Without expanding, show that a-b is a factor of |A| if [latex]A=\left[\begin{array}{lll}1 & 1 & 1 \\a & b & c \\a^2 & b^2 & c^2\end{array}\right][/latex]

Question 8.9.3: Without expanding, show that a-b is a factor of |A| if A= [1......

Precalculus Functions and Graphs [1893897]

Without expanding, show that a-b is a factor of |A| if

A=\left[\begin{array}{lll}1 & 1 & 1 \\a & b & c \\a^2 & b^2 & c^2\end{array}\right]

Question Data is a breakdown of the data given in the question above.

Matrix A: [1 1 1; a b c; a^2 b^2 c^2]

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Step 1:

We are given a matrix A with three rows and three columns. We need to find the determinant of this matrix.

Step 2:

We can simplify the calculation by using the property of determinants that states if we add a multiple of one column to another column, the determinant remains the same. In this case, we subtract the first column from the second column.

Step 3:

By subtracting the first column from the second column, we get a new matrix with the same determinant as the original matrix.

Step 4:

Next, we notice that the first column of the new matrix has a common factor of (a-b). We can factor out (a-b) from the first column.

Step 5:

By factoring out (a-b) from the first column, we get a new matrix where the first column is (a-b) times another matrix.

Step 6:

Finally, we see that the determinant of the original matrix A is equal to (a-b) times the determinant of the new matrix. Therefore, (a-b) is a factor of the determinant of matrix A.