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Zorluk: Çok zorMatrices and Determinants

Given the matrices A=(x142)A = \begin{pmatrix} x & 1 \\ 4 & 2 \end{pmatrix} and B=(231x)B = \begin{pmatrix} 2 & 3 \\ 1 & x \end{pmatrix}, what is the sum of all real values of xx for which the matrix C=AB−BTC = AB - B^T is singular?

  1. 52\frac{5}{2}Cevap
  2. B
    132\frac{13}{2}
  3. C
    11
  4. D
    −52-\frac{5}{2}

Cevap

The sum of all real values of xx is 52\frac{5}{2}.
Evaluating ABAB yields (2x+14x102x+12)\begin{pmatrix} 2x + 1 & 4x \\ 10 & 2x + 12 \end{pmatrix} and subtracting BT=(213x)B^T = \begin{pmatrix} 2 & 1 \\ 3 & x \end{pmatrix} gives matrix C=(2x−14x−17x+12)C = \begin{pmatrix} 2x - 1 & 4x - 1 \\ 7 & x + 12 \end{pmatrix}. Setting det⁡(C)=(2x−1)(x+12)−7(4x−1)=0\det(C) = (2x - 1)(x + 12) - 7(4x - 1) = 0 results in the quadratic equation 2x2−5x−5=02x^2 - 5x - 5 = 0. Since the discriminant Δ=65>0\Delta = 65 > 0, both roots are real, and their sum is given by −ba=52-\frac{b}{a} = \frac{5}{2}.

Adım Adım Çözüm

1
Compute the matrix product ABAB
AB=(x142)(231x)=(2x+14x102x+12)AB = \begin{pmatrix} x & 1 \\ 4 & 2 \end{pmatrix} \begin{pmatrix} 2 & 3 \\ 1 & x \end{pmatrix} = \begin{pmatrix} 2x + 1 & 4x \\ 10 & 2x + 12 \end{pmatrix}
Matrix multiplication requires taking the dot product of rows of the first matrix with columns of the second matrix.
2
Find the transpose of matrix BB, denoted BTB^T
BT=(213x)B^T = \begin{pmatrix} 2 & 1 \\ 3 & x \end{pmatrix}
The transpose of a matrix is formed by swapping its rows and columns.
3
Compute matrix C=AB−BTC = AB - B^T
C=(2x+14x102x+12)−(213x)=(2x−14x−17x+12)C = \begin{pmatrix} 2x + 1 & 4x \\ 10 & 2x + 12 \end{pmatrix} - \begin{pmatrix} 2 & 1 \\ 3 & x \end{pmatrix} = \begin{pmatrix} 2x - 1 & 4x - 1 \\ 7 & x + 12 \end{pmatrix}
Perform element-wise subtraction of matrix BTB^T from matrix ABAB.
4
Set the determinant of matrix CC to zero for singularity
det⁡(C)=(2x−1)(x+12)−7(4x−1)=2x2+23x−12−(28x−7)=2x2−5x−5=0\det(C) = (2x - 1)(x + 12) - 7(4x - 1) = 2x^2 + 23x - 12 - (28x - 7) = 2x^2 - 5x - 5 = 0
A matrix is singular if and only if its determinant is zero.
5
Find the sum of the real roots of the quadratic equation 2x2−5x−5=02x^2 - 5x - 5 = 0
Discriminant Δ=(−5)2−4(2)(−5)=65>0\Delta = (-5)^2 - 4(2)(-5) = 65 > 0. By Vieta's formula, x1+x2=−ba=−−52=52x_1 + x_2 = -\frac{b}{a} = -\frac{-5}{2} = \frac{5}{2}.
Since the discriminant is positive, two distinct real roots exist, and their sum is given by −ba-\frac{b}{a}.

Anahtar Kavram

Singular matrix condition, matrix multiplication, transpose operations, and Vieta's formulas.
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Konular

Matrices and Determinants

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