The remainder when the determinant left|begin | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) To find the remainder when the determinant $D$ is divided by $5$,we evaluate the entries modulo $5$.$2014 \equiv -1 \pmod{5}$,$2015 \equiv 0 \pmod

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) To find the remainder when the determinant $D$ is divided by $5$,we evaluate the entries modulo $5$.$2014 \equiv -1 \pmod{5}$,$2015 \equiv 0 \pmod{5}$,$2016 \equiv 1 \pmod{5}$,$2017 \equiv 2 \pmod{5}$,$2018 \equiv 3 \equiv -2 \pmod{5}$,$2019 \equiv 4 \equiv -1 \pmod{5}$,$2020 \equiv 0 \pmod{5}$,$2021 \equiv 1 \pmod{5}$,$2022 \equiv 2 \pmod{5}$.Substituting these values into the determinant modulo $5$:$D \equiv \left|\begin{array}{ccc} (-1)^{2014} & 0^{2015} & 1^{2016} \ 2^{2017} & (-2)^{2018} & (-1)^{2019} \ 0^{2020} & 1^{2021} & 2^{2022} \end{array}ight| \pmod{5}$$D \equiv \left|\begin{array}{ccc} 1 & 0 & 1 \ 2^{2017} & 2^{2018} & -1 \ 0 & 1 & 2^{2022} \end{array}ight| \pmod{5}$Expanding along the first row:$D \equiv 1(2^{2018} \cdot 2^{2022} - (-1)(1)) - 0 + 1(2^{2017} \cdot 1 - 0) \pmod{5}$$D \equiv 2^{4040} + 1 + 2^{2017} \pmod{5}$Using Fermat's Little Theorem,$a^4 \equiv 1 \pmod{5}$ for $a$ not divisible by $5$:$2^{4040} = (2^4)^{1010} \equiv 1^{1010} \equiv 1 \pmod{5}$$2^{2017} = (2^4)^{504} \cdot 2^1 \equiv 1^{504} \cdot 2 \equiv 2 \pmod{5}$Thus,$D \equiv 1 + 1 + 2 = 4 \pmod{5}$.The remainder is $4$.

The remainder when the determinant $\left|\begin{array}{lll} 2014^{2014} & 2015^{2015} & 2016^{2016} \\ 2017^{2017} & 2018^{2018} & 2019^{2019} \\ 2020^{2020} & 2021^{2021} & 2022^{2022} \end{array}\right|$ is divided by $5$ is

Similar Questions

The value of $x$ obtained from the equation $\left| \begin{array}{ccc} x + \alpha & \beta & \gamma \\ \gamma & x + \beta & \alpha \\ \alpha & \beta & x + \gamma \end{array} \right| = 0$ is:

Evaluate the determinant: $\left|\begin{array}{ccc}2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0\end{array}\right|$

If $x, y, z$ are in arithmetic progression with common difference $d$,$x \neq 3d$,and the determinant of the matrix $\begin{bmatrix} 3 & 4\sqrt{2} & x \\ 4 & 5\sqrt{2} & y \\ 5 & k & z \end{bmatrix}$ is zero,then the value of $k^2$ is ..... .

$\left|\begin{array}{lll}125 & 5 & 25 \\ 343 & 7 & 49 \\ 729 & 9 & 81\end{array}\right|=$ (in $!$)

If matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$,then $|A|^{-1}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module