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

If $a,b,c$ are different and $\left| {\,\begin{array}{*{20}{c}}a&{{a^2}}&{{a^3} – 1}\\b&{{b^2}}&{{b^3} – 1}\\c&{{c^2}}&{{c^3} – 1}\end{array}\,} \right| = 0$, then

medium

Let $M$ be a $3 \times 3$ invertible matrix with real entries and let $I$ denote the $3 \times 3$ identity matrix. If $M ^{-1}=\operatorname{adj}(\operatorname{adj} M )$, then which of the following statement is/are $ALWAYS TRUE$ ?

$(A)$ $M=I$ $(B)$ $\operatorname{det} M =1$ $(C)$ $M ^2= I$ $(D)$ $(\operatorname{adj} M)^2=I$

medium

(IIT-2020)

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

By using properties of determinants, show that:

$\left|\begin{array}{ccc}y+k & y & y \\ y & y+k & y \\ y & y & y+k\end{array}\right|=k^{2}(3 x+k)$

medium

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

easy

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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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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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

By using properties of determinants, show that:

$\left|\begin{array}{ccc}y+k & y & y \\ y & y+k & y \\ y & y & y+k\end{array}\right|=k^{2}(3 x+k)$