Algebra 2 Matrix Operations

Algebra 2 Matrix Operations Chain Reaction Activity Created by Mary Hinton, Hillwood High School2009-2010: SPI 3103.2.2 Compute with all real and complex numbers.Materials Needed: Four game cards per group, per round; paper; pencilsInstructions to the teacher for making activity: Copy game cards and separate by cutting them apart. Optional: paste game cards onto colored index cards, using a different color for each part of each round (card 1: blue; card 2: yellow; card 3: green; card 4: orange) and laminateInstructions for conducting the activity: 1. Divide class into groups of 4. 2. Students should be seated in rows and given cards in numerical order: Card 1, Card 2, Card 3, Card 4. 3. Calculators are optional 4. Have the answer key on hand during the activity.Directions to the students: 1. The student who holds card 1 will evaluate the expression. 2. Student 1 will pass his solution to the student who holds card 2, who uses that result to evaluate his/her expression. 3. This process continues until the last student evaluates his/her expression and writes in on a white board or calls it out. 4. The first group to arrive at the correct results wins the round!*Please contact me if there are any corrections needed for this document…or any suggestions [email protected] Answer KeyX= A + B Y = ½ X Z = CY Final answer: -3Z Round 1 4 2 Z = 28  84 X =   Y =   6 3 Round 2  4   2  Z =  3 9 X =   Y =    6  3 Round 3 2  4 1  2  5 1 15 3  X =   Y =   Z =     2 2  1 1   5 4   15 12 Round 4  2 8  1 4   3 8    9  24 Z =       56 12 168  36 X = 10 0  Y =  5 0       6  8  3  4 Round 5 X = Y = 2 14  6 42 Z = 10 14  5  7 2 14  6 42          8 0   4 0 Round 1 Round 2 1 3  0   4  A =   B =   C = 5 6 A =   B =   C6 5 2 4 1  5Round 3  1 1 1  3  2 3  A =   B =   C =   1 1  3 1   3  2Round 4  0 6    2 2       2 1 0  A =  3 4  B =  13  4 C =    3 7  6  5  9 11 1 Round 5 1  3  9 11 2 2 A =   B =   C =   5  7 13 7  2 2Round 1 Card 1 Round 2 Card 11 3  0   4  A =   B =   A =   B =   2 4 1  5X= A + B X= A + BRound 1 Card 2 Round 2 Card 2Y = ½ X Y = ½ XRound 1 Card 3 Round 2 Card 3C = 5 6 C= 6 5Z = CY Z = CYRound 1 Card 4 Round 2 Card 4Final answer: Final answer: -3Z -3Z Round 4 Card 1 Round 3 Card 1  0 6    2 2   1 1 1  3     A = B = A =  3 4 B = 13  4 1 1  3 1           5  9 11 1     X= A + B X= A + BRound 3 Card 2 Round 4 Card 2Y = ½ X Y = ½ XRound 3 Card 3 Round 4 Card 3 2 3   2 1 0  C =   C =    3  2  3 7  6Z = CY Z = CYRound 3 Card 4 Round 4 Card 4Final answer: Final answer: -3Z -3Z Round 5 Card 11  3  9 11 A =   B =   5  7 13 7 X= A + BRound 5 Card 2Y = ½ XRound 5 Card 32 2 C =   2 2Z = CYRound 5 Card 4Final answer: -3Z Cards (alternative form)Round 1 Card 1 Round 2 Card 11 3 A = B = C = 5 6  0   4  2 4 A =   B =   C6 5     1  5X= A + B X= A + BRound 1 Card 2 Round 2 Card 21 3  0   4  A =   B =   C = 5 6 A =   B =   C6 5 2 4 1  5Y = ½ X Y = ½ XRound 1 Card 3 Round 2 Card 31 3  0   4  A =   B =   C = 5 6 A =   B =   C6 5 2 4 1  5Z = CY Z = CYRound 1 Card 4 Round 2 Card 4 1 3 A = B = C = 5 6 2 4  0   4      A =   B =   C6 5 1  5 Final answer: -3Z Final answer: -3Z Round 3 Card 1 Round 4 Card 1 1 1 1  3  0 6    2 2  A = B = 1 1  3 1          A =  3 4  B =  13  4  2 3   5  9 11 1  C =    3  2  2 1 0  C =    3 7  6 X= A + B X= A + BRound 3 Card 2 Round 4 Card 2  1 1 1  3 A =   B =    0 6    2 2  1 1  3 1      A =  3 4 B = 13  4  2 3      C =    5  9 11 1   3  2  2 1 0  C =   Y = ½ X  3 7  6Y = ½ XRound 3 Card 3 Round 4 Card 3  1 1 1  3 A =   B =    0 6    2 2  1 1  3 1      A =  3 4 B = 13  4  2 3      C =    5  9 11 1   3  2  2 1 0  C =   Z = CY  3 7  6Z = CYRound 3 Card 4 Round 4 Card 4 1 1 1  3  0 6    2 2  A = B = 1 1  3 1          A =  3 4  B =  13  4 C =  2 3   5  9 11 1  C =    3  2  2 1 0     3 7  6 Final answer: -3Z Final answer: -3Z Round 5 Card 1 Round 5 Card 21  3 1  3 A =   A =   5  7 5  7  9 11  9 11 B =   B =   13 7  13 7  2 2 2 2 C =   C =   2 2 2 2X= A + B Y = ½ XRound 5 Card 3 Round 5 Card 41  3 1  3 A =   A =   5  7 5  7  9 11  9 11 B =   B =   13 7  13 7  2 2 2 2 C =   C =   2 2 2 2 Final answer: Z = CY -3Z

Algebra 2 Matrix Operations

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