Unit - 2B

Matrices

Q1) Perform all the basic operations for the following matrices.

A =     B =    C =    D =

A1) From the matrices we can know that b, d are compatible. This means we calculate B+D, B-D, D-B, but we cannot add or subtract any other pair of matrices.

The matrix cT = is a 3 2 matrix, so we could add or subtract CT to from B and D

Addition and subtraction of matrices:

B + D = + = =

Similarly, we do subtraction as follow,

B – D = - = = 

We do scalar multiplication for the above matrices.

-3A = =

4C =  =

Q2) Find the scalar multiplication for the following:

A2)

   =

Q3) Solve the equations

4x+7y-9=0

5x-8y+15=0

A3) Given equation can be written in matrix form as A = , X= ,

B =

Given system can be written as: AX=B, where x= A-1 B

Let us find determinant :|A| =  4*(-8)-(5*7) = -32-35 = -67 So, solution exist.

Minor and co-factor of matrix A are:

Cofactor matrix =

X =

X =   and Y =

Q4) Solve the system of equations:

2x+y+3z = 1, x+z =2, 2x+y+z= 3

A4) Given equation can be written in matrix form as A =

X = ,  B = 

Given system of equations can be written as AX = B where X = A-1B

Let us find determinant :|A| = 2(0-1)-1(1-2) +3(1-0) = -2+1+3 = 2.

So, solution exist.

Cofactor = and Adj(A) =

X = 3, Y = -2, Z = -1.

Q5) Reduce the following matrix into normal form and find its rank,

A5) Let A =

Apply we get

A 

Apply we get

A 

Apply

A 

Apply

A 

Apply 

A 

Hence the rank of matrix A is 2 i.e. .

Q6) Reduce the following matrix into normal form and find its rank,

A6) Let A =   

Apply and

A

Apply

A

Apply

A

Apply

A

Apply

A

Hence the rank of the matrix A is 2 i.e. .

Q7) Reduce the following matrix into normal form and find its rank,

A7) Let A =

Apply

Apply

 Apply

Apply 

Apply and

Apply

Hence the rank of matrix A is 2 i.e. .

Q8) Determinants of a Matrix

If A = find |A|

A8) 1 -2 + (-3)

Therefore, determinant value = -33

Q9) Find the determinants of the following matrix.

A =

A9) 2*det -(-3) + 1*det

 =

= 2(0+4)+3(10+1)+1(8)

= 8+3(11)+8

= 49.

Q10) Use Cramer’s rule to solve each system of equations.

2x+y = 10…. (1)

3x-2y = 8…. (2)

A10) we rewrite the given system of equations in the matrix form.

   D= = 2(-2)-1(3) = -7

 D 0, so the system is consistent.

Now we solve for each variable by replacing the co-efficients of thet variable with the constants as shown below.

X =

Therefore, the solution is (4, 2).

Q11) Use Cramer’s rule to solve each system of equations.

3x - 2y = -5 ….  (1)

9x-6y = -15 …. (2)

A11) we rewrite the given system of equations in the matrix form.

   D= = 3(-6)-9(-2) = 0

D = 0, so the system is either inconsistent or dependent.

Checking the numerators for X and Y to see if either is 0.

Now we solve for each variable by replacing the co-efficients of the variable with the constants as shown below.

X =

Since at least one numerator is 0, the system is dependent and has infinitely many solutions.

Q12) Given 33 Rectangular Matrix 

A12) The augumented matrix is as follows

After applying the Gauss-Jordan elimination method:

The inverse of a matrix is as follows,

Q13) Find the inverse of

A=    

A13)

Step 1: Adjoin the identity matrix to the right side of A:

Step 2: Apply row operations to this matrix until the left side is reduced to I. The computations are:

              R2  R2-R1  , R3 R3-R1

             R3  R3 + 2R2

           R1R1 -3R3 ,  R2 R2+3R3

     R1R1-2R2

Step 3: Conclusion: The inverse matrix is:

A-1 =

Q14) Find the inverse of matrix A given below:

A14)   Let A = IA

Or

Thus, the inverse of matrix A is given by I =

Therefore,

Q15) x+y+z=5

2+3y=5z=8

4x+5z=2

A15) So for the given above set of linear equations the Guass-jordan method represents as the following

1    1    1 5

2     3    5 8

4     0     5 2

The augmented matrix is.,

1     1     1

2      3    5

4      0     5

Now we reduce the above matrix in gauss Jordan form.

2    3    5 r2      r1

1     1    1 r1       r2

4     0    5

1    2     4 r2 - 2r1

1    1     1

4     0     5

1     2    4

1     1     1  r3 – r1

3     -2     1

So here represents the gauss Jordan method by having a unique number one diagonally in the matrix.

Q16) x+y+z=3

2x+3y=7z=0

X+3y-2z=17

A16) The given set of linear equations can be written as.,

1    1   1   3

2     3   7   0

1     3   -2   17

The augmented matrix o

For the given system of equations is.,

1    1   1

2    3    7

1     3    -2

1      1     1

7       3      2 c1 c2

-2      3     1

 1     1       1

5       1       0 r2             r2 - 2r1

-2      3       1

So, here the Guass-jordan method completes by making all the diagonal elements to unit number.

Q17) 2X+Y=4

X-3Y=9

X+4Y=-5

A17) The augmented matrix is.,

      2     1      4

       1    -3     9  r1   r2

       1    4     -5 

        1   -3   9

        2     1    4 -2r1+r2 & -r1+r3

        1     4    -5

         1     -3    9

         0      7   -14    ( 1/7) r2  &( 1/7) r3

         0       7   -14

       1    -3    9

        0   1    -2 -1r2+r3

       0     1     -2

           1     0      3

0     1     -2

0      0      0

Here the final matrix we have reduced to echelon form and it also represents system of equations with x=3, y=-2.