Skip Navigation LinksHome > Matrix Algebra calculators

Educational Level Secondary school, High school and College
Program Purpose Provide step by step solutions of your problems using online calculators (online solvers)
Problem Source Your textbook, etc

1.1 Matrix operations
1. Addition/Subtraction of two matrix
2. Multiplication of two matrix
3. Power of a matrix
4. Transpose of a matrix
5. Deteminant of a matrix
6. Adjoint of a matrix
7. Inverse of a matrix
8. Prove that any two matrix expression is equal or not
9. Minor of a matrix
10. Cofactor of a matrix
11. Trace of a matrix

1.2 Matrix operations
1. Reduce matrix
2. Rank of matrix
3. Characteristic polynomial
4. Eigenvalue
5. Eigenvector
6. Triangular Matrix
7. LU Decomposition
8. Diagonal Matrix

2. Find inverse of a matrix using
1. Simple inverse matrix
2. Gauss elimination (method of reduction)

3. Solve linear equation of any number of variables using
1. Inverse Matrix method
2. Cramer's Rule method
3. Gauss Elimination (Jordan) method
4. Gauss Elimination (Back Substitution) method
5. Gauss Sield method
6. Gauss Jacobi method
1. Addition of two matrix
1. A + B
2. A - B
3. B - A
4. A + B + C
5. m A
6. n B
7. m A + n B

Find `A + B` ...
`A=[[3,1,1],[-1,2,1],[1,1,1]]`,`B=[[5,0,-2],[7,-6,0],[1,1,2]]`


Solution:
`A + B` = 
`3``1``1`
`-1``2``1`
`1``1``1`
 + 
`5``0``-2`
`7``-6``0`
`1``1``2`
 = 
`8``1``-1`
`6``-4``1`
`2``2``3`
2. Multiplication of two matrix
1. A × B
2. B × A
3. m × A
4. n × B
5. I × A
6. Am × Bn
7. Bm × An

Find `A × B` ...
`A=[[3,1,1],[-1,2,1],[1,1,1]]`,`B=[[5,0,-2],[7,-6,0],[1,1,2]]`


Solution:
`A×B`=
`3``1``1`
`-1``2``1`
`1``1``1`
×
`5``0``-2`
`7``-6``0`
`1``1``2`


=
`3×5+1×7+1×1``3×0+1×-6+1×1``3×-2+1×0+1×2`
`-1×5+2×7+1×1``-1×0+2×-6+1×1``-1×-2+2×0+1×2`
`1×5+1×7+1×1``1×0+1×-6+1×1``1×-2+1×0+1×2`


=
`15+7+1``0-6+1``-6+0+2`
`-5+14+1``0-12+1``2+0+2`
`5+7+1``0-6+1``-2+0+2`


=
`23``-5``-4`
`10``-11``4`
`13``-5``0`
 
3. Power of a matrix
1. A2
2. A3
3. Am
4. A2 × A
5. Am × A
6. Am × An

Find `A^2` ...
`A=[[3,1,1],[-1,2,1],[1,1,1]]`


Solution:
`A^2`=`A×A`=
`3``1``1`
`-1``2``1`
`1``1``1`
×
`3``1``1`
`-1``2``1`
`1``1``1`


=
`3×3+1×-1+1×1``3×1+1×2+1×1``3×1+1×1+1×1`
`-1×3+2×-1+1×1``-1×1+2×2+1×1``-1×1+2×1+1×1`
`1×3+1×-1+1×1``1×1+1×2+1×1``1×1+1×1+1×1`


=
`9-1+1``3+2+1``3+1+1`
`-3-2+1``-1+4+1``-1+2+1`
`3-1+1``1+2+1``1+1+1`


=
`9``6``5`
`-4``4``2`
`3``4``3`


`A^2` = 
`3``1``1`
`-1``2``1`
`1``1``1`
2
 = 
`9``6``5`
`-4``4``2`
`3``4``3`
4. Transpose of a matrix
1. A'
2. B'
3. (A × B)'
4. B' × A'
5. (B × A)'
6. A' × B'
7. A + A'
8. A - A'
9. (A + B)'
10. A' + B'

