Question

Find the real numbers \(x\) and \(y\) such that det \(A=0\) if: a. \(A=\left[\begin{array}{lll}0 & x & y \\ y & 0 & x \\ x & y & 0\end{array}\right]\) $$ \text { b. } A=\left[\begin{array}{rrr} 1 & x & x \\ -x & -2 & x \\ -x & -x & -3 \end{array}\right] $$ $$ \begin{array}{l} \text { c. } A=\left[\begin{array}{rrrr} 1 & x & x^{2} & x^{3} \\ x & x^{2} & x^{3} & 1 \\ x^{2} & x^{3} & 1 & x \\ x^{3} & 1 & x & x^{2} \end{array}\right] \\ \text { d. } A=\left[\begin{array}{llll} x & y & 0 & 0 \\ 0 & x & y & 0 \\ 0 & 0 & x & y \\ y & 0 & 0 & x \end{array}\right] \end{array} $$

Short answer

Part a: \(x = -y\); Part b: No real solutions; Part c: \(x = 1\); Part d: \(x = 0\) or \(x = \pm y\).

Step-by-step solution

Find Determinant of Matrix A (Part a)

For matrix A: \[ A = \begin{bmatrix} 0 & x & y \ y & 0 & x \ x & y & 0 \end{bmatrix} \] We will use the cofactor expansion formula to find the determinant. Det(A) is: \[ \det(A) = 0(0 \cdot 0 - x \cdot y) - x(y \cdot 0 - x \cdot x) + y(y \cdot y - x \cdot 0) \] Simplifying this, we get: \[ \det(A) = -x(-x^2) + y(y^2) = x^3 + y^3 = 0 \] This simplifies to \( x^3 + y^3 = 0 \). Since the sum of cubes is zero, \( x = -y \).

Find Determinant of Matrix A (Part b)

For matrix A: \[ A = \begin{bmatrix} 1 & x & x \ -x & -2 & x \ -x & -x & -3 \end{bmatrix} \] Expand along the first row: \[ \det(A) = 1((-2)(-3) - x(-x)) - x(-x(-3) - x(-x)) + x((-x)(-2) - (-x)(-x)) \] Simplifying gives us \[ \det(A) = 1(6 - x^2) + x(3x + x^2) + x(2x - x^2) \] \[ = 6 - x^2 + 3x^2 + 2x - x^2 \] \[ = 6 + 2x^2 + 2x = 0 \] Simplify to find \(2x^2 + 2x + 6 = 0\). Solving this quadratic equation yields no real solutions, so there are no real numbers \((x)\) making the determinant zero.

Find Determinant of Matrix A (Part c)

For matrix A: \[ A = \begin{bmatrix} 1 & x & x^2 & x^3 \ x & x^2 & x^3 & 1 \ x^2 & x^3 & 1 & x \ x^3 & 1 & x & x^2 \end{bmatrix} \] This matrix is a circulant matrix. The determinant is difficult to find directly, but a known result for determinants of such matrices yields: \( \det(A) = (1 - x + x^2 - x^3)^2 + (x^3 - x^2 + x - 1)^2 \). We solve: \(1 - x + x^2 - x^3 = 0 \). Factorize to find \((1 - x)(1 + x + x^2) = 0\). Hence \(x = 1\) or \(x = \frac{-1 \pm \sqrt{-3}}{2}\), which does not exist for reals. So \(x = 1\).

Find Determinant of Matrix A (Part d)

For matrix A: \[ A = \begin{bmatrix} x & y & 0 & 0 \ 0 & x & y & 0 \ 0 & 0 & x & y \ y & 0 & 0 & x \end{bmatrix} \] This is a block diagonal matrix. The determinant of block diagonal matrices is the product of their diagonal blocks (quadratic here): \( \det(P) = x(x \cotimes\ x - y \cotimes\ y)(1) = x[(x^2 - y^2] = 0 \). Thus, either \(x = 0\) or \(x^2 = y^2\), meaning \(x = \pm y\).

Conclusion for All Parts

After analyzing matrices one by one, we derive the following conditions for each: - Part a: \( x = -y \).- Part b: No real solutions.- Part c: \( x = 1 \).- Part d: \( x = 0 \) or \( x = \pm y \).

Key concepts

Determinant Calculation

The determinant of a matrix is a special number that can be calculated from a square matrix. It provides insight into the properties of the matrix, including whether the matrix is invertible. A determinant of zero indicates that the matrix is singular and does not have an inverse.

Determinant calculation involves specific rules that depend on the dimension of the matrix. For a 2x2 matrix, the determinant formula is simple: given a matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the determinant \( \det(A) \) is \( ad - bc \).

For larger matrices, like 3x3 or 4x4, the process becomes more complex and often uses methods like cofactor expansion. This ensures a systematic approach to solve. It is essential to have a keen understanding of these methods to tackle matrices of higher dimensions effectively.

Cofactor Expansion

Cofactor expansion, also known as Laplace's expansion, is a method used to compute the determinant of a matrix by expanding along any row or column. This technique is particularly useful for larger matrices, such as 3x3 or more.

Here's how it works: consider a 3x3 matrix \( A = \begin{bmatrix} a & b & c \ d & e & f \ g & h & i \end{bmatrix} \). To find the determinant, you can expand along the first row:

• Calculate the minor for each element in the row, which is the determinant of the smaller matrix that remains if you remove the row and column containing the element.
• Apply a checkerboard pattern of plus and minus signs (starting with plus in the top left). The formula becomes \( a(ei - fh) - b(di - fg) + c(dh - eg) \).

Cofactor expansion is flexible because you can choose any row or column that appears to simplify the calculations. Picking rows or columns with zeros can make computations easier and faster.

Block Diagonal Matrix

A block diagonal matrix is an interesting concept where the matrix is partitioned into smaller block matrices along its diagonal. These smaller blocks are square matrices, and the off-diagonal blocks are filled with zeros.

For example, consider a block diagonal matrix \( A = \begin{bmatrix} A_1 & 0 \ 0 & A_2 \end{bmatrix} \). To find the determinant of a block diagonal matrix, you simply multiply the determinants of its diagonal blocks. Thus, \( \det(A) = \det(A_1) \cdot \det(A_2) \).

• This property makes calculating determinants of large matrices easier when they are in block form.
• Block diagonal matrices often arise in applications like quantum mechanics and control theory where systems can be decoupled.

Understanding how to work with block diagonal matrices can offer computational advantages and insights into matrix properties.

Circulant Matrix

Circulant matrices are a special type of matrix where each row is a right cyclic shift of the previous row. This unique structure provides interesting properties and simplifies some mathematical operations.

Consider a 4x4 circulant matrix, where each subsequent row is a shift of the previous. The determinant of a circulant matrix does not follow straightforward arithmetic methods, but instead, it utilizes known results or transformation techniques.

In many cases, the eigenvalues of a circulant matrix can be directly computed using the discrete Fourier transform, offering an elegant and efficient way to find its determinant.

• These matrices often appear in various applications, including signal processing and solving differential equations.
• They are highly symmetric and characterized by having predictable patterns, which allow a more analytical approach.

By understanding circulant matrices, one can leverage their patterns to potentially simplify complex computations.