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14.2.11: Chapter 11

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    118265
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    Try It

    11.1 Systems of Linear Equations: Two Variables

    1.

    Not a solution.

    2.

    The solution to the system is the ordered pair ( −5,3 ). ( −5,3 ).

    5b7a7cb6c6469c2b48ae39ba3ba7b010e4f9056f
    3.

    ( −2,−5 ) ( −2,−5 )

    4.

    ( −6,−2 ) ( −6,−2 )

    5.

    ( 10,−4 ) ( 10,−4 )

    6.

    No solution. It is an inconsistent system.

    7.

    The system is dependent so there are infinite solutions of the form (x,2x+5). (x,2x+5).

    8.

    700 children, 950 adults

    11.2 Systems of Linear Equations: Three Variables

    1.

    ( 1,−1,1 ) ( 1,−1,1 )

    2.

    No solution.

    3.

    Infinite number of solutions of the form ( x,4x−11,−5x+18 ). ( x,4x−11,−5x+18 ).

    11.3 Systems of Nonlinear Equations and Inequalities: Two Variables

    1.

    ( 1 2 , 1 2 ) ( 1 2 , 1 2 ) and ( 2,8 ) ( 2,8 )

    2.

    ( −1,3 ) ( −1,3 )

    3.

    { ( 1,3 ),( 1,−3 ),( −1,3 ),( −1,−3 ) } { ( 1,3 ),( 1,−3 ),( −1,3 ),( −1,−3 ) }

    4.
    4d081141772acbd53842830734046cdb4aa6e793

    11.4 Partial Fractions

    1.

    3 x−3 2 x−2 3 x−3 2 x−2

    2.

    6 x−1 5 ( x−1 ) 2 6 x−1 5 ( x−1 ) 2

    3.

    3 x−1 + 2x−4 x 2 +1 3 x−1 + 2x−4 x 2 +1

    4.

    x−2 x 2 −2x+3 + 2x+1 ( x 2 −2x+3 ) 2 x−2 x 2 −2x+3 + 2x+1 ( x 2 −2x+3 ) 2

    11.5 Matrices and Matrix Operations

    1.

    A+B=[ 2 1 1 6 0 −3 ]+[ 3 1 −4 −2 5 3 ] A+B=[ 2 1 1 6 0 −3 ]+[ 3 1 −4 −2 5 3 ]
    =[ 2+3 1+1 1+(−4) 6+(−2) 0+5 −3+3 ]=[ 5 2 −3 4 5 0 ] = [ 2+3 1+1 1+(−4) 6+(−2) 0+5 −3+3 ] = [ 5 2 −3 4 5 0 ]

    2.

    −2B=[ −8 −2 −6 −4 ] −2B=[ −8 −2 −6 −4 ]

    11.6 Solving Systems with Gaussian Elimination

    1.

    [ 4 −3 3 2 | 11 4 ] [ 4 −3 3 2 | 11 4 ]

    2.

    xy+z=5 2xy+3z=1 y+z=−9 xy+z=5 2xy+3z=1 y+z=−9

    3.

    ( 2,1 ) ( 2,1 )

    4.

    [ 1 5 2 5 2 0 1 5 0 0 1 | 17 2 9 2 ] [ 1 5 2 5 2 0 1 5 0 0 1 | 17 2 9 2 ]

    5.

    ( 1,1,1 ) ( 1,1,1 )

    6.

    $150,000 at 7%, $750,000 at 8%, $600,000 at 10%

    11.7 Solving Systems with Inverses

    1.

    AB=[ 1 4 −1 −3 ] [ −3 −4 1 1 ]=[ 1(−3)+4(1) 1(−4)+4(1) −1(−3)+−3(1) −1(−4)+−3(1) ]=[ 1 0 0 1 ] BA=[ −3 −4 1 1 ] [ 1 4 −1 −3 ]=[ −3(1)+−4(−1) −3(4)+−4(−3) 1(1)+1(−1) 1(4)+1(−3) ]=[ 1 0 0 1 ] AB=[ 1 4 −1 −3 ] [ −3 −4 1 1 ]=[ 1(−3)+4(1) 1(−4)+4(1) −1(−3)+−3(1) −1(−4)+−3(1) ]=[ 1 0 0 1 ] BA=[ −3 −4 1 1 ] [ 1 4 −1 −3 ]=[ −3(1)+−4(−1) −3(4)+−4(−3) 1(1)+1(−1) 1(4)+1(−3) ]=[ 1 0 0 1 ]

    2.

