Question

[tex] \left|\begin{array}{ccc}x&1&y+z\\y&1&z+x\\z&1&x+y\end{array}\right| .........,Select Proper option from the given options.

(a) x+y+z

(b) (x+y)(y+z)(z+x)

(c) 3

(d) 0
[/tex]

Answer 1

$Given : \left[\begin{array}{ccc}x&1&y + z\\y&1&z + x\\z&1&x + y\end{array}\right]$

We know that:

$=> det\left[\begin{array}{ccc}a&b&c\\d&e&f\\g&h&i\end{array}\right] = a.det\left[\begin{array}{ccc}e&f\\h&i\\\end{array}\right] - b.det\left[\begin{array}{ccc}d&f\\g&i\\\end{array}\right] + c.det\left[\begin{array}{ccc}d&e\\g&h\\\end{array}\right]$

$=> x.det\left[\begin{array}{ccc}1&z + x\\1&x + y\\\end{array}\right] - 1.det\left[\begin{array}{ccc}y&z + x\\z&x+y\\\end{array}\right] + (y+z).det\left[\begin{array}{ccc}y&1\\z&1\\\end{array}\right]$

$=> x[1.(x + y) - (z + x) .1] - 1[y.(x + y) - (z + x).z]+(y+z)[y.1 - y.z]$

$=> x[x + y - z - x] - 1[xy + y^2 - z^2 - xz] + (y + z)(y - z)$

$=> x^2 + xy - xz - x^2 - xy - y^2 + z^2 + zx + y^2 - yz + yz - z^2$

$= > 0$

Therefore, the answer is 0 - Option (D).

Hope this helps!

Answer 2

Hello,

Answer : Option D (0)

Solution:

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

Apply elementary column operation:

Apply C1 -> C1 + C3
$\left|\begin{array}{ccc}x+y+z&1&y+z\\x+y+z&1&z+x\\x+y+z&1&x+y\end{array}\right|$

Now as you can seen that (x +y+ z) is common in first Column

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


Now as we know that, if any two rows or columns are identical,then Determinant is zero.

Here you can see that Column 1 = Column 2

So, Determinant is zero.

= (x+y+z) (0)

$\left|\begin{array}{ccc}x&1&y+z\\y&1&z+x\\z&1&x+y\end{array}\right|$ = 0

Hope it helps you