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What vector in \({\mathbb{R}^{\bf{3}}}\) has homogeneous coordinates \(\left( {\frac{1}{2}, - \frac{1}{4},\frac{1}{8},\frac{1}{{24}}} \right)\)?

Short Answer

Expert verified

The vector is \(\left( {12, - 6,3} \right)\).

Step by step solution

01

State the homogeneous coordinate of the vector

Generally, for the vector\(\left( {x,y,z} \right)\), thehomogeneous coordinates are\(\left( {X,Y,Z,H} \right)\), where\(H \ne 0\).

The vector entries can be obtained as shown below:

\(x = \frac{X}{H}\), \(y = \frac{Y}{H}\), and \(z = \frac{Z}{H}\)

02

Obtain the vector that has homogeneous coordinates

Compare thehomogeneous coordinates\(\left( {\frac{1}{2}, - \frac{1}{4},\frac{1}{8},\frac{1}{{24}}} \right)\)with the generalhomogeneous coordinates \(\left( {X,Y,Z,H} \right)\)to get\(X = \frac{1}{2}\),\(Y = - \frac{1}{4}\),\(Z = \frac{1}{8}\), and\(H = \frac{1}{{24}}\).

Now, obtain the vector entries, as shown below:

\(\begin{array}{l}x = \frac{{1/2}}{{1/24}}\\x = 12\end{array}\),

\(\begin{array}{l}y = \frac{{ - 1/4}}{{1/24}}\\y = - 6\end{array}\),

And

\(\begin{array}{l}z = \frac{{1/8}}{{1/24}}\\z = 3\end{array}\)

Thus, the vector is \(\left( {12, - 6,3} \right)\).

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Most popular questions from this chapter

[M] For block operations, it may be necessary to access or enter submatrices of a large matrix. Describe the functions or commands of your matrix program that accomplish the following tasks. Suppose A is a \(20 \times 30\) matrix.

  1. Display the submatrix of Afrom rows 15 to 20 and columns 5 to 10.
  2. Insert a \(5 \times 10\) matrix B into A, beginning at row 10 and column 20.
  3. Create a \(50 \times 50\) matrix of the form \(B = \left[ {\begin{array}{*{20}{c}}A&0\\0&{{A^T}}\end{array}} \right]\).

[Note: It may not be necessary to specify the zero blocks in B.]

Show that \({I_n}A = A\) when \(A\) is \(m \times n\) matrix. (Hint: Use the (column) definition of \({I_n}A\).)

Give a formula for \({\left( {ABx} \right)^T}\), where \({\bf{x}}\) is a vector and \(A\) and \(B\) are matrices of appropriate sizes.

A useful way to test new ideas in matrix algebra, or to make conjectures, is to make calculations with matrices selected at random. Checking a property for a few matrices does not prove that the property holds in general, but it makes the property more believable. Also, if the property is actually false, you may discover this when you make a few calculations.

38. Use at least three pairs of random \(4 \times 4\) matrices Aand Bto test the equalities \({\left( {A + B} \right)^T} = {A^T} + {B^T}\) and \({\left( {AB} \right)^T} = {A^T}{B^T}\). (See Exercise 37.) Report your conclusions. (Note:Most matrix programs use \(A'\) for \({A^{\bf{T}}}\).

In Exercise 9 mark each statement True or False. Justify each answer.

9. a. In order for a matrix B to be the inverse of A, both equations \(AB = I\) and \(BA = I\) must be true.

b. If A and B are \(n \times n\) and invertible, then \({A^{ - {\bf{1}}}}{B^{ - {\bf{1}}}}\) is the inverse of \(AB\).

c. If \(A = \left( {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right)\) and \(ab - cd \ne {\bf{0}}\), then A is invertible.

d. If A is an invertible \(n \times n\) matrix, then the equation \(Ax = b\) is consistent for each b in \({\mathbb{R}^{\bf{n}}}\).

e. Each elementary matrix is invertible.

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