0
$\begingroup$

Find the number k such that:

$$det\begin{bmatrix} 3a_1 & 2a_1 + a_2 - a_3 & a_3\\\ 3b_1 & 2b_1 + b_2 - b_3 & b_3\\\ 3c_1 & 2c_1 + c_2 - c_3 & c_3\end{bmatrix}$$

$$ = k \bullet det\begin{bmatrix} a_1 & a_2 & a_3\\\ b_1 & b_2 & b_3\\\ c_1 & c_2 & c_3\end{bmatrix}$$

Ive been working on this question for a while now, and I still cant seem to get it out. What I first thought was the best option would be to expand out the first matrix to get a value multiplied by the second matrix (this value would be k). However, I kept having trouble, and eventually tried a new approach. I tried to compute both determinates to find the value of k. Nevertheless, I was still unable to answer the question. Any help or hints would be much appreciated. Thanks in advance.

$\endgroup$
1
  • 1
    $\begingroup$ Do you know how column and row operations affect the determinant? What column operations can be performed on the second matrix to get the first? $\endgroup$
    – xxxxxxxxx
    May 28, 2019 at 23:36

3 Answers 3

2
$\begingroup$

Performing elementary column operations, $$det\begin{bmatrix} 3a_1 & 2a_1 + a_2 - a_3 & a_3\\\ 3b_1 & 2b_1 + b_2 - b_3 & b_3\\\ 3c_1 & 2c_1 + c_2 - c_3 & c_3\end{bmatrix}=$$

$$det\begin{bmatrix} 3a_1 & 2a_1 + a_2 & a_3\\\ 3b_1 & 2b_1 + b_2 & b_3\\\ 3c_1 & 2c_1 + c_2 & c_3\end{bmatrix}=$$

$$det\begin{bmatrix} 3a_1 & a_2 & a_3\\\ 3b_1 & b_2 & b_3\\\ 3c_1 & c_2 & c_3\end{bmatrix}=$$

$$3det\begin{bmatrix} a_1 & a_2 & a_3\\\ b_1 & b_2 & b_3\\\ c_1 & c_2 & c_3\end{bmatrix}$$

$\endgroup$
2
$\begingroup$

Hint: The most concise solution is to notice that $$\begin{bmatrix} 3a_1 & 2a_1 + a_2 - a_3 & a_3\\\ 3b_1 & 2b_1 + b_2 - b_3 & b_3\\\ 3c_1 & 2c_1 + c_2 - c_3 & c_3\end{bmatrix} = \begin{bmatrix} a_1 & a_2 & a_3\\\ b_1 & b_2 & b_3\\\ c_1 & c_2 & c_3\end{bmatrix} \begin{bmatrix} 3 & 2 & 0\\\ 0 & 1 & 0\\\ 0 & -1 & 1\end{bmatrix}$$ and then apply the property that $\det(AB) = \det(A)\det(B)$.

Alternatively, you can perform column operations to get from the first matrix to the second, and keep track of how those operations change the determinant.

$\endgroup$
0
$\begingroup$

Hint:

The determinant is multilinear, alternate function w.r.t. the columns. Hence, when you add to a column a linear combination of the other column, you don't change the determinant. Whan you multiply a column by a scalar, you multiply the determinant by this scalar. When you swap two columns, you change its sign, and, more generally, when you permute columns, you multiply the determinant by the signature of the permutation.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .