If the determinant left| begin{array}{ccc} a+ | vedclass

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

Question

(B) The property of determinants states that if each element of a column (or row) is the sum of $n$ terms,the determinant can be expressed as the sum

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) The property of determinants states that if each element of a column (or row) is the sum of $n$ terms,the determinant can be expressed as the sum of $n$ determinants.In the given determinant,each column consists of the sum of $2$ terms.Column $1$ has $2$ terms $(a+p, b+q, c+r)$.Column $2$ has $2$ terms $(1+x, m+y, n+z)$.Column $3$ has $2$ terms $(u+f, v+g, w+h)$.Since there are $3$ columns and each column has $2$ terms,the total number of determinants formed is $2 	imes 2 	imes 2 = 2^3 = 8$.Therefore,$K = 8$.

If the determinant $\left| \begin{array}{ccc} a+p & 1+x & u+f \\ b+q & m+y & v+g \\ c+r & n+z & w+h \end{array} \right|$ splits into exactly $K$ determinants of order $3$,each element of which contains only one term,then the value of $K$ is:

Similar Questions

The product of a matrix and its transpose is an identity matrix. The value of the determinant of this matrix is:

$\left|\begin{array}{ccc}x+y & y+z & z+x \\ z & x & y \\ 1 & 1 & 1\end{array}\right|=$ . . . . . . .

For non-zero $a, b, c$,if $\Delta = \begin{vmatrix} 1 + a & 1 & 1 \\ 1 & 1 + b & 1 \\ 1 & 1 & 1 + c \end{vmatrix} = 0$,then the value of $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = $

Without expanding the determinant,prove that $\left|\begin{array}{lll}a & a^{2} & b c \\ b & b^{2} & c a \\ c & c^{2} & a b\end{array}\right|=\left|\begin{array}{lll}1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3}\end{array}\right|$

$\left|\begin{array}{ccc}1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3\end{array}\right|=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module