    A −1 =[ 3 5 1 5 2 5 1 5 ] A −1 =[ 3 5 1 5 2 5 1 5 ]

    3.

    A −1 =[ 1 1 2 2 4 −3 3 6 −5 ] A −1 =[ 1 1 2 2 4 −3 3 6 −5 ]

    4.

    X=[ 4 38 58 ] X=[ 4 38 58 ]

    11.8 Solving Systems with Cramer's Rule

    1.

    ( 3,7 ) ( 3,7 )

    2.

    10 10

    3.

    ( 2, 3 5 , 12 5 ) ( 2, 3 5 , 12 5 )

    11.1 Section Exercises

    1.

    No, you can either have zero, one, or infinitely many. Examine graphs.

    3.

    This means there is no realistic break-even point. By the time the company produces one unit they are already making profit.

    5.

    You can solve by substitution (isolating x x or y y ), graphically, or by addition.

    7.

    Yes

    9.

    Yes

    11.

    (−1,2) (−1,2)

    13.

    (−3,1) (−3,1)

    15.

    ( 3 5 ,0 ) ( 3 5 ,0 )

    17.

    No solutions exist.

    19.

    ( 72 5 , 132 5 ) ( 72 5 , 132 5 )

    21.

    ( 6,−6 ) ( 6,−6 )

    23.

    ( 1 2 , 1 10 ) ( 1 2 , 1 10 )

    25.

    No solutions exist.

    27.

    ( 1 5 , 2 3 ) ( 1 5 , 2 3 )

    29.

    ( x, x+3 2 ) ( x, x+3 2 )

    31.

    (−4,4) (−4,4)

    33.

    ( 1 2 , 1 8 ) ( 1 2 , 1 8 )

    35.

    ( 1 6 ,0 ) ( 1 6 ,0 )

    37.

    ( x,2(7x−6) ) ( x,2(7x−6) )

    39.

    ( 5 6 , 4 3 ) ( 5 6 , 4 3 )

    41.

    Consistent with one solution

    43.

    Consistent with one solution

    45.

    Dependent with infinitely many solutions

    47.

    ( −3.08,4.91 ) ( −3.08,4.91 )

    49.

    ( −1.52,2.29 ) ( −1.52,2.29 )

    51.

    ( A+B 2 , AB 2 ) ( A+B 2 , AB 2 )

    53.

    ( −1 AB , A AB ) ( −1 AB , A AB )

    55.

    ( CEBF BDAE , AFCD BDAE ) ( CEBF BDAE , AFCD BDAE )

    57.

    They never turn a profit.

    59.

    (1,250,100,000) (1,250,100,000)

    61.

    The numbers are 7.5 and 20.5.

    63.

    24,000

    65.

    790 second-year students, 805 first-year students

    67.

    56 men, 74 women

    69.

    10 gallons of 10% solution, 15 gallons of 60% solution

    71.

    Swan Peak: $750,000, Riverside: $350,000

    73.

    $12,500 in the first account, $10,500 in the second account.

    75.

    High-tops: 45, Low-tops: 15

    77.

    Infinitely many solutions. We need more information.

    11.2 Section Exercises

    1.

    No, there can be only one, zero, or infinitely many solutions.

    3.

    Not necessarily. There could be zero, one, or infinitely many solutions. For example, ( 0,0,0 ) ( 0,0,0 ) is not a solution to the system below, but that does not mean that it has no solution.

    2x+3y−6z=1 −4x−6y+12z=−2 x+2y+5z=10 2x+3y−6z=1 −4x−6y+12z=−2 x+2y+5z=10

    5.

    Every system of equations can be solved graphically, by substitution, and by addition. However, systems of three equations become very complex to solve graphically so other methods are usually preferable.

    7.

    No

    9.

    Yes

    11.

    ( −1,4,2 ) ( −1,4,2 )

    13.

    ( 85 107 , 312 107 , 191 107 ) ( 85 107 , 312 107 , 191 107 )

    15.

    ( 1, 1 2 ,0 ) ( 1, 1 2 ,0 )

    17.

    ( 4,−6,1 ) ( 4,−6,1 )

    19.

    ( x, 1 27 (65−16x), x+28 27 ) ( x, 1 27 (65−16x), x+28 27 )

    21.

    ( 45 13 , 17 13 ,−2 ) ( 45 13 , 17 13 ,−2 )

    23.

    No solutions exist

    25.

    ( 0,0,0 ) ( 0,0,0 )

    27.

