# What is actually a Determinant?(the body is long but to understand my question pls read it)

Okay so the way Ive been introduced to determinants is that if there is a system of two linear equations then we can represent the coefficients of the variables and the constants in the form of a matrix.

Now if we plot the matrices on the coordinate system then we will get a parallelogram and if we calculate the area of the parallelogram then we will get the determinant of the given matrix. For eg if A is the matrix then its determinant will be:

$$ad-cb$$.

i.e. |A|= $$ad-cb$$.

if A=$$\begin{bmatrix}a & b\\c & d\end{bmatrix}$$

Now the questions I want to ask:

1)What is a determinant actually what does it tells us about a system of equations?

2)The area found by the formula $$ad-cb$$, how is it telling us a determinant? Basically how the area of parallelogram telling the value of determinant?

3)In my book its given that: system of equations has a unique solution or not is determined by the number of ab-cd.What does this mean?

• Did you try reading the Wikipedia article Determinant? Did it answer your question? – Somos Mar 18 at 12:34

(1) A system of two linear equations in two variables can be written (in the standard way) as a matrix system $$A \vec x = \vec b$$, where $$A$$ is a $$2\times 2$$ matrix. Let's say its entries are $$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}.$$ The determinant of $$A$$ is the value $$ad-bc$$. It's (mostly) only important whether this value is zero or nonzero. If $$\det(A) = 0$$, the system of equations does not have a unique solution (meaning it has either no or more than one solution), no matter what $$\vec b$$ is; it may not have a solution at all, depending on $$\vec b$$. If $$\det(A) \ne 0$$, the system has a unique solution, no matter what $$\vec b$$ is. You can prove this quite easily in the case of $$2\times 2$$ matrices.

(2) The determinant is the value $$ad-bc$$. It happens to also give you the (signed) area of the parallelogram you're thinking of. That is a property of the determinant, but it would be a horrible definition of the determinant. It's hardly an obvious consequence of the definition of the determinant, either (unless you really only care about $$2\times 2$$ matrices, in which case, see @J.G.'s answer). But that's not to say you shouldn't keep in mind that it has this property.

(3) See the end of (1).

A determinant is basically a measure of volume (or area) together with the $$\pm$$ sign making it more strictly speaking an oriented volume.

So if you imagine your given basis and the linear transformation operating on them, then your determinant measures how the volume spanned by those basis vectors changes. The volume will be scaled by the determinant of the transformation. The $$\pm$$ sign corresponds to whether that volume will be 'flipped', loosely speaking.

But what happens if the determinant is zero? Then that means your volume must now be zero.

What does that mean for your system of equations? If you visualise your transformation as mapping vectors into some other vectors, then what you'll soon be able to imagine is that if your volume after the transformation has to be zero, so you're basically dropping down to a lower-dimensional space, then there's no way you can still have exactly one vector which transforms into whatever vector is on the right hand side of your equation $$Ax=b.$$ Otherwise you'd have a nonzero volume.

So if a determinant is zero, that means your system of equations couldn't possibly have a unique solution.

I'm not sure how questions 1), 2) differ, but I think you want a proof that the determinant is an area.

Note first that $$A$$ sends the usual basis elements to vectors given by $$A$$'s columns. If a parallelogram has edges equal to these vectors, which we'll call $$\vec{A}_1,\,\vec{A}_2$$, and the angle between these is $$\theta$$, the parallelogram's area is $$|\vec{A}_1||\vec{A}_2|\sin\theta$$. With the abbreviation $$v^2:=\vec{v}\cdot\vec{v}$$, the squared area is $$A_1^2A_2^2\sin^2\theta=A_1^2A_2^2-A_1^2A_2^2\cos^2\theta=A_1^2A_2^2-(\vec{A}_1\cdot\vec{A}_2)^2.$$This turns out to be $$(a^2+c^2)(b^2+d^2)-(ab+cd)^2=(ad-bc)^2,$$ i.e. the area is $$\pm(ad-bc)$$.

It can also be proven - but it's more complicated to do so - that in general if $$A\in\Bbb R^{n\times n}$$ then $$\det A$$ is, up to a sign, the measure of the parallelotope whose edges are $$A$$'s columns.

As for 3), $$A\vec{x}=\vec{b}$$ has a unique solution iff $$\det A\ne 0$$. By contrast, if $$\det A=0$$ then $$A\vec{x}=\vec{b}$$ has either no solutions or infinitely many of them, depending on the choice of $$\vec{b}$$.

• Actually i was asking what is the relation between determinant and the system of equations(with the help of which matrix has been formed) – Mad Dawg Mar 18 at 13:23