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Chapter 2: Problem 17

Show that \(A=\left[\begin{array}{rrr}1 & 0 & 2 \\ 2 & -1 & 3 \\ 4 & 1 & 8\end{array}\right]\) and \(B=\left[\begin{array}{rrr}-11 & 2 & 2 \\ -4 & 0 & 1 \\\ 6 & -1 & -1\end{array}\right]\) are inverses.

Short Answer

Expert verified
We have matrices A and B: \[A = \begin{bmatrix} 1 & 0 & 2 \\ 2 & -1 & 3 \\ 4 & 1 & 8 \end{bmatrix},\quad B = \begin{bmatrix} -11 & 2 & 2 \\ -4 & 0 & 1 \\ 6 & -1 & -1 \end{bmatrix}\] Let's multiply them: \(AB=C\), and simplify the product: \[C=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0\end{bmatrix}\] Since the result is not the identity matrix, matrices A and B are not inverses of each other.

Step by step solution

01

Write down matrices A and B

Matrix A is given by: \[A = \begin{bmatrix} 1 & 0 & 2 \\ 2 & -1 & 3 \\ 4 & 1 & 8 \end{bmatrix}\] Matrix B is given by: \[B = \begin{bmatrix} -11 & 2 & 2 \\ -4 & 0 & 1 \\ 6 & -1 & -1 \end{bmatrix}\]
02

Multiply matrices A and B

To multiply two matrices, we take the dot product of each row of the first matrix with each column of the second matrix. If \(A_{i,j}\) are the matrix elements of A, and \(B_{i,j}\) are the matrix elements of B, the product \(AB=C\) has the elements: \[C_{i,j} = \sum_k A_{i, k} B_{k, j}\] Let's calculate the product AB: \[ \begin{bmatrix} (1)(-11) + (0)(-4) + (2)(6) & (1)(2) + (0)(0) + (2)(-1) & (1)(2) + (0)(1) + (2)(-1) \\ (2)(-11) + (-1)(-4) + (3)(6) & (2)(2) + (-1)(0) + (3)(-1) & (2)(2) + (-1)(1) + (3)(-1) \\ (4)(-11) + (1)(-4) + (8)(6) & (4)(2) + (1)(0) + (8)(-1) & (4)(2) + (1)(1) + (8)(-1) \end{bmatrix} \]
03

Simplify the product

Now, we simplify the product matrix: \[ \begin{bmatrix} (-11) + (12) & 2 - 2 & 2 - 2 \\ (-22) + 4 + 18 & 4 - 3 & 4 - 3 \\ (-44) - 4 + 48 & 8 - 8 & 8 - 8 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}\]
04

Check if AB is the identity matrix

Comparing the result to the identity matrix, we see the result fails to be exactly the identity matrix: \[I=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\] The result is not the identity matrix, hence matrices A and B are not inverses of each other.

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Most popular questions from this chapter

Let \(c_{0}, c_{1}, \ldots, c_{n}\) be distinct scalars from an infinite field \(F\). Define \(\mathrm{T}: \mathrm{P}_{n}(F) \rightarrow \mathrm{F}^{n+1}\) by $\mathrm{T}(f)=\left(f\left(c_{0}\right), f\left(c_{1}\right), \ldots, f\left(c_{n}\right)\right)\(. Prove that \)\mathrm{T}$ is an isomorphism. Hint: Use the Lagrange polynomials associated with \(c_{0}, c_{1}, \ldots, c_{n}\).

Determine which of the following matrices are normal: \(A=\left[\begin{array}{cc}3+4 i & 1 \\ i & 2+3 i\end{array}\right]\) and \(B=\left[\begin{array}{cc}1 & 0 \\ 1-i & i\end{array}\right]\)

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.

A differential equation $$ y^{(n)}+a_{n-1} y^{(n-1)}+\cdots+a_{1} y^{(1)}+a_{0} y=x $$ is called a nonhomogeneous linear differential equation with constant coefficients if the \(a_{i}\) 's are constant and \(x\) is a function that is not identically zero.(a) Prove that for any \(x \in{C}^{\infty}\) there exists $y \in{C}^{\infty}\( such that \)y$ is a solution to the differential equation. Hint: Use Lemma 1 to Theorem \(2.32\) to show that for any polynomial \(p(t)\), the linear operator $p(\mathrm{D}): \mathrm{C}^{\infty} \rightarrow \mathrm{C}^{\infty}$ is onto. (b) Let \(V\) be the solution space for the homogeneous linear equation $$ y^{(n)}+a_{n-1} y^{(n-1)}+\cdots+a_{1} y^{(1)}+a_{0} y=0 . $$ Prove that if \(z\) is any solution to the associated nonhomogeneous linear differential equation, then the set of all solutions to the nonhomogeneous linear differential equation is $$ \\{z+y: y \in \mathrm{V}\\} . $$

Let \(A\) be an arbitrary \(2 \times 2\) (real) orthogonal matrix. (a) Prove: If \((a, b)\) is the first row of \(A,\) then \(a^{2}+b^{2}=1\) and $$A=\left[\begin{array}{rr} a & b \\ -b & a \end{array}\right] \quad \text { or } \quad A=\left[\begin{array}{rr} a & b \\ b & -a \end{array}\right].$$ (b) Prove Theorem 2.7: For some real number \(\theta,\) $$A=\left[\begin{array}{rr} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right] \quad \text { or } \quad A=\left[\begin{array}{rr} \cos \theta & \sin \theta \\ \sin \theta & -\cos \theta \end{array}\right].$$
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