\,} \right|$ then the value of $\frac{{{D_1}}}{{{D_2}}}$ where $b \ne 0$ and $ad \ne bc$, is

Solution

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