Let S=left{A=left(begin{array}{lll}0 & 1 & c 1 & a & d 1 & b & eend{array}ri

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3 and 4 .Determinants and Matrices

medium

Let $S=\left\{A=\left(\begin{array}{lll}0 & 1 & c \\ 1 & a & d \\ 1 & b & e\end{array}\right): a, b, c, d, e \in\{0,1\}\right.$ and $\left.|A| \in\{-1,1\}\right\}$, where $|A|$ denotes the determinant of $A$. Then the number of elements in $S$ is. . . . .

$10$

$15$

$16$

$20$

(IIT-2024)

Solution

$|A|=0(a e-b d)-1(e-d)+c(b-a)$

$=c(b-a)+(d-e)$

$|A| \in\{-1,1\}$ and $a, b, c, d, e \in\{0,1\}$

Case-$I$

$c =0 \quad d =1, e =0, a , b \in(0,1)$

$d =0, e =1$

$a b b d e$

$2 \downarrow$

$2.21 \quad 2 \rightarrow 8 \text { cases }$

Case-$II$

$c =1 \quad b =1, a =0, \quad d =0, e =0, d =1, e =1$

$b =0, a =1, \quad d =0, e =0, d =1, e =1$

$b =0, a =0, \quad d =1, e =0$

$d =1, a =1, \quad d =1, e =0$

$d =0, e =1$

$\rightarrow 8$ cases

$\Rightarrow$ Total $16$ cases

Standard 12

Mathematics

If $\omega $ is cube root of unity, then root of the equation $\left| {\begin{array}{*{20}{c}}
{x + 2}&\omega &{{\omega ^2}} \\
\omega &{x + 1 + {\omega ^2}}&1 \\
{{\omega ^2}}&1&{x + 1 + \omega }
\end{array}} \right| = 0$ is

normal

If the system of equations $\alpha x+y+z=5, x+2 y+$ $3 z=4, x+3 y+5 z=\beta$ has infinitely many solutions, then the ordered pair $(\alpha, \beta)$ is equal to:

medium

(JEE MAIN-2022)

If the following system of linear equations

$2 x+y+z=5$

$x-y+z=3$

$x+y+a z=b$

has no solution, then :

hard

(JEE MAIN-2021)

$\left| {\,\begin{array}{*{20}{c}}1&1&1\\a&b&c\\{{a^3}}&{{b^3}}&{{c^3}}\end{array}\,} \right| = $

medium

Evaluate $\left|\begin{array}{rr}2 & 4 \\ -1 & 2\end{array}\right|$

easy

Let $S=\left\{A=\left(\begin{array}{lll}0 & 1 & c \\ 1 & a & d \\ 1 & b & e\end{array}\right): a, b, c, d, e \in\{0,1\}\right.$ and $\left.|A| \in\{-1,1\}\right\}$, where $|A|$ denotes the determinant of $A$. Then the number of elements in $S$ is. . . . .

Solution

Similar Questions

If $\omega $ is cube root of unity, then root of the equation $\left| {\begin{array}{*{20}{c}} {x + 2}&\omega &{{\omega ^2}} \\ \omega &{x + 1 + {\omega ^2}}&1 \\ {{\omega ^2}}&1&{x + 1 + \omega } \end{array}} \right| = 0$ is

If the system of equations $\alpha x+y+z=5, x+2 y+$ $3 z=4, x+3 y+5 z=\beta$ has infinitely many solutions, then the ordered pair $(\alpha, \beta)$ is equal to:

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If $\omega $ is cube root of unity, then root of the equation $\left| {\begin{array}{*{20}{c}}
{x + 2}&\omega &{{\omega ^2}} \\
\omega &{x + 1 + {\omega ^2}}&1 \\
{{\omega ^2}}&1&{x + 1 + \omega }
\end{array}} \right| = 0$ is

If the following system of linear equations

$2 x+y+z=5$

$x-y+z=3$

$x+y+a z=b$

has no solution, then :