Given the matrix P = \begin{pmatrix} x & 1 & 2 \\ 3 & 0 & x \\ 1 & 2 & 1 \end{pmatrix}, fo… — JAMB UTME

Given the matrix $P = \begin{pmatrix} x & 1 & 2 \\ 3 & 0 & x \\ 1 & 2 & 1 \end{pmatrix}$ , for what positive value of $x$ is the determinant of $P$ equal to $3$ ?

Answer: 2

Answer

The positive value of

x

2

Expanding the determinant of

P

along the first row yields

\det(P) = -2x^2 + x + 9

. Setting

\det(P) = 3

leads to

-2x^2 + x + 9 = 3

, which rearranges to

2x^2 - x - 6 = 0

. Factoring gives

(2x + 3)(x - 2) = 0

, leading to roots

x = -1.5

and

x = 2

. The positive value is

2

Step-by-Step Solution

Calculate the determinant of matrix

P

in terms of

x

\det(P) = -2x^2 + x + 9

Expanding along the first row gives

x(0 \cdot 1 - 2x) - 1(3 \cdot 1 - x \cdot 1) + 2(3 \cdot 2 - 0 \cdot 1) = -2x^2 + x + 9

Set up the quadratic equation using

\det(P) = 3

2x^2 - x - 6 = 0

Subtracting

3

from both sides gives

-2x^2 + x + 6 = 0

, which simplifies to

2x^2 - x - 6 = 0

Solve the quadratic equation for

x

x = 2 \text{ or } x = -1.5

Factoring

(2x + 3)(x - 2) = 0

yields

x = -\frac{3}{2}

and

x = 2

Select the positive solution as required by the problem statement

x = 2

The problem specifically requests the positive real value of

x

Key Concept

3x3 Determinant Evaluation and Quadratic Solving

Question

Answer

Step-by-Step Solution

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