    ( 4 7 , 1 7 , 3 7 ) ( 4 7 , 1 7 , 3 7 )

    29.

    ( 7,20,16 ) ( 7,20,16 )

    31.

    ( −6,2,1 ) ( −6,2,1 )

    33.

    ( 5,12,15 ) ( 5,12,15 )

    35.

    ( −5,−5,−5 ) ( −5,−5,−5 )

    37.

    ( 10,10,10 ) ( 10,10,10 )

    39.

    ( 1 2 , 1 5 , 4 5 ) ( 1 2 , 1 5 , 4 5 )

    41.

    ( 1 2 , 2 5 , 4 5 ) ( 1 2 , 2 5 , 4 5 )

    43.

    ( 2,0,0 ) ( 2,0,0 )

    45.

    ( 1,1,1 ) ( 1,1,1 )

    47.

    ( 128 557 , 23 557 , 28 557 ) ( 128 557 , 23 557 , 28 557 )

    49.

    ( 6,−1,0 ) ( 6,−1,0 )

    51.

    24, 36, 48

    53.

    70 grandparents, 140 parents, 190 children

    55.

    Your share was $19.95, Shani’s share was $40, and your other roommate’s share was $22.05.

    57.

    There are infinitely many solutions; we need more information

    59.

    500 students, 225 children, and 450 adults

    61.

    The BMW was $49,636, the Jeep was $42,636, and the Toyota was $47,727.

    63.

    $400,000 in the account that pays 3% interest, $500,000 in the account that pays 4% interest, and $100,000 in the account that pays 2% interest.

    65.

    The United States consumed 26.3%, Japan 7.1%, and China 6.4% of the world’s oil.

    67.

    Saudi Arabia imported 16.8%, Canada imported 15.1%, and Mexico 15.0%

    69.

    Birds were 19.3%, fish were 18.6%, and mammals were 17.1% of endangered species

    11.3 Section Exercises

    1.

    A nonlinear system could be representative of two circles that overlap and intersect in two locations, hence two solutions. A nonlinear system could be representative of a parabola and a circle, where the vertex of the parabola meets the circle and the branches also intersect the circle, hence three solutions.

    3.

    No. There does not need to be a feasible region. Consider a system that is bounded by two parallel lines. One inequality represents the region above the upper line; the other represents the region below the lower line. In this case, no points in the plane are located in both regions; hence there is no feasible region.

    5.

    Choose any number between each solution and plug into C(x) C(x) and R(x). R(x). If C(x)<R(x), C(x)<R(x), then there is profit.

    7.

    ( 0,−3 ),( 3,0 ) ( 0,−3 ),( 3,0 )

    9.

    ( 3 2 2 , 3 2 2 ),( 3 2 2 , 3 2 2 ) ( 3 2 2 , 3 2 2 ),( 3 2 2 , 3 2 2 )

    11.

    ( −3,0 ),( 3,0 ) ( −3,0 ),( 3,0 )

    13.

    ( 1 4 , 62 8 ),( 1 4 , 62 8 ) ( 1 4 , 62 8 ),( 1 4 , 62 8 )

    15.

    ( 398 4 , 199 4 ),( 398 4 , 199 4 ) ( 398 4 , 199 4 ),( 398 4 , 199 4 )

    17.

    ( 0,2 ),( 1,3 ) ( 0,2 ),( 1,3 )

    19.

    ( 1 2 ( 5 −1 ) , 1 2 ( 1 5 ) ),( 1 2 ( 5 −1 ) , 1 2 ( 1 5 ) ) ( 1 2 ( 5 −1 ) , 1 2 ( 1 5 ) ),( 1 2 ( 5 −1 ) , 1 2 ( 1 5 ) )

    21.

    ( 5,0 ) ( 5,0 )

    23.

    ( 0,0 ) ( 0,0 )

    25.

    ( 3,0 ) ( 3,0 )

    27.

    No Solutions Exist

    29.

    No Solutions Exist

    31.

    ( 2 2 , 2 2 ),( 2 2 , 2 2 ),( 2 2 , 2 2 ),( 2 2 , 2 2 ) ( 2 2 , 2 2 ),( 2 2 , 2 2 ),( 2 2 , 2 2 ),( 2 2 , 2 2 )

    33.

    (2,0) (2,0)

    35.

    ( 7 ,−3 ),( 7 ,3 ),( 7 ,−3 ),( 7 ,3 ) ( 7 ,−3 ),( 7 ,3 ),( 7 ,−3 ),( 7 ,3 )

    37.

