Question

NO LINKS OR ANSWERING QUESTIONS YOU DON'T KNOW!! Part 2.

Answer 1

9514 1404 393

Explanation:

Definition

The determinant of a square matrix is a single number that is computed (recursively) as the sum of products of the elements of a row or column and the determinants of their cofactors. The determinant of a single element is the value of that element.

The cofactor of an element in an n by n matrix is the (n-1) by (n-1) matrix that results when the row and column of that element are deleted. The "appropriate sign" of the element is applied to the cofactor matrix. The "appropriate sign" of an element is positive if the sum of its row and column numbers is even, negative otherwise. (Rows and columns are considered to be numbered 1 to n in an n by n matrix.)

Uses

The inverse of a square matrix is the transpose of the cofactor matrix, divided by the determinant. Hence if the determinant is zero, the inverse matrix is undefined. This means any system of equations the matrix might represent will have no distinct solution. (There may be zero solutions, or there may be an infinite number of solutions. The determinant by itself cannot tell you which.)

Cramer's Rule for the solution of linear systems of equations specifies that the value of any given variable is the ratio of the determinants of two matrices. The numerator matrix is the original matrix with the coefficients of the variable replaced by the constants in the standard-form equations; the denominator matrix is the original coefficient matrix. This rule lets you solve a system of 3 equations in 3 variables by computing 3+1 = 4 determinants, for example.

Answer 2

Let's look at an example.

If we wanted to solve this system of equations

$\begin{cases}2x-y = 2\\x+y = 7\end{cases}$

Then it's equivalent to solving this matrix equation

$\begin{bmatrix}2 & -1\\1 & 1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}2\\7\end{bmatrix}$

We can then further condense that into the form

$Aw = B$

Where,

$A = \begin{bmatrix}2 & -1\\1 & 1\end{bmatrix}\\\\w = \begin{bmatrix}x\\y\end{bmatrix}\\\\B = \begin{bmatrix}2\\7\end{bmatrix}$

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To solve the matrix equation Aw = B, we could compute the inverse matrix $A^{-1}$ and left-multiply both sides by this to isolate w.

So we'd go from $Aw=B$ to $w = A^{-1}*B$. The order of multiplication is important.

For any 2x2 matrix of the form

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

its inverse is

$P^{-1} = \frac{1}{ad-bc}\begin{bmatrix}d & -b\\-c & a\end{bmatrix}$

Notice the expression ad-bc in the denominator of that fractional term outside. This $ad-bc$ expression represents the determinant of matrix P. Some books may use the notation "det" to mean "determinant"

$P^{-1} = \frac{1}{\det(P)}\begin{bmatrix}d & -b\\-c & a\end{bmatrix}$

or you may see it written as

$P^{-1} = \frac{1}{|P|}\begin{bmatrix}d & -b\\-c & a\end{bmatrix}$

Those aren't absolute value bars, even if they may look like it.

Based on that, we can see that the determinant must be nonzero in order to compute the inverse of the matrix. Consequently, the determinant must be nonzero in order for Aw = B to have one solution.

If the determinant is 0, then we have two possibilities:

1. There are infinitely many solutions (aka the system is dependent)
2. There are no solutions (the system is inconsistent)

So a zero determinant would have to be investigated further as to which outcome would occur.

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Let's return to the example and compute the inverse (if possible).

$A = \begin{bmatrix}2 & -1\\1 & 1\end{bmatrix}\\\\A^{-1} = \frac{1}{2*1 - (-1)*1}\begin{bmatrix}1 & 1\\-1 & 2\end{bmatrix}\\\\A^{-1} = \frac{1}{3}\begin{bmatrix}1 & 1\\-1 & 2\end{bmatrix}$

In this case, the inverse does exist.

This further leads to

$w = A^{-1}*B\\\\w = \frac{1}{3}\begin{bmatrix}1 & 1\\-1 & 2\end{bmatrix}*\begin{bmatrix}2\\7\end{bmatrix}\\\\w = \frac{1}{3}\begin{bmatrix}1*2+1*7\\-1*2+2*7\end{bmatrix}\\\\w = \frac{1}{3}\begin{bmatrix}9\\12\end{bmatrix}\\\\w = \begin{bmatrix}(1/3)*9\\(1/3)*12\end{bmatrix}\\\\w = \begin{bmatrix}3\\4\end{bmatrix}\\\\\begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}3\\4\end{bmatrix}$

This shows that the solution is (x,y) = (3,4).

As the other person pointed out, you could use Cramer's Rule to solve this system. Cramer's Rule will involve using determinants and you'll be dividing over determinants. So this is another reason why we cannot have a zero determinant.