What's the determinant of matrix B using the determinant of matrix A

Question

The problem states: "Let A be a square matrix of order n and detA = 6. Let matrix B be another matrix, which is the result of matrix A, after all the elements of a single row have been replaced with the elements of another row in the matrix A, multiplied by 8. What's detB equal to?"

My suggestion is that the result is -48, because first we replace two rows, which means the determinant's sign changes, so it becomes -6, and after that we multiple that row by 8, which means we have to multiply the whole determinant by 8, resulting in -48.

Is that right?

Answer 1

Hint:

Take a look at one such pair of matrices. Set $$A=\begin{bmatrix}2 & 0\\0&3\end{bmatrix}$$

Now, we replace the first row with $8\times$ second row to get

$$B=\begin{bmatrix}0 & 24\\0&3\end{bmatrix}$$

Now, ask yourself:

• What is the determinant of $B$ (and is it $48$?)
• When is the determinant of $B$ equal to what it is in my example? (think linear dependence)
• Does the same thing happen in al matrices?

Answer 2

Let's try:

$$ \left( \begin{array}{cc} 1 & 0\\0&6\end{array}\right)$$

has determinant 6

now let's replace the first row by the second row multiplied by 8

$$ \left( \begin{array}{cc} 0 & 48\\0&6\end{array}\right)$$

has determinant 0

can you see why this will always be the case?

also you may be confusing "replace" and "swap" if you swap the rows then indeed the sign changes. If you swap and multiply:

$$ \left( \begin{array}{cc} 0 & 48\\1&0\end{array}\right)$$

indeed the determinant is -48 (this is not what I would have understood from the question though)