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aguilar (haa832) – Section 3.1 – tsishchanka – (54175) This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001

1. f (−2) = 20

2. E = 16 3. E = 17

10.0 points

Find the value of f (−2) when      3 −3   2  −3  f (x) =  x +  3  −2 3 

1. E = 19

4. E = 18  3  x. −2 

2. f (−2) = 14 3. f (−2) = 18 correct

5. E = 20 correct Explanation: For a 2 × 2 determinant,   a b     c d  = ad − bc . Thus

4. f (−2) = 22

    2 3 2 4     − E = 2 1 4 3 1

5. f (−2) = 16

= 2(8 − 3) − (2 − 12) .

Explanation: For any 2 × 2 determinant   a b   = ad − bc .    c d 

Consequently, E = 20 .

Thus

     3 −3   2  −3  x +  f (x) =   3  −2 3 

 3  x −2 

= ((3) (3) − (−2) (−3)) x2

+ ((−3) (−2) − (3) (3)) x . Consequently, f (x) = 3x2 − 3x , and so

keywords: matrix, determinant 003

10.0 points

By evaluating the determinant, express    1 x x2   f (x) =  0 −4 −2   −1 2 0 

as a quadratic function in x. f (−2) = 18 . 1. f (x) = −4 + 2x + 4x2

keywords: determinant 002

10.0 points

Evaluate the expression     2 3 2 4 . − E = 2  1 4 3 1

2. f (x) = 4 + 2x − 4x2 correct 3. f (x) = −4 + 4x − 2x2 4. f (x) = 4 − 2x − 4x2 5. f (x) = −4 − 2x + 4x2

1

aguilar (haa832) – Section 3.1 – tsishchanka – (54175) 6. f (x) = 4 − 4x + 2x2 Explanation: For this 3 × 3 determinant, use expansion by minors along the top row:        −4 −2   0 −2   0 −4        2 f (x) =  − x+  2 0   −1 0  −1 2  x .

2

keywords: matrix, determinant, quadratic function, expansion by minors 005

10.0 points

Find the value of the determinant   1   D =  −3   1

Evaluating the 2 × 2 determinants, we thus see that f (x) = 4 + 2x − 4x2 .

2 −1 3

 −1    −3  .  −2 

1. D = −1 keywords: matrix, determinant, quadratic function, expansion by minors

2. D = −3 3. D = 3

004

10.0 points

By evaluating the determinant, express    1 x x2   f (x) =  1 −4 0   −4 3 2

4. D = 1 correct 5. D = −5 Explanation: For any 3 × 3 determinant

as a quadratic function in x.

2

1. f (x) = −8 − 13x − 2x

2. f (x) = −8 + 2x − 13x2 3. f (x) = 8 + 2x + 13x2

 A    a1  a 2

4. f (x) = 8 + 13x + 2x2 5. f (x) = −8 − 2x − 13x2 correct 6. f (x) = 8 − 2x + 13x2 Explanation: For this 3 × 3 determinant, use expansion by minors along the top row:        1 −4   −4 0   1 0   2  x+  −  f (x) =   −4 3  x . 3 2   −4 2  Evaluating the 2 × 2 determinants, we thus see that

Thus

B b1 b2

  C  b1    c1  = A    b2 c2 

  a1  −B   a2

 c1   c2 

   a1 c1   + C   a2 c2

   1 2 −1      D =  −3 −1 −3     1 3 −2      −3  −1 −3     =  −2   1  3 −2 

 b1  . b2 

  −3   −3  − −2   1

= (−1)(−2) − (3)(−3)

− 2 ((−3)(−2) − (1)(−3)) f (x) = −8 − 2x − 13x2 .

− ((−3)(3) − (1)(−1)) .

 −1   3 

aguilar (haa832) – Section 3.1 – tsishchanka – (54175) Consequently,

Consequently,

D = −5x + 13y + z .

