(i)△=\(\begin{vmatrix} x+4 & 2x & 2x\\ 2x & x+4 & 2x \\2x &2x&x+4\end{vmatrix}\)
Applying R1 → R1 + R2 + R3, we have:
△=\(\begin{vmatrix} 5x+4 & 5x+4 & 5x+4\\ 2x & x+4 & 2x \\2x &2x&x+4\end{vmatrix}\)
=(5x+4)\(\begin{vmatrix} 1 & 1 & 1\\ 2x & x+4 & 2x \\2x &2x&x+4\end{vmatrix}\)
Applying C2 → C2 − C1, C3 → C3 − C1, we have:
△=(5x+4)I\(\begin{vmatrix} 1 & 0 & 0\\ 2x & -x+4 & 0 \\2x &0&-x+4\end{vmatrix}\)
=(5x+4)(4-x)(4-x)\(\begin{vmatrix} 1 & 0 & 0\\ 2x & 1 & 0 \\2x &0&1\end{vmatrix}\)
Expanding along C3, we have:
△=(5x+4)(4-x)2\(\begin{vmatrix} 1 & 0 \\ 2x & 1 \end{vmatrix}\)
=(5x+4)(4-x)2
(ii)△=\(\begin{vmatrix} y+k & y & y\\ y & y+k & y \\y &y&y+k\end{vmatrix}\)
Applying R1 → R1 + R2 + R3, we have:
△=\(\begin{vmatrix} 3y+k & 3y+k & 3y+k\\ y & y+k & y \\y &y&y+k\end{vmatrix}\)
=(3y+k)\(\begin{vmatrix}1 & 1 & 1\\ y & y+k & y \\y &y&y+k\end{vmatrix}\)
Applying C2 → C2 − C1 and C3 → C3 − C1, we have:
△=(3y+k)I\(\begin{vmatrix}1 & 0 & 0\\ y & k & 0 \\y &0&k\end{vmatrix}\)
=k2(3y+k)I\(\begin{vmatrix}1 & 0 & 0\\ y & 1 & 0 \\y &0&1\end{vmatrix}\)
Expanding along C3, we have:
△=k2(3y+k)\(\begin{vmatrix} 1 & 0 \\ y & 1 \end{vmatrix}\)=k2(3y+k)
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