Determinant of a matrix

**BAdhi** · November 9th 2010, 06:25 PM

if $A$ is a $nxn$ matrix and $X$ is a column vector where

$AX=0$

can anyone tell me why we take $|A|$ should be equal to 0 when $X$ is not equal to 0

**pickslides** · November 9th 2010, 06:40 PM

Originally Posted by BAdhi

if $AX=0$

can anyone tell me why we take $|A|$ should be equal to 0 when $X$ is not equal to 0

For this equation to be satisfied then $A$ must have at least one all zero row vector. As a consequence of this $|A|=0$

**BAdhi** · November 10th 2010, 07:13 PM

pickslides, can you tell me why A must have at least one all zero row vector to satisfy this equation?

**ineedyourhelp** · November 10th 2010, 07:22 PM

Firstly,

When you have 2 things multiply by each other and the product is zero, if the first (x) is not zero, the second (A) must be zero. Determinant, loosely speaking, is finding the magnitude of the matrix A.

As for why at least one all zero row vector, this is not always true. A matrix with at least one all zero row vector will gives determinant zero, but having determinant zero does not mean it has one all zero row vector (eg 2by2 matrix with all element 1).

**Drexel28** · November 10th 2010, 07:24 PM

Originally Posted by BAdhi

pickslides, can you tell me why A must have at least one all zero row vector to satisfy this equation?

So, let's clarify. For simplicity's sake (guessing at the level of course you are in) we have that $A\in\text{Mat}_n\left(\mathbb{R}\right)$ . Thus, the question is asking, if $\begin{bmatrix}x_1\\ \vdots\\ x_n\end{bmatrix}\in\mathbb{R}^n$ is non-zero and $A\begin{bmatrix}x_1\\ \vdots\\ x_n\end{bmatrix}=\begin{bmatrix}0\\ \vdots\\ 0\end{bmatrix}$ why $\det A=0$ ? What it wasn't? Then $A$ would be invertible right? What would happen then if we multiplied both sides of $A\begin{bmatrix}x_1\\ \vdots\\ x_n\end{bmatrix}=\begin{bmatrix}0\\ \vdots\\ 0\end{bmatrix}$ ?

**Drexel28** · November 10th 2010, 07:26 PM

Originally Posted by ineedyourhelp

When you have 2 things multiply by each other and the product is zero, if the first (x) is not zero, the second (A) must be zero. Determinant, loosely speaking, is finding the magnitude of the matrix A.

Be careful with what you're saying here. Is this universally true?

**ineedyourhelp** · November 10th 2010, 07:32 PM

First sentence or second sentence?

**Drexel28** · November 10th 2010, 07:56 PM

Originally Posted by ineedyourhelp

First sentence or second sentence?

I mean, you didn't qualify in what context you were speaking. The function $\mathbf{1}_{\mathbb{R}^-}$ and $\mathbf{1}_{[0,\infty)]$ are both non-zero but their product is zero.

**ineedyourhelp** · November 10th 2010, 09:07 PM

Sorry my bad. Yes the product of 2 non zero matrix can be zero. see (Multiplication of Matrices) if need example. In the original question, we need the determinant of A to be zero for AX=0.

One of the way of calculating det, is to reduce it to reduced echelon form. Now we need det(A)=0 if AX=0 for nonzero x.

In order for det(A)=0, one of the element must be 0, as long as at least one of the element in the main diagonal is zero in reduced echelon form. In order for 1 element in the main diagonal to be zero after reducing, at least 1 row or 1 column must be zeroes in the original matrix.

Another way for calculating det is to use the cofactor expansion in which we will choose the most convenient row/column, which in this case, is the all-zero vector row/column, which will give us det(A)=0 as well.

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