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Lesson: Cramer's Rule

Sample Question Videos

Worksheet • 17 Questions • 1 Video

Q1:

Use determinants to solve the system

  • A π‘₯ = 3 , 𝑦 = βˆ’ 2 5 , 𝑧 = βˆ’ 2 1
  • B π‘₯ = 4 , 𝑦 = βˆ’ 4 , 𝑧 = 3 2
  • C π‘₯ = 3 5 , 𝑦 = βˆ’ 5 , 𝑧 = βˆ’ 6 5
  • D π‘₯ = βˆ’ 3 , 𝑦 = 2 5 , 𝑧 = 2 1
  • E π‘₯ = 1 9 , 𝑦 = βˆ’ 2 1 , 𝑧 = βˆ’ 1 3

Q2:

Use determinants to solve the system

  • A π‘₯ = 1 , 𝑦 = 0 , 𝑧 = 0
  • B π‘₯ = 9 , 𝑦 = βˆ’ 1 , 𝑧 = 7 2
  • C π‘₯ = βˆ’ 5 7 1 3 , 𝑦 = 0 , 𝑧 = 6 9 1 3
  • D π‘₯ = βˆ’ 1 , 𝑦 = 0 , 𝑧 = 0
  • E π‘₯ = βˆ’ 1 1 9 , 𝑦 = βˆ’ 4 0 5 7 , 𝑧 = 1 0 1 9

Q3:

Nabil is solving simultaneous equations using Cramer’s rule. He writes down

What does he write down for Ξ” ?

  • A | | 9 βˆ’ 8 βˆ’ 3 βˆ’ 2 | |
  • B | | βˆ’ 9 9 2 βˆ’ 3 | |
  • C | | βˆ’ 9 βˆ’ 8 2 βˆ’ 2 | |
  • D | | βˆ’ 8 βˆ’ 9 βˆ’ 2 2 | |
  • E | | 9 βˆ’ 9 βˆ’ 3 2 | |

Q4:

Solve the simultaneous equations

  • A π‘₯ = βˆ’ 6 , 𝑦 = 5 , 𝑧 = 7
  • B π‘₯ = βˆ’ 6 , 𝑦 = 7 , 𝑧 = 6
  • C π‘₯ = βˆ’ 7 , 𝑦 = 5 , 𝑧 = 9
  • D π‘₯ = 7 0 2 5 3 , 𝑦 = βˆ’ 4 0 3 5 3 , 𝑧 = 7
  • E π‘₯ = 6 6 3 4 7 , 𝑦 = 5 , 𝑧 = βˆ’ 3 7 1 4 7

Q5:

Use determinants to solve the system

  • A π‘₯ = 2 1 2 9 , 𝑦 = 1 2 9
  • B π‘₯ = 1 , 𝑦 = 1
  • C π‘₯ = 7 9 , 𝑦 = 1 2 7
  • D π‘₯ = 1 2 9 , 𝑦 = 2 1 2 9
  • E π‘₯ = βˆ’ 2 1 2 9 , 𝑦 = βˆ’ 1 2 9

Q6:

Use determinants to solve the system

  • A π‘₯ = 9 1 1 , 𝑦 = βˆ’ 2 3 1 1
  • B π‘₯ = 5 , 𝑦 = βˆ’ 2 3
  • C π‘₯ = 9 2 9 , 𝑦 = βˆ’ 2 3 2 9
  • D π‘₯ = βˆ’ 2 3 1 1 , 𝑦 = 9 1 1
  • E π‘₯ = βˆ’ 9 1 1 , 𝑦 = 2 3 1 1

Q7:

11 sketchbooks and 15 crayons cost 377 LE, and 15 sketchbooks and 16 crayons cost 474 LE. Use Cramer’s rule to find the price of a sketchbook and the price of a crayon.

  • AA sketchbook costs 22 LE and a crayon costs 9 LE.
  • BA sketchbook costs 9 LE and a crayon costs 2.70 LE.
  • CA sketchbook costs 441 LE and a crayon costs 1 078 LE.
  • DA sketchbook costs 441 LE and a crayon costs 9 LE.
  • EA sketchbook costs 22 LE and a crayon costs 1 078 LE.

