in exercises 3 - 6, use the following matrices to compute the indicated expression if it is defined.\na =…

in exercises 3 - 6, use the following matrices to compute the indicated expression if it is defined.\na = -1, 3, 1, 2, 0, 1, b = 0, 4, 2, -1, c = 3, 1, 1, 4, 5, 2, d = -1, 1, 3, 0, 5, 2, 4, 1, 2, 3, e = -1, 6, 4, 1, 3, 2\n3. (a) d + e\n(b) d - e\n(c) 5a\n(d) -7c\n(e) 2b - c\n(f) 4e - 2d\n(g) -3(d + 2e)\n(h) a - a\n(i) tr(d)\n(j) tr(d - 3e)\n(k) 4tr(7b)\n(l) tr(a)

in exercises 3 - 6, use the following matrices to compute the indicated expression if it is defined.\na = -1, 3, 1, 2, 0, 1, b = 0, 4, 2, -1, c = 3, 1, 1, 4, 5, 2, d = -1, 1, 3, 0, 5, 2, 4, 1, 2, 3, e = -1, 6, 4, 1, 3, 2\n3. (a) d + e\n(b) d - e\n(c) 5a\n(d) -7c\n(e) 2b - c\n(f) 4e - 2d\n(g) -3(d + 2e)\n(h) a - a\n(i) tr(d)\n(j) tr(d - 3e)\n(k) 4tr(7b)\n(l) tr(a)

Answer

Part 1: Conversion of numbers to binary

(1) Convert ((78)_{10}) to binary

  • Step 1: Divide by 2 (78\div2 = 39) with remainder (0) (39\div2=19) with remainder (1) (19\div2 = 9) with remainder (1) (9\div2=4) with remainder (1) (4\div2 = 2) with remainder (0) (2\div2=1) with remainder (0) (1\div2 = 0) with remainder (1)
  • Step 2: Read remainders from bottom ((78)_{10}=(1001110)_2)

(2) Convert ((121)_{10}) to binary

  • Step 1: Divide by 2 (121\div2 = 60) with remainder (1) (60\div2=30) with remainder (0) (30\div2 = 15) with remainder (0) (15\div2=7) with remainder (1) (7\div2 = 3) with remainder (1) (3\div2=1) with remainder (1) (1\div2 = 0) with remainder (1)
  • Step 2: Read remainders from bottom ((121)_{10}=(1111001)_2)

(3) Convert ((29)_{10}) to binary

  • Step 1: Divide by 2 (29\div2 = 14) with remainder (1) (14\div2=7) with remainder (0) (7\div2 = 3) with remainder (1) (3\div2=1) with remainder (1) (1\div2 = 0) with remainder (1)
  • Step 2: Read remainders from bottom ((29)_{10}=(11101)_2)

(4) Convert ((1F)_{16}) to binary

  • Recall that in hexadecimal to binary conversion, each hex digit is represented by 4 binary digits.
  • (1_{16}=0001_2) and (F_{16}=1111_2)
  • So ((1F)_{16}=(00011111)_2=(11111)_2)

(5) Convert ((3E)_{16}) to binary

  • (3_{16}=0011_2) and (E_{16}=1110_2)
  • So ((3E)_{16}=(00111110)_2=(111110)_2)

(6) Convert ((CD)_{16}) to binary

  • (C_{16}=1100_2) and (D_{16}=1101_2)
  • So ((CD)_{16}=(11001101)_2)

(7) Convert ((27)_{8}) to binary

  • In octal to binary conversion, each octal digit is represented by 3 binary digits.
  • (2_{8}=010_2) and (7_{8}=111_2)
  • So ((27)_{8}=(010111)_2=(10111)_2)

(8) Convert ((33)_{8}) to binary

  • (3_{8}=011_2)
  • So ((33)_{8}=(011011)_2=(11011)_2)

