If $X, Y$ and $Z$ are positive numbers such that $Y$ and $Z$ have respectively 1 and 0 at their unit's place and $\Delta = \left| {\begin{array}{*{20}{l}} X&4&1 \\ Y&0&1 \\ Z&1&0 \end{array}} \right|$ If $(\Delta+1)$ is divisible by 10 , then $X$ has at its unit's place

JEE MATHEMATICS MCQ: MATRICES

1 Easy

1 of 1

Application +4 | -1

X, Y

and

Z

are positive numbers such that

Y

and

Z

have respectively 1 and 0 at their unit's place and

\Delta = \left| {\begin{array}{llllllllllllllllllll} X&4&1 \\ Y&0&1 \\ Z&1&0 \end{array}} \right|

(\Delta+1)

is divisible by 10 , then

X

has at its unit's place

View answer and explanation

Correct option: C

Option C (2) is correct because the unit's place of X is 2, derived from the divisibility condition.

Practice All Chapter MCQs →

\left|\begin{array}{ccc}a & b-c & b+c \\ c+a & b & c-a \\ a-b & a+b & c\end{array}\right|=0

a, b, c

a x+b y+c=0

{x_2} - {x_3} = 1,\; - {x_1} + 2{x_3} = - 2,\;{x_1} - 2,\;{x_1} - 2{x_2} = 3

2x + 3y + 1 = 0,3x + y - 2 = 0

ax + 2y - b = 0

\left|\begin{array}{ccc}1 & \left(\alpha^{2 x}-\alpha^{-2 x}\right)^2 & \left(\alpha^{2 x}+\alpha^{-2 x}\right)^2 \\ 1 & \left(\beta^{2 x}-\beta^{-2 x}\right)^2 & \left(\beta^{2 x}+\beta^{-2 x}\right)^2 \\ 1 & \left(\gamma^{2 x}-\gamma^{-2 x}\right)^2 & \left(\gamma^{2 x}+\gamma^{-2 x}\right)^2\end{array}\right|

\omega

\left| {\begin{array}{cccccccccccccccccccc} 1&\omega &{{\omega ^2}}\\ \omega &{{\omega ^2}}&1\\ {{\omega ^2}}&1&\omega \end{array}} \right|

A

2 \times 2

A(\operatorname{adj} A)=\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]

|A|