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Solutions to various problems related to linear algebra, including the properties of determinants, cramer's rule, linear transformations, and change of basis. The problems involve finding the determinant of a skew-symmetric matrix, solving linear systems using cramer's rule, determining whether a polynomial is in the range of a linear transformation, finding a basis for the range and kernel of a linear transformation, and finding the transition matrix between two ordered bases.
Typology: Exams
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An n × n matrix A is called skew-symmetric if A
t = −A. Show that if A is skew-symmetric and n is an
odd positive integer, then A is not invertible.
By the properties of determinant, det ( A
t ) = det ( −A ) Ñ det ( A ) = det ( −A ) Ñ det ( A ) = ( − 1)
n det ( A ) Ñ
det ( A ) = −det ( A ). So, we get det ( A ) = 0 which implies that A is not invertible. Note that −A means that
EVERY ROW of A is multiplied by -1.
Solve the linear system using Cramer’s Rule:
x 1 + 2 x 2 + 3 x 3 = 6
2 x 2 + 3 x 3 = 5
x 3 = 1
x 1 =
x 2 =
x 3 =
Therefore, (1 , 1 , 1) is the solution of the given system.
For which a ′ s ∈ R the set of polynomials { p 1 , p 2 , p 3 } is linearly independent in P 2 where
p 1 ( x ) = a, p 2 ( x ) = − 2 + ( a − 4) x, p 3 ( x ) = 1 + 2 x + ( a − 1) x
2 .
Consider P 2 with the standard basis. Then p 1 → ( a, 0 , 0), p 2 → ( − 2 , a − 4 , 0), p 3 → (1 , 2 , a − 1). We get
a − 2 1
0 a − 4 2
0 0 a − 1
The set { p 1 , p 2 , p 3 } is linearly independent if and only if the columns of A are linearly independent, i.e.,
det ( A ) ̸ = 0. Assume that det( A ) = 0. Then we derive that a = 0 or a = 4 or a = 1. So the given
polynomials are linearly independent for every value of a except 0 , 1 , 4.
Suppose T : R 6 Ï R 4 is a linear transformation so that the standard matrix of T is
a. Find the range of T.
b. Find the kernel of T.
c. Find dim ( R ( T )).
d. Find dim ( N ( T )).
e. Show that the Rank Theorem holds.
(a) We reduce the given matrix:
Range ( T ) = {c 1 (1 , 2 , 1 , 3) + c 2 (1 , 1 , 0 , 2) : c 1 , c 2 ∈ R }.
(b)
Kernel ( T ) =
x 3
: xi ∈ R , i = 3 , 4 , 5 , 6
(c) dim ( R ( T )) = 2.
(d) dim ( N ( T )) = 4.
(e) dim ( R ( T )) + dim ( N ( T )) = 2 + 4 = 6 = dim (R
6 ).