Find `A'` ...
`A=[[3,1,1],[-1,2,1],[1,1,1]]`


Solution:
`A^T` = 
`3``1``1`
`-1``2``1`
`1``1``1`
T
 = 
`3``-1``1`
`1``2``1`
`1``1``1`
 
5. Deteminant of a matrix
1. | A |
2. | B |
3. | Am |
4. | Bm |
5. | A × B |
6. | B × A |
7. | Am × Bn |
8. | Bm × An |
9. | A' |
10. | A -1 |

Find `| A |` ...
`A=[[3,1,1],[-1,2,1],[1,1,1]]`


Solution:
`|A|` = 
 `3`  `1`  `1` 
 `-1`  `2`  `1` 
 `1`  `1`  `1` 


 =
 `3` × 
 `2`  `1` 
 `1`  `1` 
 `-1` × 
 `-1`  `1` 
 `1`  `1` 
 `+1` × 
 `-1`  `2` 
 `1`  `1` 


`=3 xx (2 × 1 - 1 × 1) -1 xx (-1 × 1 - 1 × 1) +1 xx (-1 × 1 - 2 × 1)`

`=3 xx (2 -1) -1 xx (-1 -1) +1 xx (-1 -2)`

`=3 xx (1) - -1 xx (-2) +1 xx (-3)`

`= 3 +2 -3`

`=2`
6. Adjoint of a matrix
1. Adj(A)
2. Adj(B)
3. Adj(Am)
4. Adj(Bm)
5. Adj(A × B)
6. Adj(B × A)
7. Adj(Am × Bn)
8. Adj(Bm × An)
9. Adj(A')
10. Adj(A -1)

Find `Adj(A)` ...
`A=[[3,1,1],[-1,2,1],[1,1,1]]`


Solution:
`Adj(A)` = 
Adj
`3``1``1`
`-1``2``1`
`1``1``1`


 = 
 + 
 `2`  `1` 
 `1`  `1` 
 - 
 `-1`  `1` 
 `1`  `1` 
 + 
 `-1`  `2` 
 `1`  `1` 
 - 
 `1`  `1` 
 `1`  `1` 
 + 
 `3`  `1` 
 `1`  `1` 
 - 
 `3`  `1` 
 `1`  `1` 
 + 
 `1`  `1` 
 `2`  `1` 
 - 
 `3`  `1` 
 `-1`  `1` 
 + 
 `3`  `1` 
 `-1`  `2` 
T


 = 
`+(2 × 1 - 1 × 1)``-(-1 × 1 - 1 × 1)``+(-1 × 1 - 2 × 1)`
`-(1 × 1 - 1 × 1)``+(3 × 1 - 1 × 1)``-(3 × 1 - 1 × 1)`
`+(1 × 1 - 1 × 2)``-(3 × 1 - 1 × (-1))``+(3 × 2 - 1 × (-1))`
T


 = 
`+(2 -1)``-(-1 -1)``+(-1 -2)`
`-(1 -1)``+(3 -1)``-(3 -1)`
`+(1 -2)``-(3 +1)``+(6 +1)`
T


 = 
`1``2``-3`
`0``2``-2`
`-1``-4``7`
T


 = 
`1``0``-1`
`2``2``-4`
`-3``-2``7`
 
7. Inverse of a matrix
1. A -1
2. B -1
3. (Am) -1
4. (Bm) -1
5. (A × B) -1
6. (B × A) -1
7. (Am × Bn) -1
8. (Bm × An) -1
9. (A') -1
10. (A -1) -1

Find `A^-1` ...
`A=[[3,1,1],[-1,2,1],[1,1,1]]`


Solution:
`|A|` = 
 `3`  `1`  `1` 
 `-1`  `2`  `1` 
 `1`  `1`  `1` 


 =
 `3` × 
 `2`  `1` 
 `1`  `1` 
 `-1` × 
 `-1`  `1` 
 `1`  `1` 
 `+1` × 
 `-1`  `2` 
 `1`  `1` 


