# Using the properties of determinants to computer for the matrix determinant.

So I have here a matrix $$X =$$ $$\begin{bmatrix}3&0&-1&0\\0&2&-4&0\\0&-4&7&0\\-3&0&1&1\end{bmatrix}$$

The questions are:

• (a) Use cofactor expansion to compute the determinant of $$X$$
• (b) Using properties of determinants, compute $$\det X.$$

I was already able to compute for $$\det X$$ using cofactor expansion but I'm having trouble with $$b.$$ I already looked at the properties of determinants and I can't seem to find any properties that I could apply to solve for it. Could you guys help me how to approach this problem?

If you replace the fourth row with its sum with the first row, you will get another matrix with the same determinant, but now that determinant is much easier to compute.

• I can't seem to determine what property is this? – user831749 Oct 3 '20 at 7:43
• It's the proporty that it says that if you add to a row (or a column) a multiple of another row (or column), that will not change the determinant. It's property 14 from this list. – José Carlos Santos Oct 3 '20 at 7:50
• I see. Thank you so much! – user831749 Oct 3 '20 at 7:52

Use the following properties:

• Adding a row (or column) of X multiplied by a scalar $$k$$ to another row (or column) of $$X$$, then the determinant will not change.
• Triangle property: If all the elements of a determinant above or below the main diagonal consist of zeros, then the determinant is equal to the product of diagonal elements.

See 1 and 2 for more properties.

• So the first property you mentioned is just saying that if you do elementary row operations on a matrix, it doesn't change its determinant? Is it right that you could do multiple and the property still holds? Also, I don't understand how will I able to apply the triangle property? Should I first perform row operations to get to the point where "all the elements of a determinant above or below the main diagonal consist of zeros"? – user831749 Oct 3 '20 at 8:12
• Yes, you apply the triangle property after the row operations. See this for the proof of the first property. – Äres Oct 3 '20 at 8:24