Solutions to Midterm 2 Practice Problems Written by Victoria Kala vtkala@math.ucsb.edu Last updated //25 Answers This page contains answers only. Detailed solutions are on the following pages. 2 7. (a) A (g) 7 5 cos θ sin θ 4. (b) T (2 ) ( 2 ) sin θ cos θ (c) ker(t ) {} (d) range(t ) span 2. (a) Linear (b) Not linear (c) Linear (d) Not linear 9 4. (a) 8 7 (b) 5 2 (c) Undefined 9 27 42 (d) 4 4 2 4 4 (e) Undefined 8 28 (f) 8 64 5. (a) det(g) (b) G does not exist 6. (a) det(j) 5 (b) J / / /5 /5 /7 /7 7. U is not a subspace 8. See detailed solution 4 5 9. (a) col(k) span 2 2 9 9 8 9 4 5 4 7 (b) null(k) span 4 5
Detailed Solutions. Let T be the linear transformation defined by the formula T (x x 2 ) (x 2 x x + x 2 x x 2 ). (a) Find the standard matrix A for the linear transformation such that T (x) Ax. Solution. Recall that the standard matrix A is given by A ( T (e ) T (e 2 ) ) where e ( ) and e 2 ( ). We need to find T ( ) and T ( ). Using the formula above we see that T ( ) ( ) and T ( ) ( ). Therefore A. Let s check: x 2 x x x 2 x + x 2 x x 2 This is the same as T (x x 2 )! (b) Find the image of (x x 2 ) (2 ). Solution. Use the formula above to find T (2 ) (or you can use matrix multiplication using the matrix in (a)): T (2 ) ( 2 ). (c) Find the kernel of T (Hint: This is the null space of A). Solution. The kernel of T is the set of all vectors x such that T (x). But since T (x) Ax then we solve Ax (this is the null space of A): R2+R R R2+R4 R4 R+R R R+R4 R4 2
R2 R R R This shows x x 2 hence x. This is the only solution to Ax hence ker(t ) null(a) {}. (d) Find the range of T (Hint: This is the column space of A). Solution. The range of T is the column space of A. The column space of A is the span of the columns: col(a) span. This is the solution. However here is another question: is this a basis of the column space? The answer is yes! We found the reduced row echelon form of A to be. The first and second column each have a pivot hence the first and second column of A are linearly independent and span the space. 2. Determine whether T : R 2 R 2 is a linear operator where (a) T (x (2x + y x Solution. Linear operator means linear transformation. For (a) (d) we need to show that: (( ) ( x w x w T + T + T and T ( ( ( x x c ct. ) Let s look at addition: x w 2x + y 2w + 2x + y + 2w + T + T + y x y w x y + w (( ) x w x + w 2(x + w) + (y + ) T + T y + (x + w) (y + ) We see that T (( ) ( x w x w + T + T and so the first property holds.
Let s look at scalar multiplication: ( ) x cx 2cx + cy T c T y cy cx cy (( x 2x + y c(2x + ct c ) x y c(x ( ) x x We see that T c ct and so the second property holds. y y Thus T is a linear transformation. (b) T (x (x + Solution. Let s look at addition: ( x w x + w + x + w + 2 T + T + y y + ( ) x w x + w x + w + T + T y y + y + Therefore T (( ) ( x w x w + T + T and so the first property doesn t hold. This is enough to show that the transformation is not linear. It also doesn t hold for scalar multiplication: ( ( x cx cx + T c T ) cy cy ( x x + cx + c ct c y cy Therefore T ( ( ( x x c ct and so the second property doesn t hold. ) Since neither property holds then T is not a linear transformation. (Note: you only need to show that one property fails so choose whichever one seems easiest to you.) (c) T (x (y Solution. Let s look at addition: ( ( x w y T + T (( ) x w T + T We see that T ( x + y ) w T x + T y ( + ) ) ( x + w y + y + y + y + y + w and so the first property holds. 4
Let s look at scalar multiplication: ( ) x cx cy T c T y cy cy ( ( x y cy ct c cy ( ) x x We see that T c ct and so the second property holds. y y Thus T is a linear transformation. (d) T (x ( x Solution. Let s look at addition: ( ( x w ) ( x ) ( w T + T + ) x + w y y + ( ) ( x w x + w ) x + w T + T y y + y + Therefore T (( ) ( x w x w + T + T and so the first property doesn t hold. This is enough to show that the transformation is not linear. It also doesn t hold for scalar multiplication: ( ( ( x cx ) cx T c T ) cy cy ( x ) ( ) x c x ct c y y c y Therefore T ( ( ( x x c ct and so the second property doesn t hold. ) Since neither property holds then T is not a linear transformation. (Note: you only need to show that one property fails so choose whichever one seems easiest to you.). Consider the matrices A 2 4 B C 2 4 2 D 5 5 2 2 4 Compute the following (where possible). If the operation is not defined explain why. (a) B 2 2B + I. 5