`=3 xx (2 × 1 - 1 × 1) -1 xx (-1 × 1 - 1 × 1) +1 xx (-1 × 1 - 2 × 1)`

`=3 xx (2 -1) -1 xx (-1 -1) +1 xx (-1 -2)`

`=3 xx (1) - -1 xx (-2) +1 xx (-3)`

`= 3 +2 -3`

`=2`


`Adj(A)` = 
Adj
`3``1``1`
`-1``2``1`
`1``1``1`


 = 
 + 
 `2`  `1` 
 `1`  `1` 
 - 
 `-1`  `1` 
 `1`  `1` 
 + 
 `-1`  `2` 
 `1`  `1` 
 - 
 `1`  `1` 
 `1`  `1` 
 + 
 `3`  `1` 
 `1`  `1` 
 - 
 `3`  `1` 
 `1`  `1` 
 + 
 `1`  `1` 
 `2`  `1` 
 - 
 `3`  `1` 
 `-1`  `1` 
 + 
 `3`  `1` 
 `-1`  `2` 
T


 = 
`+(2 × 1 - 1 × 1)``-(-1 × 1 - 1 × 1)``+(-1 × 1 - 2 × 1)`
`-(1 × 1 - 1 × 1)``+(3 × 1 - 1 × 1)``-(3 × 1 - 1 × 1)`
`+(1 × 1 - 1 × 2)``-(3 × 1 - 1 × (-1))``+(3 × 2 - 1 × (-1))`
T


 = 
`+(2 -1)``-(-1 -1)``+(-1 -2)`
`-(1 -1)``+(3 -1)``-(3 -1)`
`+(1 -2)``-(3 +1)``+(6 +1)`
T


 = 
`1``2``-3`
`0``2``-2`
`-1``-4``7`
T


 = 
`1``0``-1`
`2``2``-4`
`-3``-2``7`


`"Now, "A^(-1)=1/|A| × Adj(A)`

 = `1/(2)` ×
`1``0``-1`
`2``2``-4`
`-3``-2``7`


 = 
`1/2``0``-1/2`
`1``1``-2`
`-3/2``-1``7/2`
8. Prove that any two matrix expression is equal or not
1. A(BC) = (AB)C
2. A(B + C) = AB + AC
3. A2 - B2 = (A - B)(A + B)
4. (AB)' = B'A'
5. (AB) -1 = (B) -1 (A) -1
6. Adj(AB) = Adj(B) Adj(A)
7. A × Adj(A) = | A | × I
8. Adj(A') = Adj(A)'
9. | A -1 | = | A | -1
10. (A') -1 = (A') -1

Find `(A × B)' = B' × A'` ...
`A=[[3,1,1],[-1,2,1],[1,1,1]]`,`B=[[5,0,-2],[7,-6,0],[1,1,2]]`


Solution:
To find LHS = `(A × B)^T`

`A×B`=
`3``1``1`
`-1``2``1`
`1``1``1`
×
`5``0``-2`
`7``-6``0`
`1``1``2`


=
`3×5+1×7+1×1``3×0+1×-6+1×1``3×-2+1×0+1×2`
`-1×5+2×7+1×1``-1×0+2×-6+1×1``-1×-2+2×0+1×2`
`1×5+1×7+1×1``1×0+1×-6+1×1``1×-2+1×0+1×2`


=
`15+7+1``0-6+1``-6+0+2`
`-5+14+1``0-12+1``2+0+2`
`5+7+1``0-6+1``-2+0+2`


=
`23``-5``-4`
`10``-11``4`
`13``-5``0`


`(A × B)^T` = 
`23``-5``-4`
`10``-11``4`
`13``-5``0`
T
 = 
`23``10``13`
`-5``-11``-5`
`-4``4``0`


`:.``(A × B)^T` = 
`23``10``13`
`-5``-11``-5`
`-4``4``0`
` ->(1)`


To find RHS = `(B^T) × (A^T)`

`B^T` = 
`5``0``-2`
`7``-6``0`
`1``1``2`
T
 = 
`5``7``1`
`0``-6``1`
`-2``0``2`


`A^T` = 
`3``1``1`
`-1``2``1`
`1``1``1`
T
 = 
`3``-1``1`
`1``2``1`
`1``1``1`


`(B^T)×(A^T)`=
`5``7``1`
`0``-6``1`
`-2``0``2`
×
`3``-1``1`
`1``2``1`
`1``1``1`


=
`5×3+7×1+1×1``5×-1+7×2+1×1``5×1+7×1+1×1`
`0×3-6×1+1×1``0×-1-6×2+1×1``0×1-6×1+1×1`
`-2×3+0×1+2×1``-2×-1+0×2+2×1``-2×1+0×1+2×1`