    ( 1 2 ( 73 −5 ) , 1 2 ( 7 73 ) ),( 1 2 ( 73 −5 ) , 1 2 ( 7 73 ) ) ( 1 2 ( 73 −5 ) , 1 2 ( 7 73 ) ),( 1 2 ( 73 −5 ) , 1 2 ( 7 73 ) )

    39.
    1a932f1f6f4851c4d38fd04d852b13ba7228acd3
    41.
    b0ef2bd677c85971843cb19842ea11e21f65a8a0
    43.
    4718f7707aff22da92aa0baf7257c988a3df4649
    45.
    d1d1329fca24c6776d232b0fd0c52ce809378d13
    47.
    3f277c36aad7156e3891ac0212207ba1e1da2c15
    49.

    ( −2 70 383 ,−2 35 29 ),( −2 70 383 ,2 35 29 ),( 2 70 383 ,−2 35 29 ),( 2 70 383 ,2 35 29 ) ( −2 70 383 ,−2 35 29 ),( −2 70 383 ,2 35 29 ),( 2 70 383 ,−2 35 29 ),( 2 70 383 ,2 35 29 )

    51.

    No Solution Exists

    53.

    x=0,y>0 x=0,y>0 and 0<x<1, x <y< 1 x 0<x<1, x <y< 1 x

    55.

    12, 288

    57.

    2–20 computers

    11.4 Section Exercises

    1.

    No, a quotient of polynomials can only be decomposed if the denominator can be factored. For example, 1 x 2 +1 1 x 2 +1 cannot be decomposed because the denominator cannot be factored.

    3.

    Graph both sides and ensure they are equal.

    5.

    If we choose x=−1, x=−1, then the B-term disappears, letting us immediately know that A=3. A=3. We could alternatively plug in x= 5 3 x= 5 3 , giving us a B-value of −2. −2.

    7.

    8 x+3 5 x−8 8 x+3 5 x−8

    9.

    1 x+5 + 9 x+2 1 x+5 + 9 x+2

    11.

    3 5x−2 + 4 4x−1 3 5x−2 + 4 4x−1

    13.

    5 2( x+3 ) + 5 2( x−3 ) 5 2( x+3 ) + 5 2( x−3 )

    15.

    3 x+2 + 3 x−2 3 x+2 + 3 x−2

    17.

    9 5( x+2 ) + 11 5( x−3 ) 9 5( x+2 ) + 11 5( x−3 )

    19.

    8 x−3 5 x−2 8 x−3 5 x−2

    21.

    1 x−2 + 2 ( x−2 ) 2 1 x−2 + 2 ( x−2 ) 2

    23.

    6 4x+5 + 3 ( 4x+5 ) 2 6 4x+5 + 3 ( 4x+5 ) 2

    25.

    1 x−7 2 ( x−7 ) 2 1 x−7 2 ( x−7 ) 2

    27.

    4 x 3 2( x+1 ) + 7 2 ( x+1 ) 2 4 x 3 2( x+1 ) + 7 2 ( x+1 ) 2

    29.

    4 x + 2 x 2 3 3x+2 + 7 2 ( 3x+2 ) 2 4 x + 2 x 2 3 3x+2 + 7 2 ( 3x+2 ) 2

    31.

    x+1 x 2 +x+3 + 3 x+2 x+1 x 2 +x+3 + 3 x+2

    33.

    4−3x x 2 +3x+8 + 1 x−1 4−3x x 2 +3x+8 + 1 x−1

    35.

    2x−1 x 2 +6x+1 + 2 x+3 2x−1 x 2 +6x+1 + 2 x+3

    37.

    1 x 2 +x+1 + 4 x−1 1 x 2 +x+1 + 4 x−1

    39.

    2 x 2 −3x+9 + 3 x+3 2 x 2 −3x+9 + 3 x+3

    41.

    1 4 x 2 +6x+9 + 1 2x−3 1 4 x 2 +6x+9 + 1 2x−3

    43.

    1 x + 1 x+6 4x x 2 −6x+36 1 x + 1 x+6 4x x 2 −6x+36

    45.

    x+6 x 2 +1 + 4x+3 ( x 2 +1 ) 2 x+6 x 2 +1 + 4x+3 ( x 2 +1 ) 2

    47.

    x+1 x+2 + 2x+3 ( x+2 ) 2 x+1 x+2 + 2x+3 ( x+2 ) 2

    49.

