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Chapter 2: Problem 68

Find \(x, y, z\) such that \(A\) is symmetric, where (a) \(A=\left[\begin{array}{lll}2 & x & 3 \\ 4 & 5 & y \\ z & 1 & 7\end{array}\right]\) (b) \(A=\left[\begin{array}{rrr}7 & -6 & 2 x \\ y & z & -2 \\ x & -2 & 5\end{array}\right]\)

Short Answer

Expert verified
For matrix (a), the values of the variables are: \(x=4\), \(y=1\), and \(z=3\). For matrix (b), the values of the variables are: \(x=-2\), \(y=-6\), and \(z=-4\).

Step by step solution

01

Write down the transpose of A

Determine the transpose of A, i.e., \(A^T\). To do this, swap the rows with columns (or columns with rows): \(A^T = \left[\begin{array}{lll}2 & 4 & z \\\ x & 5 & 1 \\\ 3 & y & 7\end{array}\right]\)
02

Equate A and its transpose

In order for A to be symmetric, \(A = A^T\). So, we compare the corresponding elements of A and \(A^T\) to obtain the equations: \(4 = x \implies x = 4\) \(z = 3 \implies z = 3\) \(y = 1 \implies y = 1\) So, the values of the variables are: \(x=4\), \(y=1\), and \(z=3\). (b) \(A=\left[\begin{array}{rrr}7 & -6 & 2 x \\\ y & z & -2 \\\ x & -2 & 5\end{array}\right]\)
03

Write down the transpose of A

Determine the transpose of A, i.e., \(A^T\). To do this, swap the rows with columns (or columns with rows): \(A^T = \left[\begin{array}{rrr}7 & y & x \\\ -6 & z & -2 \\\ 2x & -2 & 5\end{array}\right]\)
04

Equate A and its transpose

In order for A to be symmetric, \(A = A^T\). So, we compare the corresponding elements of A and \(A^T\) to obtain the equations: \(y = -6 \implies y = -6\) \(x = -2 \implies x = -2\) \(2(-2) = z \implies z = -4\) So, the values of the variables are: \(x=-2\), \(y=-6\), and \(z=-4\).

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Most popular questions from this chapter

Let \(A\) be a square matrix. Show that (a) \(A+A^{H}\) is Hermitian, (b) \(A-A^{H}\) is skew-Hermitian, (c) \(\quad A=B+C,\) where \(B\) is Hermitian and \(C\) is skew-Hermitian.

Prove that "is similar to" is an equivalence relation on $\mathrm{M}_{n \times n}(F)$.

Let \(\mathrm{V}=\mathrm{P}_{n}(F)\), and let \(c_{0}, c_{1}, \ldots, c_{n}\) be distinct scalars in \(F\). (a) For \(0 \leq i \leq n\), define \(\mathrm{f}_{i} \in \mathrm{V}^{*}\) by \(\mathrm{f}_{i}(p(x))=p\left(c_{i}\right)\). Prove that \(\left\\{\mathrm{f}_{0}, \mathrm{f}_{1}, \ldots, \mathrm{f}_{n}\right\\}\) is a basis for \(\mathrm{V}^{*}\). Hint: Apply any linear combination of this set that equals the zero transformation to \(p(x)=\) \(\left(x-c_{1}\right)\left(x-c_{2}\right) \cdots\left(x-c_{n}\right)\), and deduce that the first coefficient is zero. (b) Use the corollary to Theorem \(2.26\) and (a) to show that there exist unique polynomials \(p_{0}(x), p_{1}(x), \ldots, p_{n}(x)\) such that \(p_{i}\left(c_{j}\right)=\delta_{i j}\) for \(0 \leq i \leq n\). These polynomials are the Lagrange polynomials defined in Section 1.6. (c) For any scalars \(a_{0}, a_{1}, \ldots, a_{n}\) (not necessarily distinct), deduce that there exists a unique polynomial \(q(x)\) of degree at most \(n\) such that \(q\left(c_{i}\right)=a_{i}\) for \(0 \leq i \leq n .\) In fact, $$ q(x)=\sum_{i=0}^{n} a_{i} p_{i}(x) $$ (d) Deduce the Lagrange interpolation formula: $$ p(x)=\sum_{i=0}^{n} p\left(c_{i}\right) p_{i}(x) $$ for any \(p(x) \in \mathrm{V}\). (e) Prove that $$ \int_{a}^{b} p(t) d t=\sum_{i=0}^{n} p\left(c_{i}\right) d_{i} $$ where $$ d_{i}=\int_{a}^{b} p_{i}(t) d t $$ Suppose now that $$ c_{i}=a+\frac{i(b-a)}{n} \text { for } i=0,1, \ldots, n \text {. } $$ For \(n=1\), the preceding result yields the trapezoidal rule for evaluating the definite integral of a polynomial. For \(n=2\), this result yields Simpson's rule for evaluating the definite integral of a polynomial.

Refer to the following matrices: $$A=\left[\begin{array}{rrr} 1 & -1 & 2 \\ 0 & 3 & 4 \end{array}\right], \quad B=\left[\begin{array}{rrr} 4 & 0 & -3 \\ -1 & -2 & 3 \end{array}\right], \quad C=\left[\begin{array}{rrrr} 2 & -3 & 0 & 1 \\ 5 & -1 & -4 & 2 \\ -1 & 0 & 0 & 3 \end{array}\right], \quad D=\left[\begin{array}{rr} 2 \\ -1 \\ 3 \end{array}\right].$$ Find (a) \(A^{T},\) (b) \(A^{T} B,\) (c) \(A^{T} C.\)

Let \(V\) be a finite-dimensional vector space with the ordered basis \(\beta\). Prove that \(\psi(\beta)=\beta^{* *}\), where \(\psi\) is defined in Theorem \(2.26\).
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