Question

Let \(\begin{vmatrix} x-1&2&1\\2&x-1&2\\1&x+2&x-1 \end{vmatrix}=ax^3+bx^2+cx+d\), where a,b,c and d are constants. Then the value of d is

• A. -8
• B. 6
• C. 0
• D. -6
• E. 16

Answer 1

Answer: (E)

Given determinant:

\(\begin{vmatrix} x-1 & 2 & 1 \\ 2 & x-1 & 2 \\ 1 & x+2 & x-1 \end{vmatrix} = ax^3 + bx^2 + cx + d\)

To find: Find the value of d.

Step 1: Compute the determinant of the 3x3 matrix.

The determinant of the given matrix can be computed using cofactor expansion along the first row:

\(\text{det} = (x-1) \begin{vmatrix} x-1 & 2 \\ x+2 & x-1 \end{vmatrix} - 2 \begin{vmatrix} 2 & 2 \\ 1 & x-1 \end{vmatrix} + 1 \begin{vmatrix} 2 & x-1 \\ 1 & x+2 \end{vmatrix}\)

Step 2: Compute each 2x2 determinant.

• For the first 2x2 matrix: \(\begin{vmatrix} x-1 & 2 \\ x+2 & x-1 \end{vmatrix} = (x-1)(x-1) - (2)(x+2) = x^2 - 2x + 1 - 2x - 4 = x^2 - 4x - 3\)
• For the second 2x2 matrix: \(\begin{vmatrix} 2 & 2 \\ 1 & x-1 \end{vmatrix} = (2)(x-1) - (2)(1) = 2x - 2 - 2 = 2x - 4\)
• For the third 2x2 matrix: \(\begin{vmatrix} 2 & x-1 \\ 1 & x+2 \end{vmatrix} = (2)(x+2) - (x-1)(1) = 2x + 4 - x + 1 = x + 5\)

Step 3: Substitute these values back into the determinant expression.

\(\text{det} = (x-1)(x^2 - 4x - 3) - 2(2x - 4) + 1(x + 5)\)

Step 4: Expand and simplify.

First, expand each term:

\((x-1)(x^2 - 4x - 3) = x(x^2 - 4x - 3) - (x^2 - 4x - 3) = x^3 - 4x^2 - 3x - x^2 + 4x + 3 = x^3 - 5x^2 + x + 3\)

-2(2x - 4) = -4x + 8

1(x + 5) = x + 5

Now combine all terms:

\(\text{det} = x^3 - 5x^2 + x + 3 - 4x + 8 + x + 5\)

Simplify the expression:

\(\text{det} = x^3 - 5x^2 + (x - 4x + x) + (3 + 8 + 5)\)

\(\text{det} = x^3 - 5x^2 - 2x + 16\)

Step 5: Identify the value of d.

The given determinant is equal to \(ax^3 + bx^2 + cx + d\), where \(a = 1\), \(b = -5\), \(c = -2\), and \(d = 16\).

Answer: The value of d is 16.