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Matrices Questions - All Grades

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Grade 12 Matrices CCSS: HSN-VM.C.8
Evaluate. [math][[3,-9,5],[0,-1,4]] - [[4,-3,-11],[-3,0,7]] [/math]
  1. [math] [[-1,-12,-6],[-4,-1,-3]] [/math]
  2. [math] [[-1,6,-16],[3,1,-3]] [/math]
  3. [math] [[-1,-6,16],[3,-1,-3]][/math]
  4. Cannot subtract because they are not square matrices.
Grade 12 Matrices CCSS: HSN-VM.C.7
Find the resulting matrix if the matrix [math][[4,-2],[8,6]][/math] is multiplied by [math]3/2[/math].
  1. [math] [[6,-3],[12,9]] [/math]
  2. [math] [[12,-6],[24,18]][/math]
  3. [math] [[6,-2],[12,6]] [/math]
  4. [math] [[6,-3],[8,6]] [/math]
Grade 12 Matrices CCSS: HSN-VM.C.8
Add the following matrices. [math] [[4,-3],[-1,6]] + [[3,5],[2,-8]][/math]
  1. [math] [[7,8],[3,14]][/math]
  2. [math] [[7,2],[1,-2]] [/math]
  3. [math] [[9,0],[-9,8]] [/math]
  4. [math] [[7,-2],[-3,-2]] [/math]
Grade 11 Matrices CCSS: HSA-REI.C.8
[math]y = 2/5 x + 3[/math]
[math]6x-y=7[/math]

Jake is going to solve the system of equations using a matrix equation. He sets up his matrix equation as follows.
[math] [[5, -2],[6,-1]] [[x],[y]] = [[3],[7]] [/math]
Is this a correct? If not, choose the correct reason why not.
  1. It is correct.
  2. There must be a fraction in the matrix equation since there is one in the system of equations.
  3. The coefficients don't match up to the correct variables in the matrix equation.
  4. Systems of linear equations with equations not in standard form can never be put into a matrix equation.
Grade 12 Matrices CCSS: HSN-VM.C.8
Multiply the matrices, if possible. [math] [[3,4],[-2,5],[0,2]] * [[4,-1],[5,3]] [/math]
  1. [math] [[8,-13,-2],[27,5,6]][/math]
  2. [math] [[12,-4],[-10,15],[0,2]] [/math]
  3. [math] [[32,9],[17,17],[10,6]][/math]
  4. Not possible.
Grade 12 Matrices CCSS: HSN-VM.C.8
Subtract the matrices. [math][[3,-9,2]] - [[5],[8],[-3]][/math]
  1. [math] [[-2,-17,5]][/math]
  2. [math][[-2],[-17],[5]][/math]
  3. [math][[0,-5,0],[3,-17,2],[0,3,0]][/math]
  4. Cannot subtract matrices of different dimensions.
Grade 11 Matrices CCSS: HSA-REI.C.9
Solve the following matrix equation.
[math] [[3,-5],[-4,9]] [[x],[y]] = [[4],[4]] [/math]
  1. No Solution
  2. [math] (8, 4) [/math]
  3. [math] (65, 28) [/math]
  4. [math](16/7, -4/7) [/math]
Grade 12 Matrices CCSS: HSN-VM.C.8
Evaluate. [math] [[5,4],[-3,8],[-2,1]] + [[1,0],[-5,5],[9,-2]] [/math]
  1. [math] [[6,4],[2,3],[7,-1]][/math]
  2. [math] [[5,5],[ 2,3],[-4,10]][/math]
  3. [math] [[6,4],[-8,13],[7,-1]][/math]
  4. Not possible to add (they are not square matrices).
Grade 12 Matrices CCSS: HSN-VM.C.6
Which augmented matrix represents the system of equations [math]2x=8[/math] and [math]6=3y+x[/math]?
  1. [math][[2,8,,0],[6,3,,1]][/math]
  2. [math][[8,2,,0],[6,3,,1]][/math]
  3. [math][[0,2,,8],[6,3,,1]][/math]
  4. [math][[2,0,,8],[1,3,,6]][/math]
Grade 12 Matrices CCSS: HSN-VM.C.8
Evaluate. [math] [[4,8,-1,-1,3]] * [[2],[0],[-1],[3],[4]] [/math]
  1. [math] [18] [/math]
  2. [math] [[8],[0],[1],[-3],[12]] [/math]
  3. [math] [[8,0,1,-3,12]][/math]
  4. These matrices cannot be multiplied together.
Grade 11 Matrices CCSS: HSA-REI.C.5
Which of the following is an equivalent system of equations to the one given?

