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Zorluk: OrtaMatrices and Determinants

Given that the matrix A=(x25x3)A = \begin{pmatrix} x & 2 \\ 5 & x - 3 \end{pmatrix} is a singular matrix, what is the positive value of xx?

  1. A
    22
  2. B
    33
  3. 55Cevap
  4. D
    1010

Cevap

The positive value of xx is 55.
For a matrix to be singular, its determinant must be zero. Calculating the determinant of AA yields det(A)=x(x3)(2)(5)=x23x10\det(A) = x(x - 3) - (2)(5) = x^2 - 3x - 10. Setting this to zero and factoring gives (x5)(x+2)=0(x - 5)(x + 2) = 0, leading to x=5x = 5 or x=2x = -2. The positive solution is 55.

Adım Adım Çözüm

1
Set the determinant of the matrix equal to zero.
det(A)=(x)(x3)(2)(5)=0\det(A) = (x)(x - 3) - (2)(5) = 0
By definition, a matrix is singular if and only if its determinant is zero.
2
Expand and simplify the algebraic equation.
x23x10=0x^2 - 3x - 10 = 0
Expanding x(x3)x(x - 3) gives x23xx^2 - 3x, and subtracting 1010 forms a standard quadratic equation.
3
Factor the quadratic equation to solve for xx.
(x5)(x+2)=0    x=5 or x=2(x - 5)(x + 2) = 0 \implies x = 5 \text{ or } x = -2
The factors of 10-10 that sum to 3-3 are 5-5 and +2+2.
4
Select the positive value requested by the question.
x=5x = 5
The question specifically asks for the positive value of xx.

Anahtar Kavram

Singular Matrices and Determinants
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