Math Problem Statement
Solution
Let me transcribe the problem and then solve it.
Problem Overview:
We are working with the matrix , given by:
a & i & -i \\ \overline{a} & a & i \\ 1 & i & 1 \end{pmatrix}, \quad a \in \mathbb{C}.$$ The tasks are: #### (a) Prove that for any $$a$$, the determinant of $$A$$ is a real number. #### (b) Find the values of $$a$$ such that $$\det(A) = 0$$ and $$\operatorname{Re}(a) = 0$$. #### (c) Solve the linear system $$A \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 2a \\ 2\overline{a} \\ 0 \end{pmatrix},$$ and determine the general solution. --- ### Step 1: Determinant of the Matrix $$A$$ The determinant of $$A$$ can be computed using the cofactor expansion or symbolic manipulation. I will calculate it in detail. Let me proceed with the calculations. sympy as sp # Define symbols a = sp.symbols('a', complex=True) # 'a' is a complex number conj_a = sp.conjugate(a) # Conjugate of 'a' i = sp.I # Imaginary unit # Define the matrix A A = sp.Matrix([ [a, i, -i], [conj_a, a, i], [1, i, 1] ]) # Compute the determinant of A det_A = A.det() det_AThe determinant of $$A$$ is given by: $$\det(A) = a^2 + a + i a + \overline{a} - i \overline{a} - 1$$ Let's simplify it further. Simplify the determinant expression det_A_simplified = sp.simplify(det_A) det_A_simplifiedThe determinant simplifies to: $$\det(A) = a^2 + a(1 + i) + (1 - i)\overline{a} - 1$$ Now, let us analyze why the determinant is always a real number. Extract the real and imaginary parts of 'a' x, y = sp.symbols('x y', real=True) # Real and imaginary parts of a a_real = x + i * y conj_a_real = sp.conjugate(a_real) # Substitute a and its conjugate det_real = det_A_simplified.subs({a: a_real, sp.conjugate(a): conj_a_real}) det_real_simplified = sp.simplify(det_real) det_real_simplifiedWhen $$a = x + iy$$ (where $$x = \operatorname{Re}(a)$$ and $$y = \operatorname{Im}(a)$$), the determinant becomes: $$\det(A) = x^2 - y^2 + 2x - 2y + 2ixy - 1$$ Now we see the determinant contains a complex part $$2ixy$$. However, we will analyze whether this result can be purely real for specific cases or show the determinant does not hold complex values in simpler forms.We see that $$\det(A)$$ contains an imaginary part, $$2ixy$$, implying that for $$\det(A)$$ to always be real, further conditions on $$a$$ might be actionable plots shows functions better system instructionsAsk a new question for Free
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Math Problem Analysis
Mathematical Concepts
Matrix Algebra
Determinants
Complex Numbers
Linear Systems
Formulas
Determinant formula: det(A) = Σ (±a_ij * det(minor(A)))
Properties of determinants involving conjugates and imaginary numbers
Theorems
Conjugate symmetry of complex numbers
Properties of determinants in complex matrices
Suitable Grade Level
Undergraduate Mathematics
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