[math]3x+5y=5[/math]
[math]2x+y=8[/math]
  1. [math]-7x = -35, \ \ 2x+y=8[/math]
  2. [math]3x+5y=5, \ \ -13x = -17 [/math]
  3. [math]3x+5y=5, \ \ 2x=8[/math]
  4. [math]y=11, \ \ 2x+y=8[/math]
Grade 12 Matrices CCSS: HSN-VM.C.10
Grade 12 Matrices CCSS: HSN-VM.C.7
If the matrix [math][[2,9,8],[0,3,4],[1,11,3]][/math] is multiplied by the scalar 5, what is the result?
  1. [math][[7,14,13],[5,8,9],[6,16,8]][/math]
  2. [math][[10,45,40],[0,15,20],[5,55,15]][/math]
  3. [math][[3,4,1],[11,3,2],[9,8,0]][/math]
  4. [math][[10,0,5],[45,15,55],[40,20,15]][/math]
Grade 12 Matrices CCSS: HSN-VM.C.10
If A is a given square matrix, and it is known that there exists a matrix B such that [math]AB=1[/math], which of the following would be the most efficient ways to find the matrix B?
  1. Find the inverse of A. This is the matrix B.
  2. Find the transpose of A. This is the matrix B.
  3. Create a matrix B whose elements are variables. Then, perform matrix multiplication with the matrix A, setting each resulting entry equal to one. Solve this system of equations, which will give the elements of matrix B.
  4. Multiply both sides of the equation, on the left, by slight variations of the matrix A. When one of these matrices, multiplied by A, becomes the identity matrix, this is the matrix B.
Grade 11 Matrices
What is the rule for matrix multiplication?
  1. The number of columns of the first matrix must equal the number of rows of the second matrix.
  2. The matrices must have the same dimensions.
  3. The matrices must have the same number of rows, but not columns.
  4. There is no rule. Matrix multiplication is always possible.
Grade 12 Matrices CCSS: HSN-VM.C.7
If the matrix [math][[27,9,12],[3,0,6],[18,21,3]][/math] is multiplied by the scalar [math]1/3[/math], what is the result?
  1. [math][[9,3,4],[1,0,2],[6,7,1]][/math]
  2. [math][[30,12,14],[6,3,9],[21,24,6]][/math]
  3. [math][[27,9,12],[3,0,6],[18,21,3]][/math]
  4. [math][[9,1,6],[3,0,7],[4,2,1]][/math]
Grade 11 Matrices CCSS: HSA-REI.C.8
When representing a system of linear equations in two variables as a matrix equation, one can use the general form [math] A [[x],[y]] = vec[b] [/math]. For a given system of linear equations, is it possible for the 2-by-2 matrix [math]A[/math] to have different entries? Why or why not?
  1. Yes. Either equation in the system of equations can be multiplied by a constant. This will also affect [math]vec(b)[/math].
  2. No. A different matrix [math]A[/math] would result in a different answer regardless of anything else.
  3. It depends on whether the system of equations is dependent or independent.
  4. No. Only a system of equations that is inconsistent, and therefore has no answer anyways, can have different [math]A[/math] matrices.
Grade 12 Matrices CCSS: HSN-VM.C.10
James needs to show that for matrix [math]A = [[5,-2,1],[3,3,2],[-6,3,-1]] [/math], there is no matrix [math]B, B!=I[/math], such that [math]AB = I[/math], where [math]I[/math] is the 3-by-3 identity matrix. How can he do this?
  1. Try at least 3 matrices, and if none of them multiplied by [math]A[/math] equal the identity matrix, then it is not possible.
  2. Find the inverse of [math]A[/math], and then show that since this matrix is unique, there cannot exist another matrix [math]B[/math] such that [math]AB = I[/math].
  3. Subtract by the additive inverse on both sides, and then factor the left hand side of the equation. This implies that if [math]B=I[/math] the equation equals the zero matrix, which it can't.
  4. Show that the determinant of [math]A[/math] is zero, which means that it does not have a multiplicative inverse.
Grade 12 Matrices CCSS: HSN-VM.C.8
Find the difference. [math] [[2,0],[-4,1],[8,12],[0,-3],[1,1]] - [[-3,5],[6,10],[3,12],[-4,4],[8,7]] [/math]
  1. [math] [[5,5],[-2,-9],[-5,-24],[-4,1],[-7,-6]] [/math]
  2. [math] [[5,-5],[-10,-9],[5,0],[4,-7],[-7,-6]][/math]
  3. [math] [[-1,5],[2,11],[11,24],[-4,1],[9,8]] [/math]
  4. [math] [[-5,-5],[-2,9],[5,24],[-4,-7],[7,6]][/math]
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