Prove the property a 1 a 1 1 1 1 b 1 1 1 1 c abc 1 1a

Prove the property (a) |1 + a 1 1 1 1 + b 1 1 1 1 + c| = abc (1 + 1/a + 1/b + 1/c). (b) |1 1 1 a b c a^2 b^2 c^2| = (a - b)(b - c)(c - a) Find |A^-1|. Begin by finding A^-1 and then evaluate its determinant. Verify your result by finding |A| and then applying the formula |A^-1| = 1/|A| (a) A = [1 2 1 0 -1 -2 2 3 2], (b) A = [1 2 1 0 -1 -2 1 2 3], (c) A = [1 1 2 1 0 0 0 -3 -1 3 2 1 3 -2 -1 2]. Use the determinant of the coefficient matrix to determine whether the system of linear equations has a unique solution. (a) x_1 - x_2 + x_3 = 4, 2x_1 - x_2 + x_3 = 6., 3x_1 - 2x_2 + 2x_3 = 0, (b) x_1 + x_2 - x_3 = 4, 2x_1 - x_2 + x_3 = 6, 3x_1 - 2x_2 + 2x_3 = 0.

Solution

problem 4

1)

Your matrix

Eliminate elements in the 1st column under the 1st element

Eliminate elements in the 2nd column under the 2nd element

Multiply the main diagonal elements

1 x (-1) x 2 = -2

Determinant is -2

inverse of matrix A

Your matrix

Determinant is not zero, therefore inverse matrix exists

Very detailed solution  

Write the augmented matrix

Find the pivot in the 1st column in the 1st row

Eliminate the 1st column

Find the pivot in the 2nd column in the 2nd row (inversing the sign in the whole row)

Eliminate the 2nd column

Make the pivot in the 3rd column by dividing the 3rd row by 2

Eliminate the 3rd column

There is the inverse matrix on the right

inverse is

now we will calculate det of A inverse

Your matrix

Eliminate elements in the 1st column under the 1st element

Eliminate elements in the 2nd column under the 2nd element

Multiply the main diagonal elements

(-2) x 0.5 x 0.5 = -0.5

Determinant is -0.5

so det(A) = 1/det of inverse(A)

Sign		A1	A2	A3
+	1	1	0	2
2	2	-1	3
3	1	-2	2