Solution. B is a 2 2 matrix. B 2 will also be a 2 2 matrix 2B will be a 2 2 matrix and I will be a 2 2 matrix therefore B 2 2B + I is defined and B 2 4 4 4 2B + I 2 + 2 2 2 6 6 8 2 + + 4 4 9 4 (b) A T C Solution. A is a 2 matrix and so A T will be a 2 matrix. C is also 2 matrix hence A T C is defined and T A T C 2 4 2 5 4 2 2 5 ( ) 4 2 2 5 8 7 5 2 (c) BD Solution. B is a 2 2 matrix D is a matrix. The number columns of B are not the same as the number rows of D hence BD is not defined. (d) (AC)D Solution. A is a 2 matrix C is a 2 matrix. AC is defined and will be a 6
matrix. D is a matrix and so (AC)D is defined. 5 2 (AC)D 2 4 2 5 2 4 + 2 + 6 + 5 2 + 6 4 + 2 2 + + 4 + 2 + 5 2 4 2 6 5 2 5 2 8 4 5 7 2 4 2 + 8 5 + + 2 6 + 2 + 24 5 + 2 + 24 25 + + 6 2 + 2 4 5 + 2 2 + + 4 8 + 5 + 28 9 27 42 4 4 2 4 4 (e) CB 2A Solution. C is a 2 matrix B is a 2 2 matrix. The number of columns of C is not the same as the number of rows of B so CB is undefined. Therefore CB 2A is not defined. (f) B Solution. B (B ). B will be a 2 2 matrix and will be defined since det(b) 4 2 8 and B 2. 8 4 (B ) will be a 2 2 matrix and ( ) (B ) 2 8 4 2 2 2 8 4 4 4 4 6 2 8 6 4 8 28 8 64 7
(g) CC T Solution. C is a 2 matrix and C T is a 2 matrix. The number of columns of C match the number of rows of C T so CC T is defined and T 4 2 4 2 4 2 4 5 5 5 2 5 + 6 + 4 + 4 + + 4 + 9 + + 25 2 7 7 5 cos θ sin θ 4. Find the inverse of. sin θ cos θ Solution. The inverse is well defined since cos θ sin θ sin θ cos θ cos2 θ + sin 2 θ and cos θ sin θ cos θ sin θ sin θ cos θ sin θ cos θ sin θ Let s check our solution: ( cos θ sin θ cos θ sin θ cos 2 θ + sin 2 θ sin θ cos θ sin θ cos θ sin θ cos θ + sin θ cos θ The other direction works as well. 5 4 5. Let G 4. 6 (a) Find det(g). ( cos θ sin θ cos θ Solution. Let s use cofactor expansion by crossing out the first column: 5 4 4 6 4 6 5 4 6 + ( ) 5 4 4 ( 24) ( 2 + 2) 24 + 24 ). ) sin θ cos θ + sin θ cos θ sin 2 θ + cos 2 θ 8
Thus det(g). (b) Does G exist? If so find it. Solution. No G does not exist since det(g). 6. Let J 5. 5 7 (a) Find det(j). Solution. J is a lower triangular matrix hence the determinant of J is the product of the diagonal entries: det(j) 5 7 5. (b) Does J exist? If so find it. Proof. We set up the augmented matrix ( J I ) and reduce to get ( I J ) : R R R R R2 R R4 R R4 5 5 7 R2 R R2 5 5 7 5 5 7 5 7 5 R R R4 R R4 5 5 7 R4 R2 R4 R2 R2 5 5 7 / / /5 /5 7 5 5 7 / / 5 7 9
5 R R / / /5 /5 /7 /7 Therefore J / / /5 /5. /7 /7 Let s check our solution: / / /5 /5 5 / + / / /5 + /5 /5 + /5 5/5 /7 /7 5 7 /7 + /7 /7 + /7 5/7 + 5/7 The other direction works as well. 7. Let U {(x : x 2 y } be a subset of R 2. Is U a subspace of R 2? Why or why not? Solution. You should always sketch the subset whenever possible. U can easily be sketched out (see the figure below U is the dark shaded set). We need to verify look at the three properties of a subspace to see if U is a subspace: ( ( (i) The ero vector of R 2 is. U since 2 and. ) ) If you are looking at this graphically you can clearly see that the ero vector is in the set U. ( x w (ii) We need to pick two vectors in U and add them together. Let U. Then x( 2 ( w) 2 ( y ) and. When we add the two vectors together we have x w x + w +. The inequalities will also add together and we will have y + x + w 4 and( y + ) 2. Since these inequalities are not preserved (they aren t 2 x + w and ) then U hence U is not closed under addition. y + If you are looking at this graphically you need to find two vectors that when added ( together they are no longer in the set U. One such example would be the vectors ( ) ) 2 2 and which are both in U. But their sum is which is not in U. (These 2 vectors are plotted in the figure below.)