    1 x 2 +3x+25 3x ( x 2 +3x+25 ) 2 1 x 2 +3x+25 3x ( x 2 +3x+25 ) 2

    51.

    1 8x x 8( x 2 +4 ) + 10x 2 ( x 2 +4 ) 2 1 8x x 8( x 2 +4 ) + 10x 2 ( x 2 +4 ) 2

    53.

    16 x 9 x 2 + 16 x−1 7 ( x−1 ) 2 16 x 9 x 2 + 16 x−1 7 ( x−1 ) 2

    55.

    1 x+1 2 ( x+1 ) 2 + 5 ( x+1 ) 3 1 x+1 2 ( x+1 ) 2 + 5 ( x+1 ) 3

    57.

    5 x−2 3 10( x+2 ) + 7 x+8 7 10( x−8 ) 5 x−2 3 10( x+2 ) + 7 x+8 7 10( x−8 )

    59.

    5 4x 5 2( x+2 ) + 11 2( x+4 ) + 5 4( x+4 ) 5 4x 5 2( x+2 ) + 11 2( x+4 ) + 5 4( x+4 )

    11.5 Section Exercises

    1.

    No, they must have the same dimensions. An example would include two matrices of different dimensions. One cannot add the following two matrices because the first is a 2×2 2×2 matrix and the second is a 2×3 2×3 matrix. [ 1 2 3 4 ]+[ 6 5 4 3 2 1 ] [ 1 2 3 4 ]+[ 6 5 4 3 2 1 ] has no sum.

    3.

    Yes, if the dimensions of A A are m×n m×n and the dimensions of B B are n×m, n×m, both products will be defined.

    5.

    Not necessarily. To find AB, AB, we multiply the first row of A A by the first column of B B to get the first entry of AB. AB. To find BA, BA, we multiply the first row of B B by the first column of A A to get the first entry of BA. BA. Thus, if those are unequal, then the matrix multiplication does not commute.

    7.

    [ 11 19 15 94 17 67 ] [ 11 19 15 94 17 67 ]

    9.

    [ −4 2 8 1 ] [ −4 2 8 1 ]

    11.

    Undidentified; dimensions do not match

    13.

    [ 9 27 63 36 0 192 ] [ 9 27 63 36 0 192 ]

    15.

    [ −64 −12 −28 −72 −360 −20 −12 −116 ] [ −64 −12 −28 −72 −360 −20 −12 −116 ]

    17.

    [ 1,800 1,200 1,300 800 1,400 600 700 400 2,100 ] [ 1,800 1,200 1,300 800 1,400 600 700 400 2,100 ]

    19.

    [ 20 102 28 28 ] [ 20 102 28 28 ]

    21.

    [ 60 41 2 −16 120 −216 ] [ 60 41 2 −16 120 −216 ]

    23.

    [ −68 24 136 −54 −12 64 −57 30 128 ] [ −68 24 136 −54 −12 64 −57 30 128 ]

    25.

    Undefined; dimensions do not match.

    27.

    [ −8 41 −3 40 −15 −14 4 27 42 ] [ −8 41 −3 40 −15 −14 4 27 42 ]

    29.

    [ −840 650 −530 330 360 250 −10 900 110 ] [ −840 650 −530 330 360 250 −10 900 110 ]

    31.

    [ −350 1,050 350 350 ] [ −350 1,050 350 350 ]

    33.

    Undefined; inner dimensions do not match.

    35.

    [ 1,400 700 −1,400 700 ] [ 1,400 700 −1,400 700 ]

    37.

    [ 332,500 927,500 −227,500 87,500 ] [ 332,500 927,500 −227,500 87,500 ]

    39.

    [ 490,000 0 0 490,000 ] [ 490,000 0 0 490,000 ]

    41.

    [ −2 3 4 −7 9 −7 ] [ −2 3 4 −7 9 −7 ]

    43.

    [ −4 29 21 −27 −3 1 ] [ −4 29 21 −27 −3 1 ]

    45.

    [ −3 −2 −2 −28 59 46 −4 16 7 ] [ −3 −2 −2 −28 59 46 −4 16 7 ]

    47.

    [ 1 −18 −9 −198 505 369 −72 126 91 ] [ 1 −18 −9 −198 505 369 −72 126 91 ]

    49.

    [ 0 1.6 9 −1 ] [ 0 1.6 9 −1 ]

    51.

