Question

X1+2x2-3x3=8
x1+3x1-5x3=11
2x1+5x2-8x3=8
Determine:
1. The rank of the coefficient matrix
2. The adjoint of the coefficient matric
3. The inverse of the coefficient matric
4. The general solution

Answer 1

To determine the rank of the coefficient matrix, we can set up the augmented matrix of the system of equations and perform row reduction:

[ 1 2 -3 | 8 ]
[ 1 3 -5 | 11 ]
[ 2 5 -8 | 8 ]

Performing row operations to put the matrix into row-echelon form:

[ 1 2 -3 | 8 ]
[ 0 1 -2 | 3 ]
[ 0 1 -2 | -8 ]

Performing further row operations to obtain reduced row-echelon form:

[ 1 2 -3 | 8 ]
[ 0 1 -2 | 3 ]
[ 0 0 0 | -11 ]

From this reduced row-echelon form, we can see that there are two leading 1's, which means the rank of the coefficient matrix is 2.

To find the adjoint of the coefficient matrix, we need to find the cofactor matrix and then take its transpose. The cofactor matrix is found by taking the determinants of the minors of each element of the coefficient matrix, multiplied by (-1)^(i+j), where i and j are the row and column indices, respectively.

[ 1 2 -3 ]
[ 1 3 -5 ]
[ 2 5 -8 ]

Taking determinants of minors:

C11 = [ 3 -5 ]
[ 5 -8 ]
= (3)(-8) - (-5)(5) = -24 + 25 = 1

C12 = [ 1 -5 ]
[ 2 -8 ]
= (1)(-8) - (-5)(2) = -8 + 10 = 2

C13 = [ 1 3 ]
[ 2 5 ]
= (1)(5) - (3)(2) = 5 - 6 = -1

C21 = [ 2 -3 ]
[ 5 -8 ]
= (2)(-8) - (-3)(5) = -16 + 15 = -1

C22 = [ 1 -3 ]
[ 2 -8 ]
= (1)(-8) - (-3)(2) = -8 + 6 = -2

C23 = [ 1 2 ]
[ 2 5 ]
= (1)(5) - (2)(2) = 5 - 4 = 1

C31 = [ 2 -3 ]
[ 1 3 ]
= (2)(3) - (-3)(1) = 6 + 3 = 9

C32 = [ 1 -3 ]
[ 1 3 ]
= (1)(3) - (-3)(1) = 3 + 3 = 6

C33 = [ 1 2 ]
[ 1 3 ]
= (1)(3) - (2)(1) = 3 - 2 = 1

Therefore, the cofactor matrix is:

[ 1 2 -1 ]
[ -1 -2 1 ]
[ 9 6 1 ]

Taking the transpose of the cofactor matrix, we get the adjoint of the coefficient matrix:

[ 1 -1 9 ]
[ 2 -2 6 ]
[ -1 1 1 ]

To find the inverse of the coefficient matrix, we can use the formula:

A^(-1) = (1/det(A)) * adj(A),

where det(A) is the determinant of the coefficient matrix.

Calculating the determinant of the coefficient matrix:

det(A) = 1(3(-8)-(-5)(5)) - 2(1(-8)-(-5)(2)) + 3(1(5)-3(2))
= 1(-29) - 2(-10) + 3(-1)
= -29 + 20 - 3
= -12.

Now we can calculate the inverse:

A^(-1) = (1/-12) * [ 1 -1 9 ]
[ 2 -2 6 ]
[ -1 1 1 ]

= [ -1/12 1/12 -3/4 ]
[ -1/6 1/6 -1/2 ]
[ 1/12 -1/12 1/4 ]

Finally, to find the general solution of the system, we can express the system as a matrix equation:

[ x1 ]
[ x2 ]
[ x3 ] = [ 1 2 -3 ]^(-1) * [ 8 ]
[ 11 ]
[ 8 ]

Multiplying the inverse and the augmented matrix:

[ x1 ]
[ x2 ]
[ x3 ] = [ -1/12 1/12 -3/4 ][ 8 ]
[ 11 ]
[ 8 ]

= [ (-1/12)(8) + (1/12)(11) + (-3/4)(8) ]
[ (-1/6)(8) + (1/6)(11) + (-1/2)(8) ]
[ (1/12)(8) + (-1/12)(11) + (1/4)(8) ]

= [ -19/6 ]
[ -11/6 ]
[ 17/6 ]

Therefore, the general solution is:

x1 = -19/6
x2 = -11/6
x3 = 17/6