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Chapter 6: Problem 1

Find the matrix associated with each of the following quadratic forms: (a) \(3 x^{2}-5 x y+y^{2}\) (b) \(2 x^{2}+3 y^{2}+z^{2}+x y-2 x z+3 y z\) (c) \(x^{2}+2 y^{2}+z^{2}+x y-2 x z+3 y z\)

Short Answer

Expert verified
The associated matrices for each quadratic form are: (a) \[ \begin{bmatrix} 3 & -\frac{5}{2} \\ -\frac{5}{2} & 1 \end{bmatrix} \] (b) \[ \begin{bmatrix} 2 & \frac{1}{2} & -1 \\ \frac{1}{2} & 3 & \frac{3}{2} \\ -1 & \frac{3}{2} & 1 \end{bmatrix} \] (c) \[ \begin{bmatrix} 1 & \frac{1}{2} & -1 \\ \frac{1}{2} & 2 & \frac{3}{2} \\ -1 & \frac{3}{2} & 1 \end{bmatrix} \]

Step by step solution

01

Identifying Matrix Coefficients

For this quadratic form, the coefficients for x^2, y^2, and xy are 3, 1, and -5 respectively.
02

Forming the Symmetric Matrix

The associated matrix for this quadratic form will be a 2x2 symmetric matrix, with the coefficients for x^2 and y^2 as diagonal entries and half the coefficient for xy in the off-diagonal entries: \[ \begin{bmatrix} 3 & -\frac{5}{2} \\ -\frac{5}{2} & 1 \end{bmatrix} \] #(b) 2x^2 + 3y^2 + z^2 + xy - 2xz + 3yz#
03

Identifying Matrix Coefficients

For this quadratic form, the coefficients for x^2, y^2, z^2, xy, xz, and yz are 2, 3, 1, 1, -2, and 3 respectively.
04

Forming the Symmetric Matrix

The associated matrix for this quadratic form will be a 3x3 symmetric matrix, with the coefficients for x^2, y^2, and z^2 as diagonal entries and half the coefficients for xy, xz, and yz in the off-diagonal entries: \[ \begin{bmatrix} 2 & \frac{1}{2} & -1 \\ \frac{1}{2} & 3 & \frac{3}{2} \\ -1 & \frac{3}{2} & 1 \end{bmatrix} \] #(c) x^2 + 2y^2 + z^2 + xy - 2xz + 3yz#
05

Identifying Matrix Coefficients

For this quadratic form, the coefficients for x^2, y^2, z^2, xy, xz, and yz are 1, 2, 1, 1, -2, and 3 respectively.
06

Forming the Symmetric Matrix

The associated matrix for this quadratic form will be a 3x3 symmetric matrix, with the coefficients for x^2, y^2, and z^2 as diagonal entries and half the coefficients for xy, xz, and yz in the off-diagonal entries: \[ \begin{bmatrix} 1 & \frac{1}{2} & -1 \\ \frac{1}{2} & 2 & \frac{3}{2} \\ -1 & \frac{3}{2} & 1 \end{bmatrix} \]

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Most popular questions from this chapter

Show that any \(3 \times 3\) matrix of the form \\[ \left(\begin{array}{lll} a & 1 & 0 \\ 0 & a & 1 \\ 0 & 0 & b \end{array}\right) \\] is defective.

For each of the following, find all possible values of the scalar \(\alpha\) that make the matrix defective or show that no such values exist: (a) \(\left(\begin{array}{lll}1 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & \alpha\end{array}\right)\) (b) \(\left(\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 0 & 0 & \alpha\end{array}\right)\) (c) \(\left(\begin{array}{rrr}1 & 2 & 0 \\ 2 & 1 & 0 \\ 2 & -1 & \alpha\end{array}\right)\) \((\mathbf{d})\left(\begin{array}{rrr}4 & 6 & -2 \\ -1 & -1 & 1 \\ 0 & 0 & \alpha\end{array}\right)\) (e) \(\left(\begin{array}{ccc}3 \alpha & 1 & 0 \\ 0 & \alpha & 0 \\ 0 & 0 & \alpha\end{array}\right)\) (f) \(\left(\begin{array}{ccc}3 \alpha & 0 & 0 \\ 0 & \alpha & 1 \\ 0 & 0 & \alpha\end{array}\right)\) (g) \(\left(\begin{array}{ccc}\alpha+2 & 1 & 0 \\ 0 & \alpha+2 & 0 \\ 0 & 0 & 2 \alpha\end{array}\right)\) (h) \(\left(\begin{array}{ccc}\alpha+2 & 0 & 0 \\ 0 & \alpha+2 & 1 \\ 0 & 0 & 2 \alpha\end{array}\right)\)

Let \(A\) be an \(n \times n\) matrix with an eigenvalue \(\lambda\) of multiplicity \(n .\) Show that \(A\) is diagonalizable if and only if \(A=\lambda I\)

For each of the following, factor the given matrix into a product \(L D L^{T}\), where \(L\) is lower triangular with 1 's on the diagonal and \(D\) is a diagonal matrix: (a) \(\left(\begin{array}{rr}4 & 2 \\ 2 & 10\end{array}\right)\) (b) \(\left(\begin{array}{rr}9 & -3 \\ -3 & 2\end{array}\right)\) \((\mathrm{c})\left(\begin{array}{rrr}16 & 8 & 4 \\ 8 & 6 & 0 \\ 4 & 0 & 7\end{array}\right)\) (d) \(\left(\begin{array}{rrr}9 & 3 & -6 \\ 3 & 4 & 1 \\ -6 & 1 & 9\end{array}\right)\)

Let \(A\) be a diagonalizable \(n \times n\) matrix. Prove that if \(B\) is any matrix that is similar to \(A,\) then \(B\) is diagonalizable.
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