Q8:

Use determinants to solve the system

  • A π‘₯ = βˆ’ 4 3 9 , 𝑦 = 2 9 2 6
  • B π‘₯ = 8 , 𝑦 = βˆ’ 8
  • C π‘₯ = βˆ’ 4 1 5 , 𝑦 = 2 9 1 0
  • D π‘₯ = 2 9 2 6 , 𝑦 = βˆ’ 4 3 9
  • E π‘₯ = 4 3 9 , 𝑦 = βˆ’ 2 9 2 6

Q9:

Use determinants to solve the system

  • A π‘₯ = 1 1 1 6 , 𝑦 = 2 3 3 2
  • B π‘₯ = 6 , 𝑦 = 1 4
  • C π‘₯ = 1 1 1 4 , 𝑦 = 2 3 2 8
  • D π‘₯ = 2 3 3 2 , 𝑦 = 1 1 1 6
  • E π‘₯ = βˆ’ 1 1 1 6 , 𝑦 = βˆ’ 2 3 3 2

Q10:

Use Cramer’s rule to solve the simultaneous equations

  • A π‘₯ = 2 , 𝑦 = 3 2 , 𝑧 = 5 2
  • B π‘₯ = βˆ’ 2 1 1 0 , 𝑦 = 8 5 , 𝑧 = 5 2
  • C π‘₯ = 1 5 2 , 𝑦 = 1 0 , 𝑧 = 2 5 2
  • D π‘₯ = 1 , 𝑦 = 3 2 , 𝑧 = 2 5
  • E π‘₯ = 2 1 5 , 𝑦 = 1 1 0 , 𝑧 = 2 2 5

Q11:

Use determinants to solve the system

  • A π‘₯ = βˆ’ 2 , 𝑦 = βˆ’ 1 , 𝑧 = 2
  • B π‘₯ = βˆ’ 3 , 𝑦 = βˆ’ 4 , 𝑧 = 6 5
  • C π‘₯ = 1 2 , 𝑦 = 1 4 , 𝑧 = βˆ’ 1 2
  • D π‘₯ = 2 , 𝑦 = 1 , 𝑧 = βˆ’ 2
  • E π‘₯ = βˆ’ 2 , 𝑦 = 2 3 2 , 𝑧 = βˆ’ 1 1 2

Q12:

True or False: Suppose that is a system of equations with . The solution vector is

  • ATrue
  • BFalse

Q13:

Use determinants to solve the system

  • A π‘₯ = 1 7 , 𝑦 = 1 2 7
  • B π‘₯ = 6 , 𝑦 = βˆ’ 1 0
  • C π‘₯ = 1 , 𝑦 = 1 2
  • D π‘₯ = 1 2 7 , 𝑦 = 1 7
  • E π‘₯ = βˆ’ 1 7 , 𝑦 = βˆ’ 1 2 7

Q14:

Use determinants to solve the system

  • A π‘₯ = 1 3 7 , 𝑦 = 1 0 4 9
  • B π‘₯ = 1 , 𝑦 = 4 7
  • C π‘₯ = βˆ’ 1 3 , 𝑦 = βˆ’ 1 0 7
  • D π‘₯ = 1 0 4 9 , 𝑦 = 1 3 7
  • E π‘₯ = βˆ’ 1 3 7 , 𝑦 = βˆ’ 1 0 4 9

Q15:

Is Cramer’s rule useful for finding solutions to systems of linear equations in which there is an infinite set of solutions?

  • Ayes
  • Bno

Q16:

Use determinants to solve the system

  • A π‘₯ = βˆ’ 2 1 , 𝑦 = 5 6 , 𝑧 = 4
  • B π‘₯ = 4 , 𝑦 = 3 , 𝑧 = 3 1 3
  • C π‘₯ = 2 1 1 1 , 𝑦 = βˆ’ 5 6 1 1 , 𝑧 = βˆ’ 8 1 1
  • D π‘₯ = 2 1 , 𝑦 = βˆ’ 5 6 , 𝑧 = βˆ’ 4
  • E π‘₯ = βˆ’ 2 7 , 𝑦 = 2 9 , 𝑧 = βˆ’ 2 6

Q17:

Use determinants to solve the system

  • A π‘₯ = βˆ’ 5 3 , 𝑦 = 2 0 3 , 𝑧 = 4 0 3
  • B π‘₯ = 5 3 , 𝑦 = βˆ’ 2 0 3 , 𝑧 = 1 0
  • C π‘₯ = 5 3 , 𝑦 = βˆ’ 2 0 3 , 𝑧 = βˆ’ 4 0 3
  • D π‘₯ = βˆ’ 9 5 3 , 𝑦 = βˆ’ 7 0 3 , 𝑧 = 1 3 0 3
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