D = 1 .

keywords: determinant

keywords: determinant 006

007

10.0 points

10.0 points

Find the value of the determinant    3 −1 −2      D =  −1 2 0  .    x y z 

Find the value of the determinant    2 1 −3      D = x y z  .   3 1 2 

1. D = 5x + 13y − z

1. D = 4x + 2y + 5z correct

2. D = 5x + 13y + z

2. D = −4x + 2y + 5z

3. D = −5x − 13y + z

3. D = −4x + 2y − 5z

4. D = −5x + 13y + z correct

4. D = −4x − 2y − 5z

5. D = −5x − 13y − z

5. D = 4x − 2y − 5z

6. D = 5x − 13y − z

6. D = 4x − 2y + 5z

Explanation: For any 3 × 3 determinant     A B C b c    1 1       a1 b1 c1  = A   b2 c2    a b c  2

Thus

2

3

2

2

 a  1 − B  a2

  a c1   1 + C    a2  c2

Explanation: For any 3 × 3 determinant     A B C b c    1 1       a1 b1 c1  = A   b2 c2    a b c 

 b1  . b2 

   2 1 −3      D = x y z    3 1 2        x y y z   x z          −3  − = 2 3 1 1 2   3 2 

= 2 (2yz) − (2x − 3z) − 3 (1x − 3y) .

2

Thus   3   D =  −1   x  2  = 3 y

2

 a  1 −B   a2

  a c1   1 + C  a2  c2

 −1 −2    2 0   y z       −1 0   −1 0    −2 +  x    x z z

 b1  . b2 

 2   y

= 3 (2z + 0y) + (−z + 0x) − 2 (−y − 2x) .

aguilar (haa832) – Section 3.1 – tsishchanka – (54175)

4

Consequently,

Consequently,

D = 7x − 7y − 7z .

D = 4x + 2y + 5z .

009 keywords: determinant 008

10.0 points

Find the value of the determinant    3 x −3      D =  −2 y −1  .    1 z −2 

10.0 points

Find the value of the determinant    1 x −2      D =  3 y 1 .    −2 z −3 

1. D = 5x − 3y − 9z

2. D = 5x − 3y + 9z

1. D = 7x + 7y + 7z

3. D = −5x + 3y + 9z

2. D = 7x + 7y − 7z

4. D = −5x − 3y + 9z correct

3. D = −7x − 7y + 7z

5. D = 5x + 3y − 9z

4. D = −7x + 7y + 7z

6. D = −5x + 3y − 9z Explanation: For any 3 × 3 determinant     A B C    b1 c1       a1 b1 c1  = A      b2 c2  a b c 

5. D = −7x − 7y − 7z 6. D = 7x − 7y − 7z correct Explanation: For any 3 × 3 determinant    A B C    b c  1 1      a1 b1 c1  = A      b2 c2  a b c  2

Thus

2

2

2

 a  1 − B  a2

   1 x −2      D =  3 y 1     −2 z −3      3 y 1     −x =   −2  z −3 

  a c1   1 + C     a2 c2

 b1  . b2 

   3 1   − 2   −2  −3

= (−3y + z) − x (−7) − 2 (3z + 2y) .

 y   z

Thus

2

2

  a1  −B   a2

   a1 c1   + C  a2 c2 

   3 x −3      D =  −2 y −1     1 z −2      y −1   −2    = 3 − x    z −2   1

 b1  . b2 

   −2 −1    − 3  1 −2 

= 3 (−2y − z) − x (5) − 3 (−2z − y) .

Consequently, D = −5x − 3y + 9z .

 y   z

aguilar (haa832) – Section 3.1 – tsishchanka – (54175) 010

10.0 points

011

Find the value of the determinant    2 −1 x     −1 −3 y  D =  .    −3 2 z 

True or False? 1. FALSE correct 2. TRUE Explanation: The determinant of A is the product of the entries on the main diagonal of A only when A is triangular.

2. D = 11x − y − 7z 3. D = 11x − y + 7z

Consequently, the statement is

4. D = 11x + y + 7z

FALSE .

5. D = −11x + y + 7z 6. D = −11x + y − 7z Explanation: For any 3 × 3 determinant   A B C     b c  1 1      a1 b1 c1  = A      b2 c2  a b c 

Thus

2

012

   2 −1 x      D =  −1 −3 y     −3 2 z      −3 y   −1    = 2  +  2 z   −3

  a c1   1 + C   a2 c2 

   −1 y   + x   −3 z

10.0 points

When A and B are n × n matrices, then (det A)(det B) = det(AB) . True or False?

2

 a  1 − B  a2

10.0 points

When A is an n × n matrix, det(A) is the product of the diagonal entries in A.

1. D = −11x − y − 7z correct

2

5

1. TRUE correct  b1  . b2 

 −3   2 

2. FALSE Explanation: This is a standard property of n × n matrices. It can be checked by direct calculation when n = 2 and established in general by expressing a square matrix in terms of elementary matrices and its row reduced echelon form. Consequently, the statement is TRUE .

= 2 (−3z − 2y) + (−z + 3y ) + x (−11) .

