Given the 3 \times 3 matrix A = \begin{pmatrix} x & 2 & 1 \\ 3 & 1 & 2 \\ 1 & x & 0 \end{p… — JAMB UTME

Given the $3 \times 3$ matrix $A = \begin{pmatrix} x & 2 & 1 \\ 3 & 1 & 2 \\ 1 & x & 0 \end{pmatrix}$ , find the positive value of $x$ for which $\det(A) = -2$ .

Answer: 2.5

Answer

The positive value of

x

is 2.5.

Expanding the determinant of matrix

A

along the third row gives

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

. Setting this equal to

-2

yields

2x^2 - 3x - 5 = 0

. Factoring gives

(2x - 5)(x + 1) = 0

, yielding solutions

x = 2.5

and

x = -1

. Taking the positive value gives

x = 2.5

Step-by-Step Solution

Calculate the determinant of matrix

A

in terms of

x

\det(A) = -2x^2 + 3x + 3

Expanding along the third row simplifies computation because of the zero entry.

Set the determinant expression equal to

-2

and rearrange terms

2x^2 - 3x - 5 = 0

Setting

-2x^2 + 3x + 3 = -2

forms a standard quadratic equation.

Factorize the quadratic equation to find the roots

x = 2.5

x = -1

Factoring

(2x - 5)(x + 1) = 0

yields two real solutions.

Filter for the positive value requested in the stem

x = 2.5

The question specifically requires the positive value of

x

Key Concept

Determinant of a 3x3 Matrix and Quadratic Equation Solving

Estimated Time:2m 30s

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