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- Let T : R^3 ā R^3 be the linear transformation such that
T
ļ£» , T
ļ£» (^) , and T
(a) Write down a matrix A such that T (x) = Ax (10 points).
A =
(b) Find an inverse to A or say why it doesnāt exist. If you canāt figure out part (a), use
the matrix
ļ£». (20 points)
Aā^1 =
(c) Explain why the product Aā^1
ļ£» (^) should equal
ļ£». Check if your inverse matrix
satisfies this property. (10 points)
We have, Aā^1
ļ£» = Aā^1 A
ļ£» = I 3 Ć 3
ļ£». In our particular case, we
just note that (1)(13) + (3)(ā4) + (0)(1) = 1, (1)(18) + (ā6)(3) + (0)(2) = 0 and (1)(ā12) + (3)(4) + (0)(ā1) = 0.
1
- Suppose M is a 3 Ć 4 matrix. If the system of equations corresponding to M x = 0 has two free variables, is it possible that the linear transformation
x ā M x
is surjective (recall surjective means onto) or injective (recall injective means one-to-one)? Why or why not? (15 points)
Since M has two free variables, it has two columns without pivots. Since it has four columns, it must have only 2 pivots. Therefore we note that M doesnāt have a pivot in every column, and so the associated linear transform is NOT injective. We likewise note that M doesnāt have a pivot in every row, and so we know that the associated linear is NOT surjective.
- Suppose that {v 1 , v 2 , v 3 } is a linearly independent set of vectors in R^6. Further suppose that w is a vector in R^6 such that
w is not an element of Span(v 1 , v 2 , v 3 ).
Is it true that {v 1 , v 2 , v 3 , w} is linearly independent? Why or why not? (15 points)
It is linearly independent. Suppose that c 1 v 1 + c 2 v 2 + c 3 v 3 + c 4 w = 0 for some constants ci. We want to show that these constants must all be zero (which will show what we want). We consider two cases. Case 1: c 4 6 = 0. But then we can write w = ā cc 41 v 1 + ā cc 42 v 2 + ā cc 43 v 3 which is impossible since w is not in the span of {v 1 , v 2 , v 3 }. Case 2: c 4 = 0. Then we have that c 1 v 1 + c 2 v 2 + c 3 v 3 = 0 which implies that c 1 , c 2 , and c 3 must be zero as well. Therefore, the only possibility is that all the ci are zero which proves what we wanted.
(EC) Suppose that A is an invertible 6 Ć 6 matrices and B is an invertible 4 Ć 4 matrix. If
C =
[
06 Ć 4 A
B 04 Ć 6
]
then find Cā^1 as a partitioned matrix in terms of Aā^1 and Bā^1. Explicitly state the sizes of the various pieces of the partitioned Cā^1. Here 0 6 Ć 4 means the matrix with all entries zero and with 6 rows and 4 columns. (5 points)
First we check that the matrix C really is square. Since A has 6 columns, we see that C has 6 + 4 = 10 columns. Since B has 4 rows, we see that C has 4 + 6 = 10 rows. Therefore C is square. So break up Cā^1 into
D = Cā^1 =
[
D 11 D 12
D 21 D 22
]
where D 11 is 4 Ć 6, D 12 is 4 Ć 4, D 21 is 6 Ć 6 and D 22 is 6 Ć 4. We know that
CD = I 10 Ć 10 =
[
I 6 Ć 6 06 Ć 4
04 Ć 6 I 4 Ć 4
]
Therefore we have the following four equations,
(1) (^06) Ć 4 D 11 + AD 21 = I 6 Ć 6
(2) BD 11 + 0 4 Ć 6 D 21 = 0 4 Ć 6
(3) (^06) Ć 4 D 12 + AD 22 = 0 6 Ć 4
(4) BD 12 + 0 4 Ć 6 D 22 = I 4 Ć 4
We first consider equation (2). This is just BD 11 = 0 4 Ć 6. Since B is invertible, we see that D 11 = Bā^104 Ć 6 = 0 4 Ć 6. Likewise when considering equation (3), we see that AD 22 = 0 6 Ć 4 , and therefore, since A is invertible, D 22 = Aā^106 Ć 4 = 0 6 Ć 4. Equation (1) says that AD 21 = I 6 Ć 6 and so D 21 = Aā^1 since D 21 is square. Likewise, equation (4) says that BD 12 = I 4 Ć 4 and so D 12 = Bā^1 since D 12 is square. In summary, D =
[
04 Ć 6 Bā^1
Aā^1 06 Ć 4
]
A quick check verifies that D is indeed is the inverse to C.