Consequently, D = −11x − y − 7z .

keywords: determinant

013

10.0 points

Compute the determinant of the following elementary matrix   1 0 0 0 1 0. 0 k 1

aguilar (haa832) – Section 3.1 – tsishchanka – (54175)

Also since the matrix is triangular, the determinant is the product of the diagonal entries:    1 0 0    0 1 0  = (1)(1)(1) = 1 .    k 0 1

1. 1 − k 2. 0 3. 1 + k 4. 1 correct

015

5. k Explanation: A cofactor expansion along row 1 gives   1 0 0      1 0 0 1 0  = 1      k 1 = 1 0 k 1 

Also since terminant is tries:  1  0  0

6

the matrix is triangular, the dethe product of the diagonal en0 1 k

014

 0  0  = (1)(1)(1) = 1 1 10.0 points

Compute the determinant of the elementary matrix   1 0 0 0 1 0 . k 0 1 1. k − 1 2. 1 correct 3. 0 4. 1 + k 5. k Explanation: A cofactor expansion along row 1 gives    1 0 0       0 1 0  = 1  1 0 = 1 .  0 1    k 0 1

10.0 points

Compute the determinant of the elementary matrix   1 0 0 0 k 0. 0 0 1 1. 0 2. k correct 3. 1 + k 4. 1 5. k − 1 Explanation: A cofactor expansion along row 1 gives    1 0 0       0 k 0  = 1  k 0  = 1 .  0 1   0 0 1 

Also since the matrix is triangular, the determinant is the product of the diagonal entries:    1 0 0    0 k 0  = (1)(k)(1) = 1 .    0 0 1 016

10.0 points

Compute the determinant of the elementary matrix   0 0 1 0 1 0 . 1 0 0 1. 0

aguilar (haa832) – Section 3.1 – tsishchanka – (54175) 2. 2

018

4. −2 5. 1

b d + kb



is obtained from

Explanation: A cofactor expansion along row 1 gives     0 0 1       0 1 0  = 1  0 1  = −1 .   1 0 1 0 0  017

A =



a c

b d



by adding k times row 1 to row 2, then det[B] = det[A] .

10.0 points

When the matrix B =

True or False? 

a b c d



c d a b



is obtained from

1. FALSE 2. TRUE correct

A =



by interchanging rows, then det[B] = det[A] .

Explanation: As 2 × 2 matrices,    a  b  det[B] =  c + ka d + kb 

= a(d + kb) − b(c + ka) = ad − bc ,

True or False?

while

1. TRUE 2. FALSE correct Explanation: As 2 × 2 matrices,    a b   = ad − bc, det[B] =  c d

while

Thus

10.0 points

When the matrix  a B = c + ka

3. −1 correct

7

 c det[A] =  a

Thus

 a  det[A] =  c

det[B] = det[A] . Consequently, the statement is

 d  = bc − ad. b

det[B] = − det[ A] . Consequently, the statement is FALSE .

 b  = ad − bc. d

TRUE .

019

10.0 points

When the matrix  a B = c + ka

b d + kb



aguilar (haa832) – Section 3.1 – tsishchanka – (54175)

8

is obtained from A =



a c

b d



by adding k times row 1 to row 2, then det[B] = det[A] . True or False?

2. FALSE Explanation: Expanding by co-factors along row 1, we see that   ka11 ka 12 ka13 det  a21 a22 a23  = ka11 det[A11] a31 a32 a33 − ka12 det[A12 ] + ka 13 det[A13]

1. FALSE

= k(a11 det[A11] − a12 det[A12 ] + a13 det[A13 ]). 2. TRUE correct Explanation: As 2 × 2 matrices,    a  b  det[B] =  c + ka d + kb

= a(d + kb) − b(c + ka) = ad − bc ,

while

Thus

 a  det[A] =  c

 b  = ad − bc. d

det[B] = det[A] .

On the other hand, det[A] = a11 det[A11 ] − a12 det[A12] + a13 det[A13] . Thus 

ka11 det  a21 a31

ka 12 a22 a32

 ka13 a23  = k det[A] , a33

so the row operation scaling a row also scales the determinant of the new matrix by the same amount. Consequently, the statement is

Consequently, the statement is

TRUE .

