Question

If $a=i-k$ , $b=xi+j+(1-x)k$ and $c=yi+xj+(1+x-y)k$ .Then $\text{ }\\!\\![\\!\\!\text{ a b c }\\!\\!]\\!\\!\text{ }$ depends on:
$\text{1}\text{.Neither x nor y}$
$\text{2}\text{.Both x and y}$
$\text{3}\text{.only x}$
$\text{4}\text{.only y}$

Answer 1

Firstly we will write all the vectors that are given that are the vectors of $\text{a,b}$ and $\text{c}$ . Then we will apply the determinant method and after that we will replace column $3$ by adding column $3$ and column $1$ then we will replace column $3$ by subtracting column $2$ from column $3$ then solve the determinant to check the correct option among them.

Complete step-by-step solution:
The physical quantities are of two types. First which are completely known if their magnitudes are known and second which are completely known only when their magnitude as well as their direction are known. First type of quantities which are known by their magnitude only are called scalars and the second type of quantity which are known by magnitude and direction both are called vectors.
Matrix is an ordered rectangular array of numbers that may be real or complex in horizontal and vertical lines called rows and columns respectively. Matrix plays an important role in various branches of mathematics, electrical engineering, genetic and sociology etc. the word matrix was first used by British mathematician J.J Silverstor in $\text{1850}$ . Another British mathematician Arthur Cayley formulated the general theory of matrix in $\text{1857}$ . Matrix is useful in every branch of science and engineering.
The determinant of a matrix of order three can be determined by expressing it in terms of second order determinants which is known as expansion of a determinant along a row or a column. There are six ways of expanding a determinant of order three corresponding to each of three rows and three columns and each way gives the same value.
Now according to the question:
We have given the vectors of $\text{a,b}$ and $\text{c}$ where:
$a=i-k$
$b=xi+j+(1-x)k$
$c=yi+xj+(1+x-y)k$
Hence, the determinant of $\text{ }\\!\\![\\!\\!\text{ a b c }\\!\\!]\\!\\!\text{ }$ will be:

1 & 0 & -1 \\\ x & 1 & 1-x \\\ y & x & 1+x-y \\\ \end{matrix} \right|$$ Replace $${{C}_{3}}\to {{C}_{3}}+{{C}_{1}}$$ we will get $$~\Rightarrow \text{ }\\!\\![\\!\\!\text{ a b c }\\!\\!]\\!\\!\text{ }=\left| \begin{matrix} 1 & 0 & -1+1 \\\ x & 1 & 1-x+x \\\ y & x & 1+x-y+y \\\ \end{matrix} \right|$$ $$~\Rightarrow \text{ }\\!\\![\\!\\!\text{ a b c }\\!\\!]\\!\\!\text{ }=\left| \begin{matrix} 1 & 0 & 0 \\\ x & 1 & 1 \\\ y & x & 1+x \\\ \end{matrix} \right|$$ Now replacing $${{C}_{3}}\to {{C}_{3}}-{{C}_{2}}$$ we will get $$~\Rightarrow \text{ }\\!\\![\\!\\!\text{ a b c }\\!\\!]\\!\\!\text{ }=\left| \begin{matrix} 1 & 0 & 0-0 \\\ x & 1 & 1-1 \\\ y & x & 1+x-x \\\ \end{matrix} \right|$$ $$~\Rightarrow \text{ }\\!\\![\\!\\!\text{ a b c }\\!\\!]\\!\\!\text{ }=\left| \begin{matrix} 1 & 0 & 0 \\\ x & 1 & 0 \\\ y & x & 1 \\\ \end{matrix} \right|$$ Solving the determinant we will get: $$\Rightarrow \text{ }\\!\\![\\!\\!\text{ a b c }\\!\\!]\\!\\!\text{ }=1[1-x(0)]+0+0$$ So,$$\text{ }\\!\\![\\!\\!\text{ a b c }\\!\\!]\\!\\!\text{ }=1$$ **Hence $$\text{ }\\!\\![\\!\\!\text{ a b c }\\!\\!]\\!\\!\text{ }$$ neither depends on $$\text{x}$$ nor on $$\text{y}$$. Therefore we can say that option $$(1)$$ is correct.** **Note:** We must remember that the determinant of a diagonal matrix is equal to the product of its diagonal elements and the determinant of an identity matrix is always $$1$$ and the determinant of upper triangular matrix and lower triangular matrix is equal to the product of its diagonal elements.