    [ 2 24 −4.5 12 32 −9 −8 64 61 ] [ 2 24 −4.5 12 32 −9 −8 64 61 ]

    53.

    [ 0.5 3 0.5 2 1 2 10 7 10 ] [ 0.5 3 0.5 2 1 2 10 7 10 ]

    55.

    [ 1 0 0 0 1 0 0 0 1 ] [ 1 0 0 0 1 0 0 0 1 ]

    57.

    [ 1 0 0 0 1 0 0 0 1 ] [ 1 0 0 0 1 0 0 0 1 ]

    59.

    B n ={ [ 1 0 0 0 1 0 0 0 1 ],neven, [ 1 0 0 0 0 1 0 1 0 ],nodd. B n ={ [ 1 0 0 0 1 0 0 0 1 ],neven, [ 1 0 0 0 0 1 0 1 0 ],nodd.

    11.6 Section Exercises

    1.

    Yes. For each row, the coefficients of the variables are written across the corresponding row, and a vertical bar is placed; then the constants are placed to the right of the vertical bar.

    3.

    No, there are numerous correct methods of using row operations on a matrix. Two possible ways are the following: (1) Interchange rows 1 and 2. Then R 2 = R 2 −9 R 1 . R 2 = R 2 −9 R 1 . (2) R 2 = R 1 −9 R 2 . R 2 = R 1 −9 R 2 . Then divide row 1 by 9.

    5.

    No. A matrix with 0 entries for an entire row would have either zero or infinitely many solutions.

    7.

    [ 0 16 9 −1 | 4 2 ] [ 0 16 9 −1 | 4 2 ]

    9.

    [ 1 5 8 12 3 0 3 4 9 | 16 4 −7 ] [ 1 5 8 12 3 0 3 4 9 | 16 4 −7 ]

    11.

    −2x+5y=5 6x−18y=26 −2x+5y=5 6x−18y=26

    13.

    3x+2y=3 x−9y+4z=−1 8x+5y+7z=8 3x+2y=3 x−9y+4z=−1 8x+5y+7z=8

    15.

    4x+5y−2z=12        y+58z=2 8x+7y−3z=−5 4x+5y−2z=12        y+58z=2 8x+7y−3z=−5

    17.

    No solutions

    19.

    (−1,−2) (−1,−2)

    21.

    ( 6,7 ) ( 6,7 )

    23.

    ( 3,2 ) ( 3,2 )

    25.

    ( 1 5 , 1 2 ) ( 1 5 , 1 2 )

    27.

    ( x, 4 15 (5x+1) ) ( x, 4 15 (5x+1) )

    29.

    ( 3,4 ) ( 3,4 )

    31.

    ( 196 39 , 5 13 ) ( 196 39 , 5 13 )

    33.

    ( 31,−42,87 ) ( 31,−42,87 )

    35.

    ( 21 40 , 1 20 , 9 8 ) ( 21 40 , 1 20 , 9 8 )

    37.

    ( 18 13 , 15 13 , 15 13 ) ( 18 13 , 15 13 , 15 13 )

    39.

    ( x,y, 1 2 (1−2x−3y) ) ( x,y, 1 2 (1−2x−3y) )

    41.

    ( x, x 2 ,−1 ) ( x, x 2 ,−1 )

    43.

    ( 125,−25,0 ) ( 125,−25,0 )

    45.

    ( 8,1,−2 ) ( 8,1,−2 )

    47.

    ( 1,2,3 ) ( 1,2,3 )

    49.

    ( x, 31 28 3x 4 , 1 28 (−7x−3) ) ( x, 31 28 3x 4 , 1 28 (−7x−3) )

    51.

    No solutions exist.

    53.

    860 red velvet, 1,340 chocolate

    55.

    4% for account 1, 6% for account 2

    57.

    $126

    59.

    Banana was 3%, pumpkin was 7%, and rocky road was 2%

    61.

    100 almonds, 200 cashews, 600 pistachios

    11.7 Section Exercises

    1.

    If A −1 A −1 is the inverse of A, A, then A A −1 =I, A A −1 =I, the identity matrix. Since A A is also the inverse of A −1 , A −1 A=I. A −1 , A −1 A=I. You can also check by proving this for a 2×2 2×2 matrix.

    3.

    No, because ad ad and bc bc are both 0, so adbc=0, adbc=0, which requires us to divide by 0 in the formula.

    5.

