△=\(\begin{vmatrix} b+c & q+r & y+z\\ c+a & r+p & z+x\\ a+b&p+q&x+y \end{vmatrix}\)
=\(\begin{vmatrix} b+c & q+r & y+z\\ c+a & r+p & z+x\\ a&p&x \end{vmatrix}\)+\(\begin{vmatrix} b+c & q+r & y+z\\ c+a & r+p & z+x\\ b&q&y \end{vmatrix}\)
=△1+△2(say) ....(1)
Now △1= \(\begin{vmatrix} b+c & q+r & y+z\\ c+a & r+p & z+x\\ a&p&x \end{vmatrix}\)
Applying R2 → R2 − R3, we have:
△1=\(\begin{vmatrix} b+c & q+r & y+z\\ c & r & z\\ a&p&x \end{vmatrix}\)
Applying R1 → R1 − R2, we have:
△1=\(\begin{vmatrix} b & q & y\\ c & r & z\\ a&p&x \end{vmatrix}\)
Applying R1 ↔R3 and R2 ↔R3, we have:
△1=(-1)2\(\begin{vmatrix}a&p&x\\b&q&y\\c&r&z\end{vmatrix}=\begin{vmatrix}a&p&x\\b&q&y\\c&r&z\end{vmatrix}\) ….....(2)
△2=\(\begin{vmatrix} b+c & q+r & y+z\\ c+a & r+p & z+x\\ b&q&y \end{vmatrix}\)
Applying R1 → R1 − R3, we have:
△2=\(\begin{vmatrix} c & r & z\\ c+a & r+p & z+x\\ b&q&y \end{vmatrix}\)
Applying R2 → R2 − R1, we have:
△2=\(\begin{vmatrix} c & r & z\\ a & p & x\\ b&q&y \end{vmatrix}\)
Applying R1 ↔R2 and R2 ↔R3, we have:
△2=(-1)2\(\begin{vmatrix} a & p& x\\ b & q & y\\ c&r&z \end{vmatrix}\)= \(\begin{vmatrix} a & p& x\\ b & q & y\\ c&r&z \end{vmatrix}\) ...(3)
From (1), (2), and (3), we have:
△=2\(\begin{vmatrix} a & p& x\\ b & q & y\\ c&r&z \end{vmatrix}\) Hence, the given result is proved.
Determine whether each of the following relations are reflexive, symmetric, and transitive.
Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.
Evaluate the determinants in Exercises 1 and 2.
\(\begin{vmatrix}2&4\\-5&-1\end{vmatrix}\)
Evaluate the determinants in Exercises 1 and 2.
(i) \(\begin{vmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{vmatrix}\)
(ii) \(\begin{vmatrix}x^2&-x+1&x-1\\& x+1&x+1\end{vmatrix}\)
Using properties of determinants,prove that:
\(\begin{vmatrix} x & x^2 & 1+px^3\\ y & y^2 & 1+py^3\\z&z^2&1+pz^3 \end{vmatrix}\)\(=(1+pxyz)(x-y)(y-z)(z-x)\)
Using properties of determinants,prove that:
\(\begin{vmatrix} 3a& -a+b & -a+c\\ -b+a & 3b & -b+c \\-c+a&-c+b&3c\end{vmatrix}\)\(=3(a+b+c)(ab+bc+ca)\)
Read More: Properties of Determinants