x (iii) We need to pick a vector in U and multiply it by a scalar. Let U and c R. ( y ) cx Then x 2 and y. If we multiply c by our vector we have. If c then our cy inequalities will be cx 2c and cy c; if c < then our inequalities will be cx 2c and cy c. Since these inequalities are not preserved then U is not closed under scalar multiplication. If you are looking at this graphically you need to find a vector and a scalar such that a scalar multiplied by this vector will no ( longer ) be ( in) the set U. One such example would be the vector and the scalar 5: 5 U. (These vectors are plotted in 5 the figure below.) Above we showed that U fails under addition and scalar multiplication therefore U is not a subspace. Showing just one of these fails is enough to show that U is not a subspace. 8. Let v ( 2 ) v 2 (2 9 ) v ( 4). Show that the set S {v v 2 v } is a basis for R. Solution. To show that S is a basis we need to show that (i) all the vectors in S are linearly independent and (ii) the vectors in S span the space which in this case is R. (i) Linearly Independent: We need to take an arbitrary linear combination and set it equal to the ero vector: c v + c 2 v 2 + c v.
We set up an augmented matrix with our vectors as the columns and reduce to solve for c c 2 c : 2 2 9 2R+R2 R2 2 5 R+R R 2 5 4 4 2 2R+R2 R2 2 2 R R 2R2+R R 2 2 This is enough for us to solve the system. This shows that c c 2 c. This proves that the set of vectors in S are linearly independent. (ii) Span: In the previous part we showed that the matrix formed by v v 2 v has a pivot in every row. This implies that these vectors span R. You can also use the fact that a set of n linearly independent vectors spans R n. 9. Let K 4 2 5 4 2 6 9 8 2 2 6 9 9 7 4 2 5 4. (a) Find a basis for the column space of K. Solution. The column space of K is the spanning set of all the columns of K: 4 2 5 4 col(k) span 2 2 6 6 9 9 8 9 2 7 4 2 5 4 However we are asked to find the basis of col(k) which means we only want to find the linearly independent vectors in this set. We will do this by reducing K: 4 2 5 4 4 2 5 4 2 6 9 8 2 2 6 9 9 7 2R+R2 R2 2 6 2 6 9 9 7 4 2 5 4 4 2 5 4 4 2 5 4 4 2 5 4 2R+R R 2 6 R+R R 2 6 4 2 5 4 2
4 2 5 4 4 2 5 4 R2+R R 2 6 5 2R+R2 R2 4 5 4 2 2 4 7 5R+R R 4 5 4R2+R R 4 5 The columns with pivots are the st rd and 5th columns. This means that the st rd and 5th columns of K will form the basis of the column space of K: 4 5 col(k) span 2 2 9 9 8 9 4 5 (b) Find a basis for the null space of K. Solution. We need to find all the vectors x such that Kx. We do this by reducing the augmented matrix ( K ). We already reduced the matrix in part (a): 4 7 4 5 The columns with pivots are associated with x x x 5. The free variables are x 2 x 4 x 6. We set x 2 s x 2 t x u for s t u R. We now solve for x x x 5 in terms of the free variables. From the third row we have: From the second row we have: From the first row we have: x 5 + 5x 6 x 5 5u x + x 4 + 4x 6 x t 4u x x 2 4x 4 7x 6 x s + 4t + 7u We write our solution out in parametric form: x s + 4t + 7u 4 7 x 2 s x x 4 t 4u t s + t + 4 u x 5 5u 5 x 6 u
This is the spanning set for null space of K: null(k) span 4 7 4 5 4