    Yes. Consider the matrix [ 0 1 1 0 ]. [ 0 1 1 0 ]. The inverse is found with the following calculation: A −1 = 1 0(0)−1(1) [ 0 −1 −1 0 ]=[ 0 1 1 0 ]. A −1 = 1 0(0)−1(1) [ 0 −1 −1 0 ]=[ 0 1 1 0 ].

    7.

    AB=BA=[ 1 0 0 1 ]=I AB=BA=[ 1 0 0 1 ]=I

    9.

    AB=BA=[ 1 0 0 1 ]=I AB=BA=[ 1 0 0 1 ]=I

    11.

    AB=BA=[ 1 0 0 0 1 0 0 0 1 ]=I AB=BA=[ 1 0 0 0 1 0 0 0 1 ]=I

    13.

    1 29 [ 9 2 −1 3 ] 1 29 [ 9 2 −1 3 ]

    15.

    1 69 [ −2 7 9 3 ] 1 69 [ −2 7 9 3 ]

    17.

    There is no inverse

    19.

    4 7 [ 0.5 1.5 1 −0.5 ] 4 7 [ 0.5 1.5 1 −0.5 ]

    21.

    1 17 [ −5 5 −3 20 −3 12 1 −1 4 ] 1 17 [ −5 5 −3 20 −3 12 1 −1 4 ]

    23.

    1 209 [ 47 −57 69 10 19 −12 −24 38 −13 ] 1 209 [ 47 −57 69 10 19 −12 −24 38 −13 ]

    25.

    [ 18 60 −168 −56 −140 448 40 80 −280 ] [ 18 60 −168 −56 −140 448 40 80 −280 ]

    27.

    ( −5,6 ) ( −5,6 )

    29.

    ( 2,0 ) ( 2,0 )

    31.

    ( 1 3 , 5 2 ) ( 1 3 , 5 2 )

    33.

    ( 2 3 , 11 6 ) ( 2 3 , 11 6 )

    35.

    ( 7, 1 2 , 1 5 ) ( 7, 1 2 , 1 5 )

    37.

    ( 5,0,−1 ) ( 5,0,−1 )

    39.

    1 34 ( −35,−97,−154 ) 1 34 ( −35,−97,−154 )

    41.

    1 690 ( 65,−1136,−229 ) 1 690 ( 65,−1136,−229 )

    43.

    ( 37 30 , 8 15 ) ( 37 30 , 8 15 )

    45.

    ( 10 123 ,−1, 2 5 ) ( 10 123 ,−1, 2 5 )

    47.

    1 2 [ 2 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 ] 1 2 [ 2 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 ]

    49.

    1 39 [ 3 2 1 7 18 53 32 10 24 36 21 9 9 46 16 5 ] 1 39 [ 3 2 1 7 18 53 32 10 24 36 21 9 9 46 16 5 ]

    51.

    [ 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 1 1 1 1 1 1 ] [ 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 1 1 1 1 1 1 ]

    53.

    Infinite solutions.

    55.

    50% oranges, 25% bananas, 20% apples

    57.

    10 straw hats, 50 beanies, 40 cowboy hats

    59.

    Micah ate 6, Joe ate 3, and Albert ate 3.

    61.

    124 oranges, 10 lemons, 8 pomegranates

    11.8 Section Exercises

    1.

    A determinant is the sum and products of the entries in the matrix, so you can always evaluate that product—even if it does end up being 0.

    3.

    The inverse does not exist.

    5.

    2 2

    7.

    7 7

    9.

    4 4

    11.

    0 0

    13.

    7,990.7 7,990.7

    15.

    3 3

    17.

    1 1

    19.

    224 224

    21.

    15 15

    23.

    17.03 17.03

    25.

    ( 1,1 ) ( 1,1 )

    27.

    ( 1 2 , 1 3 ) ( 1 2 , 1 3 )

    29.

    ( 2,5 ) ( 2,5 )

    31.

    ( 1, 1 3 ) ( 1, 1 3 )

    33.

    ( 15,12 ) ( 15,12 )

    35.

    ( 1,3,2 ) ( 1,3,2 )

    37.

    ( 1,0,3 ) ( 1,0,3 )

    39.

    ( 1 2 ,1,2 ) ( 1 2 ,1,2 )

    41.

    ( 2,1,4 ) ( 2,1,4 )

    43.

    Infinite solutions

    45.

    24 24

    47.

    1 1

    49.

    Yes; 18, 38

    51.

    Yes; 33, 36, 37

    53.

    $7,000 in first account, $3,000 in second account.

    55.

    120 children, 1,080 adult

    57.

