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Zorluk: ZorMatrices and Determinants

Given that xx is a positive real number and the determinant of the matrix A=(x102x3014)A = \begin{pmatrix} x & 1 & 0 \\ 2 & x & 3 \\ 0 & 1 & 4 \end{pmatrix} is 22, find the value of xx.

Cevap: 2

Cevap

The positive value of xx is 22.
Expanding the determinant along the first row yields det(A)=x(4x3)1(80)=4x23x8\det(A) = x(4x - 3) - 1(8 - 0) = 4x^2 - 3x - 8. Equating this to 22 gives 4x23x10=04x^2 - 3x - 10 = 0. Factoring the quadratic gives (4x+5)(x2)=0(4x + 5)(x - 2) = 0, which yields x=2x = 2 or x=1.25x = -1.25. Because xx must be positive, the correct value is 22.

Adım Adım Çözüm

1
Expand the 3×33 \times 3 determinant along the first row
\det(A) = x(4x - 3) - 1(8 - 0) + 0 = 4x^2 - 3x - 8
Expanding along the first row leverages the zero entry to simplify calculation of the determinant.
2
Equate the determinant expression to the given determinant value
4x^2 - 3x - 10 = 0
Setting the calculated determinant equal to 22 creates a quadratic equation in terms of xx.
3
Factor the quadratic equation to find the candidate values for xx
(4x + 5)(x - 2) = 0 \implies x = 2 \text{ or } x = -1.25
Factoring determines all algebraic solutions that satisfy the determinant equation.
4
Apply the positivity constraint given in the problem statement
x = 2
The question restricts xx to positive real numbers, discarding the negative root.

Anahtar Kavram

Determinant of a 3x3 matrix and quadratic equations
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