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Zorluk: ZorMatrices and Determinants

Given two matrices A=(x223)A = \begin{pmatrix} x & 2 \\ 2 & 3 \end{pmatrix} and B=(x111)B = \begin{pmatrix} x & 1 \\ -1 & 1 \end{pmatrix}, where x>0x > 0. If the determinant of the product matrix ABAB is 2020, find the value of xx.

Cevap: 3

Cevap

The positive value of xx is 3.
Using the property det(AB)=det(A)det(B)\det(AB) = \det(A) \cdot \det(B), we compute det(A)=3x4\det(A) = 3x - 4 and det(B)=x+1\det(B) = x + 1. Equating their product to 20 gives (3x4)(x+1)=20(3x - 4)(x + 1) = 20, which simplifies to the quadratic 3x2x24=03x^2 - x - 24 = 0. Factorizing yields (3x+8)(x3)=0(3x + 8)(x - 3) = 0. Since xx must be a positive number, the correct value is x=3x = 3.

Adım Adım Çözüm

1
Calculate the determinant of matrix A
\det(A) = 3x - 4
The determinant of a 2x2 matrix \begin{pmatrix} a & b \\ c & d \end{pmatrix} is ad - bc.
2
Calculate the determinant of matrix B
\det(B) = x + 1
Applying ad - bc gives (x)(1) - (1)(-1) = x + 1.
3
Use the determinant product property det(AB) = det(A) * det(B)
(3x - 4)(x + 1) = 20
The determinant of the product of two square matrices equals the product of their individual determinants.
4
Form and solve the quadratic equation
3x^2 - x - 24 = 0, which factorizes into (3x + 8)(x - 3) = 0
Expanding (3x - 4)(x + 1) gives 3x^2 - x - 4. Subtracting 20 yields 3x^2 - x - 24 = 0.
5
Determine the positive solution for x
x = 3
Solving the factors gives x = -8/3 or x = 3. Since x must be positive (x > 0), x = 3.

Anahtar Kavram

Determinant of Matrix Product and 2x2 Determinants
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