TRUE . 021 020

10.0 points

For every 3 × 3 matrix  a11 a12 A =  a21 a22 a31 a32 and scalar k ,  ka11 det  a21 a31

ka 12 a22 a32

True or False? 1. TRUE correct

 a13 a23  a33

 ka13 a23  = k det[A] . a33

10.0 points

For every 3 × 3 matrix  a11 a12 A =  a21 a22 a31 a32 and scalar k ,  ka11 ka 12 det  a21 a22 a31 a32 True or False? 1. TRUE 2. FALSE correct

 a13 a23  a33

 ka13 a23  = k 3 det[A] . a33

aguilar (haa832) – Section 3.1 – tsishchanka – (54175) Explanation: Expanding by co-factors along row 1, we see that   ka11 ka 12 ka13 det  a21 a22 a23  = ka11 det[A11 ] a31 a32 a33

9

2. FALSE correct

Explanation: After expanding by co-factors along the first row, we see that   a12 a13 det[B] = b1 det a32 a33 − ka12 det[A12 ] + ka13 det[A13 ]     a11 a13 a11 a12 = k(a11 det[A11] − a12 det[A12] + a13 det[A13]). − b2 det + b3 det . a31 a33 a31 a32 On the other hand, On the other hand, after expanding by codet[A] = a11 det[A11] factors along the second row, we see that   − a12 det[A12] + a13 det[A13 ] . a12 a13 det[A] = −b1 det a32 a33 Thus       a11 a12 a11 a13 ka11 ka 12 ka13 + b2 det − b3 det . a31 a33 a31 a32 det  a21 a22 a23  = k det[A] , a31 a32 a33 Thus so the row operation scaling a row also scales the determinant of the new matrix by the same amount.

det[B] = − det[A] .

Consequently, the statement is FALSE .

Consequently, the statement is FALSE . 022

023

10.0 points

When the matrix  b1 B =  a11 a31

b2 a12 a32

 b3 a13  a33

a12 b2 a32



is obtained from a11 A =  b1 a31

a13 b3  a33

by interchanging rows 1 and 2, then det[B] = det[A] .

1. TRUE

The determinant of an n × n matrix A can be defined recursively in terms of the determinants of (n − 1) × (n − 1) submatrices of A. True or False? 1. FALSE 2. TRUE correct



True or False?

10.0 points

Explanation: The Minor Mjk of an n × n matrix A is the determinant Mjk = det[Ajk ] of the (n−1) ×(n−1) submatrix Ajk obtained by deleting the j th -row and k th -column from A, while the Co-factor Cjk is defined by Cjk = (−1)j+k det[Ajk ] = (−1)j+k Mjk .

aguilar (haa832) – Section 3.1 – tsishchanka – (54175) The determinant of A can then be defined in terms of Co-factors by expanding either along a row as in det[A] = aj1 Cj1 + aj2 Cj2 + . . . + ajn Cjn , or down a column as in det[A] = a1k C1k + a2k C2k + . . . + ank Cnk . Consequently, the statement is TRUE .

024

10.0 points

Compute the determinant of the matrix   −2 2 2 A =  −6 8 8  −6 8 11 1. det(A) = −16 2. det(A) = −12 correct 3. det(A) = −14 4. det(A) = −15 5. det(A) = −13 Explanation: Expanding by co-factors of the first row we see that   8 8   det(A) = −2  8 11       −6 8   −6 8      −2 + 2 −6 11  −6 8 

= (−2 × (24)) + ((−2) × (−18)) + ((2) × (0)) .

Consequently,

The (j, k)-cofactor of a matrix A is the matrix Ajk obtained by deleting from A its jth row and kth column. True or False? 1. TRUE 2. FALSE correct Explanation: The (j, k)-cofactor of a matrix A is the number Cjk defined by Cjk = (−1)j+k det[Ajk ] . Consequently, the statement is FALSE . 026

10.0 points

For an n × n matrix A the cofactor expansion of det A down the j th -column is the negative of the cofactor expansion along the j th -row. True or False? 1. FALSE correct 2. TRUE Explanation: Taking a cofactor expansion down the j th column gives det[A] = aj1 Cj1 + aj2 Cj2 + ... + ajn Cjn , while the cofactor expansion along the jthrow gives det[A] = a1j C1j + a2j C2j + ... + anj Cnj . Thus the two expansions coincide. Consequently, the statement is

det(A) = −12 .

025

10.0 points

10

FALSE . 027

10.0 points

aguilar (haa832) – Section 3.1 – tsishchanka – (54175) The determinant of a triangular matrix is always the sum of the entries on the main diagonal. True or False? 1. TRUE 2. FALSE correct Explanation: The determinant of a triangular matrix is the product of the entries on the main diagonal. Consequently, the statement is FALSE .

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