I have trouble solving the following question: Question

How I approached it initially was as follow:

1) Since the questions asks for x3 only, I replaced the x3 column with the particular solution. Step 1

2) Cramer's Rule asks of us to divide the det(Substituted Matrix in Step 1) by the det(Initial Matrix) Step 2

3) To find the det of each 4x4 matrix I did C(1,1)(det(3x3 sub Matrix)). So 1(det(3x3 sub Matrix)). I evaluated the 3x3 sub Matrix as follow: 3x3 sub Matrix The red arrows are added and then subtracted to the addition of the blue arrows. Thus det(Substituted Matrix) = 3aei + 3bfg + 2cdh - 3ceg - 2afh - 3bdi.

4) I did step 3 for the initial Matrix as well and got det(Initial Matrix) = aei + bfg + cdh - ceg - afh - bdi

Now, I am unsure how to work around the determinants I just got. I should factor out something and then cancel the rest, but I don't see what. Did I miss something?


It is clear that $\begin{bmatrix}x_1\\x_2\\x_3\\x_4 \end{bmatrix} = \begin{bmatrix}0\\0\\3\\2 \end{bmatrix}$ without Cramer's rule...

What does Cramer say?

$x_1 = \frac {\det\begin{bmatrix}0&0&0&0\\3b+2c&a&b&c\\3e+2f&d&e&f\\3h+2i&g&h&i \end{bmatrix}}{\det\begin{bmatrix}1&0&0&0\\0&a&b&c\\0&d&e&f\\0&g&h&i \end{bmatrix}} = 0$

We know this because one row is all 0's

$x_2 = \frac {\det\begin{bmatrix}1&0&0&0\\0&3b+2c&b&c\\0&3e+2f&e&f\\0&3h+2i&h&i \end{bmatrix}}{\det\begin{bmatrix}1&0&0&0\\0&a&b&c\\0&d&e&f\\0&g&h&i \end{bmatrix}} = 0$

We know this because clearly the columns are not linearly independent.

$x_3 = \frac {\det\begin{bmatrix}1&0&0&0\\0&a&3b+2c&c\\0&d&3e+2f&f\\0&g&3h+2i&i \end{bmatrix}}{\det\begin{bmatrix}1&0&0&0\\0&a&b&c\\0&d&e&f\\0&g&h&i \end{bmatrix}} = 0$

rather than calculating this out.

$\begin{bmatrix}1&0&0&0\\0&a&3b+2c&c\\0&d&3e+2f&f\\0&g&3h+2i&i \end{bmatrix} = \begin{bmatrix}1&0&0&0\\0&a&b&c\\0&d&e&f\\0&g&h&i \end{bmatrix}\begin{bmatrix}1&0&0&0\\0&2&3&0\\0&0&1&0\\0&0&0&1 \end{bmatrix}$

The determinant of the product equals the product of the determinants. And

$\det\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&3&0\\0&0&2&1 \end{bmatrix} = 2$

And $x_4$ can be approached the same way.

  • $\begingroup$ You should get $3$. The last matrix is off slightly. $\endgroup$ – Chris Custer Dec 13 '18 at 1:32
  • $\begingroup$ I think it's $\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&3&0\\0&0&2&1\end{pmatrix}$. $\endgroup$ – Chris Custer Dec 13 '18 at 1:45
  • $\begingroup$ @ChrisCuster thanks, I something came up and rushed the ending. $\endgroup$ – Doug M Dec 13 '18 at 16:38

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