=
`15+7+1``-5+14+1``5+7+1`
`0-6+1``0-12+1``0-6+1`
`-6+0+2``2+0+2``-2+0+2`


=
`23``10``13`
`-5``-11``-5`
`-4``4``0`


`:.``(B^T) × (A^T)` = 
`23``10``13`
`-5``-11``-5`
`-4``4``0`
` ->(2)`


From (1) and (2)
`:. (A × B)^T=(B^T) × (A^T)` (proved)
 
9. Minor of a matrix
1. Minor(A)

Find `"Minor"(A)` ...
`A=[[3,1,1],[-1,2,1],[1,1,1]]`


Solution:
`MINOR(A)` = 
`MINOR`
`3``1``1`
`-1``2``1`
`1``1``1`


=
 `2`  `1` 
 `1`  `1` 
 `-1`  `1` 
 `1`  `1` 
 `-1`  `2` 
 `1`  `1` 
 `1`  `1` 
 `1`  `1` 
 `3`  `1` 
 `1`  `1` 
 `3`  `1` 
 `1`  `1` 
 `1`  `1` 
 `2`  `1` 
 `3`  `1` 
 `-1`  `1` 
 `3`  `1` 
 `-1`  `2` 


=
`2 × 1 - 1 × 1``-1 × 1 - 1 × 1``-1 × 1 - 2 × 1`
`1 × 1 - 1 × 1``3 × 1 - 1 × 1``3 × 1 - 1 × 1`
`1 × 1 - 1 × 2``3 × 1 - 1 × (-1)``3 × 2 - 1 × (-1)`


=
`2 -1``-1 -1``-1 -2`
`1 -1``3 -1``3 -1`
`1 -2``3 +1``6 +1`


=
`1``-2``-3`
`0``2``2`
`-1``4``7`
10. Cofactor of a matrix
1. Cofactor(A)

Find `"Cofactor"(A)` ...
`A=[[3,1,1],[-1,2,1],[1,1,1]]`


Solution:
`COFACTOR(A)` = 
`COFACTOR`
`3``1``1`
`-1``2``1`
`1``1``1`


=
 + 
 `2`  `1` 
 `1`  `1` 
 - 
 `-1`  `1` 
 `1`  `1` 
 + 
 `-1`  `2` 
 `1`  `1` 
 - 
 `1`  `1` 
 `1`  `1` 
 + 
 `3`  `1` 
 `1`  `1` 
 - 
 `3`  `1` 
 `1`  `1` 
 + 
 `1`  `1` 
 `2`  `1` 
 - 
 `3`  `1` 
 `-1`  `1` 
 + 
 `3`  `1` 
 `-1`  `2` 


=
`+(2 × 1 - 1 × 1)``-(-1 × 1 - 1 × 1)``+(-1 × 1 - 2 × 1)`
`-(1 × 1 - 1 × 1)``+(3 × 1 - 1 × 1)``-(3 × 1 - 1 × 1)`
`+(1 × 1 - 1 × 2)``-(3 × 1 - 1 × (-1))``+(3 × 2 - 1 × (-1))`


=
`+(2 -1)``-(-1 -1)``+(-1 -2)`
`-(1 -1)``+(3 -1)``-(3 -1)`
`+(1 -2)``-(3 +1)``+(6 +1)`


=
`1``2``-3`
`0``2``-2`
`-1``-4``7`
 
11. Trace of a matrix
1. Trace(A)

Find `Trace(A)` ...
`A=[[3,1,1],[-1,2,1],[1,1,1]]`


Solution:
`trace(A)` = 
`trace`
`3``1``1`
`-1``2``1`
`1``1``1`


`=3+2+1`

`=6`
 

 
Copyright © 2017. All rights reserved.
Ad: Web Hosting By Arvixe