Math 235 Section 1 Solution of Midterm 2 Fall 2000

- (18 points)
You are given below the matrix
*A*together with its row reduced echelon form*B*a) (10 points) Find a spanning set for (i.e., a set of vectors which spans) the null space

*Null*(*A*) of*A*.**Answer:**We use the row reduced echelon form to write, in parametric form, the general solution of (i.e., the general vector in*Null*(*A*)). The free variables are: , , and . Hence, the general solution isand the three column vectors above span

*Null*(*A*).b) (8 points) Are the column spaces and equal?

**Justify**your answer carefully! (either explain why they are equal, or find a vector in one that does not belong to the other).**Answer:**No, the two column spaces are different! For example, the last column of*A*has a non-zero fourth entry, while any vector in*Col*(*B*) is a linear combination of the columns of*B*, and hence, its fourth entry is zero. - (16 points)
Let
be the identity matrix and
.
a) (5 points) Show that the matrix

*A*-5*I*is*invertible*. Note that*A*-5*I*is simply .**Answer**. Since , the matrix*A*-5*I*is invertible.b) (5 points) Express the determinant of the matrix

*A*-*tI*in terms of the scalar parameter*t*.**Answer**.c) (6 points) Show that there are precisely two values of

*t*for which the matrix*A*-*tI*is*not*invertible. Find these values of*t*.**Answer***A*-*tI*is*not*invertible if and only if its determinant vanishes. We find those values of*t*buy solving the quadratic equation . The two solutions are*t*=6 and*t*=-1. - (18 points)
Determine if the following set in is a subspace.
If it is not, find a property in the definition of a subspace which this set
violates.
If it is a subspace, find a matrix
*A*such that this set is either*Null*(*A*) or*Col*(*A*).-
**Answer**This subset of does not contain the zero vector. Hence, it is*not*a subspace. -
**Answer**We can write the general element as a linear combination

Hence, this subset is a subspace, which is equal to the column space of the matrix

**Answer:**This subset of is the general solution of the system of homogeneous linear equations It is hence a subspace which is equal to the Null space of -
- (16 points)
a) (8 points)
Compute the volume of the parallelepiped in with vertices
, , , , , , , where

**Answer:**The volume is the absolute value of the determinant of the matrix*M*with columns and .b) (8 points) Let

*T*be the linear transformation from to sending a vector to , where*A*is the matrix Compute the volume of the parallelepiped obtained as the image of the one in part (a) under the transformation*T*. Express your answer in terms of*a*,*b*and*c*.*Note:*The image is the parallelepiped with vertices , , , , , , , .**Answer:**First Method: where

*M*is the matrix in part (a).Second Method: Calculate directly the determinant.

. Hence the volume is 3|

*c*|. -
a) (6 points)
Let
*A*,*B*, and*C*be matrices satisfying the equationwith

*A*invertible. Solve for*C*in terms of*A*and*B*.**Answer:**Multiply both sides by on the*left*to get . Then multiply both sides by*A*on the*right*to get .b) (10 points) Let Compute its inverse . (

**Check**that .)**Answer:**Augment*A*by the identity matrix and row reduce:Hence,

c) (6 points) Compute the (2,3) entry of if

*A*is given in part (b) and**Answer:**Calculate the product of the second row of A times*B*. Then take the product of the resulting row vector with the third column of . - (10 points)
Let be the vector space of polynomials of degree .
Recall that a vector in is a polynomial
*p*(*t*) of the form where the coefficients are arbitrary real numbers.a) (6 points) Show that the subset

*H*of of polynomials*p*(*t*) of degree which in addition satisfyis a

*subspace*of . (The straightforward answer would include the definition of a subspace and a verification that*H*satisfies all the properties.)**Answer:**We check the three conditions for a subset of the vector space to be a*subspace*:i) The zero vector has value 0 when

*t*=0 and when*t*=1. Hence, the zero vector is in*H*.ii) (

*H*is closed under addition) If*p*and*q*are in*H*, then they satisfy equation (1). Consequently,*p*(0)=0,*p*(1)=0,*q*(0)=0, and*q*(1)=0. We need to check that their sum satisfied equation (1) as well. Indeed:(

*p*+*q*)(0)=*p*(0)+*q*(0)=0+0=0 and (*p*+*q*)(1)=*p*(1)+*q*(1)=0+0=0.iii) (

*H*is closed under scalar multiplication) We need to check that if*p*is in*H*and*c*is a scalar, then*cp*satisfy equation (1) as well. Indeed:and .

b) (4 points) Find a spanning set for

*H*.**Explain**why the set you found spans*H*.**Answer:**We can rewrite equation (1) explicitly in the formWe find the general solution, regarding , and as unknowns: is free and

If we choose , we get the polynomial . Any other polynomial in

*H*is a scalar multiple of this one. Hence,

Mon Nov 20 09:20:19 EST 2000