(9) Convert ((52)_{8}) to binary

  • (5_{8}=101_2) and (2_{8}=010_2)
  • So ((52)_{8}=(101010)_2)

Part 2: Binary Operations

(1) (10111+1101)

  • Step 1: Align the binary numbers (\ \ 10111) (+\ 01101) (add leading zero to make length equal)
  • Step 2: Add bit - by - bit from right (1 + 1=10), write (0), carry (1) (1+0 + 1=10), write (0), carry (1) (1+1+1 = 11), write (1), carry (1) (0+1+1=10), write (0), carry (1) (1+0 + 1=10)
  • Step 3: Result (10111+1101 = 100100)

(2) (1101010 - 101101)

  • Step 1: Align the binary numbers (\ \ 1101010) (-\ 0101101) (add leading zero to make length equal)
  • Step 2: Subtract bit - by - bit from right (0 - 1), borrow (1) from next bit, (10 - 1=1) (0 - 0) (after borrowing, the previous bit is reduced by 1), (0 - 0 = 0) (1-1 = 0) (0 - 1), borrow (1) from next bit, (10 - 1=1) (0 - 0) (after borrowing), (0 - 0 = 0) (1-0 = 1) (1-0 = 1)
  • Step 3: Result (1101010 - 101101=1001101)

(3) (10101+11110)

  • Step 1: Align the binary numbers (\ \ 10101) (+\ 11110)
  • Step 2: Add bit - by - bit from right (1+0 = 1) (0+1 = 1) (1+1 = 10), write (0), carry (1) (0+1+1 = 10), write (0), carry (1) (1+1+1 = 11)
  • Step 3: Result (10101+11110 = 110011)

(4) (10101 - 1111)

  • Step 1: Align the binary numbers (\ \ 10101) (-\ 01111) (add leading zero to make length equal)
  • Step 2: Subtract bit - by - bit from right (1 - 1=0) (0 - 1), borrow (1) from next bit, (10 - 1 = 1) (1-1 = 0) (after borrowing, the previous bit is reduced by 1) (0 - 1), borrow (1) from next bit, (10 - 1=1) (1-0 = 1) (after borrowing)
  • Step 3: Result (10101 - 1111=1010)

(5) (1101\times101)

  • Step 1: Recall binary multiplication rules (similar to decimal, but with only 0 and 1) (1101\times1 = 1101) (1101\times0 = 0000) (shift one position left) (1101\times1=1101) (shift two positions left)
  • Step 2: Add the results (\ \ \ \ 1101) (\ \ \ 0000) (shifted) (+1101) (shifted twice) (=1000001)

(6) (111011\div11)

  • Step 1: Recall binary division rules (similar to decimal long division) (11) (which is (3_{10})) into (11) (first two bits of (111011)) is (1), (1\times11 = 11), subtract: (11 - 11=0) Bring down the next bit: (1), (11) into (1) is (0), bring down the next bit: (1), now we have (11) into (11) is (1), (1\times11 = 11), subtract: (11 - 11 = 0) Bring down the next bit: (0), (11) into (0) is (0), bring down the next bit: (1), (11) into (1) is (0), bring down the next bit: (1), now we have (11) into (11) is (1)
  • Step 2: Result (111011\div11 = 10011)

Final Answers

Part 1:

  • ((78)_{10}=(1001110)_2)
  • ((121)_{10}=(1111001)_2)
  • ((29)_{10}=(11101)_2)
  • ((1F)_{16}=(11111)_2)
  • ((3E)_{16}=(111110)_2)
  • ((CD)_{16}=(11001101)_2)
  • ((27)_{8}=(10111)_2)
  • ((33)_{8}=(11011)_2)
  • ((52)_{8}=(101010)_2)

Part 2:

  • (10111 + 1101=100100)
  • (1101010 - 101101 = 1001101)
  • (10101+11110 = 110011)
  • (10101 - 1111=1010)
  • (1101\times101 = 1000001)
  • (111011\div11 = 10011)