Find the determinant of matrix B, using matrix A and elementary row operations.

Question

I have matrix A which is

\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{bmatrix}

with a determinant of -4; and matrix B which is

\begin{bmatrix} -9d & 8e & f-6d \\ -9a & 8b & c-6a \\ -9g & 8h & i-6g \\ \end{bmatrix}

Now, B can be reduced to A by; dividing Column 1 by -9, dividing column 2 by 8, swapping rows 1 and 2 but then I am not sure what to do with column 3 to get it by itself

Then I understand I would multiply all the coefficients together with the determinant to find the determinant of B;

so -4x-9x8x? = det B

Answer 1

Recall the effect of elementary column (or row) operations on the determinant:

1. Swapping two columns changes the sign of the determinant.

2. Multiplying a column by a nonzero constant results in the determinant being multiplied by that constant.

3. Adding a multiple of a column to another column does not change the determinant.

You described operations 1 and 2, but you are missing the third which in this case would be adding $6$ times the first column to the last column. This will get us to arrive at the matrix $A$. In each of these elementary column operations, do not forget the effect on the value of the determinant as stated above.

Therefore, the operations are $-\dfrac{1}{9}C_1$, $\dfrac{1}{8}C_2$, $R_1\leftrightarrow R_2$, and $6C_1+C_3\rightarrow C_3$, where $R_i$ and $C_i$ indicate the rows and columns, respectively.

In the end, you should get $\det B=9(8)(-4)=-288$.