Question

If \( \begin{vmatrix} x+y+2z & x & y z & y+z+2x & y z & x & z+x+2y \end{vmatrix} = k(x+y+z)^3 \), then the value of \( k \) is

• A. 1
• B. 2
• C. 4
• D. 8

Hint:

For symmetric determinants, try adding rows/columns to extract common factors like \( (x+y+z) \).

Answer 1

The Correct Option is B

We need to determine the value of \( k \) such that the determinant of the given matrix equals \( k(x+y+z)^3 \). Let's evaluate the determinant of the matrix: 

\(x+y+2z\)	\(x\)	\(y\)
\(z\)	\(y+z+2x\)	\(y\)
\(z\)	\(x\)	\(z+x+2y\)

The determinant of a 3x3 matrix \(A\) is calculated using:

\(\text{det}(A) = a(ei-fh) - b(di-fg) + c(dh-eg)\)

Let's designate the elements of the matrix:

• \(a = x+y+2z\), \(b = x\), \(c = y\)
• \(d = z\), \(e = y+z+2x\), \(f = y\)
• \(g = z\), \(h = x\), \(i = z+x+2y\)

Substituting the elements into the determinant formula gives:

\(\text{det}(A) = (x+y+2z)[(y)(z+x+2y) - (x)(y)] - (x)[(z)(z+x+2y) - (x)(y)] + (y)[(z)(x) - (z)(y+z+2x)]\)

After simplification, the determinant is found to be equal to \( 2(x+y+z)^3 \).

Thus, the value of \( k \) is \( 2 \).

This means the correct option is:

2