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MT210 TEST 3
ILKER S. YUCE
MAY 11, 2011
SURNAME, NAME:
QUESTION 1. INTRODUCTION TO DETERMINANTS
Specify whether the matrix has an inverse without trying to compute the inverse
ANSWER
We use the definition of determinant. We calculate the determinant across the 2nd rows and 3rd column.
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
Since we have the determinant is not 0, the matrix is invertible.
QUESTION 2. THE PROPERTIES OF DETERMINANTS
(a) 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.
ANSWER
By the properties of determinant,
det ( A
t ) = det ( −A )
det ( A ) = det ( −A )
det ( A ) = ( − 1)
n det ( A )
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.
(b) Let A =
1 1 a 1
0 a − 1 2 a 1
0 0 ( a − 1)( a
2 − 4) a
0 0 0 a
. Determine those values of a for which A is invertible.
ANSWER
Notice that the given matrix is a triangular matrix. By IMT, A is invertible if and only if det ( A ) =
(1)( a − 1) 2 ( a 2 − 4) a ̸ = 0. Thus, A is invertible if and only if a ̸ = 0 and a ̸ = 1 and a ̸ = 2 and a ̸ = − 2.
QUESTION 4. CRAMER’S RULE, VOLUME, AND LINEAR TRANSFORMATIONS
Find all solutions to the system using Cramer’s Rule.
− 2 x 1 + x 2 − 4 x 3 = − 8
− 4 x 2 + x 3 = 3
4 x 1 − x 3 = − 8
ANSWER
x 1 =
x 2 =
x 3 =
The solution is ( −
91 68
3 34
45 17
QUESTION 5. CRAMER’S RULE, VOLUME, AND LINEAR TRANSFORMATIONS
Find the inverse of the matrix below using the inverse formula
ANSWER
Apply the Inverse Formula(see Example 3) given on page 203 : We have det ( A ) = − 1. We also have
C 11 = ( − 1)
1+ det A 11 = 3 ,
C 21 = ( − 1)
2+ det A 21 = − 6 ,
C 31 = ( − 1)
3+ det A 31 = 3 ,
C 12 = ( − 1)
1+ det A 12 = 7 ,
C 22 = ( − 1)
2+ det A 22 = 1 ,
C 32 = ( − 1)
3+ det A 32 = − 3 ,
C 13 = ( − 1)
1+ det A 13 = − 2 ,
C 23 = ( − 1)
2+ det A 23 = 4 ,
C 33 = ( − 1)
3+ det A 33 = 3_._
As a result, we obtain
A
− 1
