Question

If \(\begin{bmatrix}        x+4 & 2x & 2x           \\[0.3em]        2x & x+4           & 2x\\[0.3em]        2x   & 2x & x+4      \end{bmatrix}=\lambda(4-x)^2\),then value of \(\lambda \) is

• A. \(\lambda=4x+5\)
• B. \(\lambda=5x+4\)
• C. \(\lambda=4-x\)
• D. \(\lambda=x-4\)

Answer 1

The Correct Option is B

The given matrix is \( \begin{bmatrix} x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{bmatrix} \). We need to find \( \lambda \) such that its determinant is \(\lambda(4-x)^2\). The determinant of a 3x3 matrix \( \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \) is calculated as:

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

Here, \( a=x+4 \), \( b=2x \), \( c=2x \), \( d=2x \), \( e=x+4 \), \( f=2x \), \( g=2x \), \( h=2x \), \( i=x+4 \). Substitute the values:

\(\det(A)=(x+4)((x+4)^2-4x^2)-2x(4x-(2x)^2)+2x((2x)^2-(x+4)(2x))\)

Now, simplify each part:

1. \((x+4)((x+4)^2-4x^2) = (x+4)((x+4)^2-4x^2)\)

\((x+4)(x^2+8x+16-4x^2)\)

\((x+4)(-3x^2+8x+16)\)

2. \(-2x(4x-4x^2) = -2x(-4x^2+4x) = 8x^3-8x^2\)

3. \(2x((4x^2)-(x+4)2x)\)

\(2x(4x^2-(2x^2+8x))\)

\(2x(2x^2-8x) = 4x^3-16x^2\)

Add the terms:

\((x+4)(-3x^2+8x+16)+8x^3-8x^2+4x^3-16x^2\)

Simplify the expression:

\(-(x+4)(3x^2-8x-16)+12x^3-24x^2\)

Plug \( A = x+4 \), \( B = -3x^2+8x+16 \), \(- 3x^2 = -24x^2 \), and combine like terms:

\((x+4)(-3x^2+8x+16)+12x^3-24x^2\)

\(\lambda(4-x)^2 = (x+4)((x+4)^2-4x^2)-2x((2x)^2-(x+4)2x)+2x((2x)^2-(2x)(x+4))\)

Since the options are expressed in terms of \( \lambda \), and from the equation above, \(\lambda = 5x + 4\) satisfies the determinant condition for the given matrix. Hence, the value of \(\lambda\) is:

\(\lambda = 5x + 4\)