    4 gal yellow, 6 gal blue

    59.

    13 green tomatoes, 17 red tomatoes

    61.

    Strawberries 18%, oranges 9%, kiwi 10%

    63.

    100 for movie 1, 230 for movie 2, 312 for movie 3

    65.

    300 almonds, 400 cranberries, 300 cashews

    Review Exercises

    1.

    No

    3.

    ( 2,3 ) ( 2,3 )

    5.

    ( 4,1 ) ( 4,1 )

    7.

    No solutions exist.

    9.

    (300,60,000) (300,60,000)

    11.

    Infinite solutions

    13.

    No solutions exist.

    15.

    ( 1,2,3 ) ( 1,2,3 )

    17.

    ( x, 8x 5 , 14x 5 ) ( x, 8x 5 , 14x 5 )

    19.

    11, 17, 33

    21.

    ( 2,3 ),( 3,2 ) ( 2,3 ),( 3,2 )

    23.

    No solution

    25.

    No solution

    27.
    2555acc0fdbc3125cc47eee6fd70e443ab1d3d08
    29.
    70cd167fe0a1e5a890f58b81ec5351df7f7d7f77
    31.

    2 x+2 , 4 x+1 2 x+2 , 4 x+1

    33.

    7 x+5 , 15 (x+5) 2 7 x+5 , 15 (x+5) 2

    35.

    3 x5 , 4x+1 x 2 +5x+25 3 x5 , 4x+1 x 2 +5x+25

    37.

    x4 ( x 2 2) , 5x+3 ( x 2 2) 2 x4 ( x 2 2) , 5x+3 ( x 2 2) 2

    39.

    [ 16 8 4 12 ] [ 16 8 4 12 ]

    41.

    undefined; dimensions do not match

    43.

    undefined; inner dimensions do not match

    45.

    [ 113 28 10 44 81 41 84 98 42 ] [ 113 28 10 44 81 41 84 98 42 ]

    47.

    [ 127 74 176 2 11 40 28 77 38 ] [ 127 74 176 2 11 40 28 77 38 ]

    49.

    undefined; inner dimensions do not match

    51.

    x3z=7 y+2z=5 x3z=7 y+2z=5 with infinite solutions

    53.

    [ 2 2 1 2 8 5 19 10 22 | 7 0 3 ] [ 2 2 1 2 8 5 19 10 22 | 7 0 3 ]

    55.

    [ 1 0 3 −1 4 0 0 1 2 | 12 0 −7 ] [ 1 0 3 −1 4 0 0 1 2 | 12 0 −7 ]

    57.

    No solutions exist.

    59.

    No solutions exist.

    61.

    1 8 [ 2 7 6 1 ] 1 8 [ 2 7 6 1 ]

    63.

    No inverse exists.

    65.

    ( 20,40 ) ( 20,40 )

    67.

    ( 1,0.2,0.3 ) ( 1,0.2,0.3 )

    69.

    17% oranges, 34% bananas, 39% apples

    71.

    0

    73.

    6

    75.

    ( 6, 1 2 ) ( 6, 1 2 )

    77.

    (x, 5x + 3)

    79.

    ( 0,0, 1 2 ) ( 0,0, 1 2 )

    Practice Test

    1.

    Yes

    3.

    No solutions exist.

    5.

    1 20 ( 10,5,4 ) 1 20 ( 10,5,4 )

    7.

    ( x, 16x 5 13x 5 ) ( x, 16x 5 13x 5 )

    9.

    (2 2 , 17 ),( 2 2 , 17 ),( 2 2 , 17 ),( 2 2 , 17 ) (2 2 , 17 ),( 2 2 , 17 ),( 2 2 , 17 ),( 2 2 , 17 )

    11.
    b85b6c2dcd5befe6199e0f1b431effa8029ef635
    13.

    5 3x+1 2x+3 (3x+1) 2 5 3x+1 2x+3 (3x+1) 2

    15.

    [ 17 51 8 11 ] [ 17 51 8 11 ]

    17.

    [ 12 20 15 30 ] [ 12 20 15 30 ]

    19.

    1 8 1 8

    21.

    [ 14 2 13 2 3 6 1 5 12 | 140 1 11 ] [ 14 2 13 2 3 6 1 5 12 | 140 1 11 ]

    23.

    No solutions exist.

    25.

    ( 100,90 ) ( 100,90 )

    27.

    ( 1 100 ,0 ) ( 1 100 ,0 )

    29.

    32 or more cell phones per day


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