For each of the following statements, write \"true\" if the statement is true, a
ID: 2876210 • Letter: F
Question
Explanation / Answer
a.FALSE The domain is R 5
explaination: The set n is called the domain of A, and m is called the codomain of T.
The notation A: n m says the domain of A is n .
If A is a 3 × 5 matrix and T is a transformation defined by T(x) = Ax, then the domain of T is R 3 .
b.True
explanation:An n x n matrix A is said to be invertible if and only if there is any n x n matrix X such that XA= I and AX= I where I is the n x n identity matrix.
d.true.
Yes, the zero vector is orthogonal to every vector. The only time that two vectors are orthogonal is when their dot product is zero.
e.false
Any system of linear equations can have 0, 1, or infinitely many solutions.
f.FALSE. The converse (every matrix transformation is a linear transformation) is true.
g.False
h. True
If T is a linear transformation, then T(0) = 0 and T(cu + dv) = cT(u) + dT(v) for all scalars c and d. and all vectors u and v in the domain of T.
j.true “the inverse of the transpose of a a matrix is the same as its transpose of the inverse.”
k.true
An n x n matrix (M) is said to be invertible if there is an n x n matrix (C) such that CM= I and MC= I where I is the n x n identity matrix.Basically M-1 M= I and M M-1 = I where M is an invertible matrix and M-1 is the inverse of M.
L.true
The obvious two are a/|a| and -a/|a|, they are both unit and proportional to a. Now assume there is a third unit vector u. Then u=ka and |u| = |k||a|. Since u is also a unit vector, |k||a| = 1 and so |k| = 1/|a| => k = ±1/|v|. Therefore the assumption of a third is a contradiction and thus only the 2 mentioned exist.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.