# When A and B are of different order given the $\det(AB)$,then calculate $\det(BA)$

Let 'A' be a $2 \times 3$ matrix where as B be a $3 \times 2$ matrix if $\det(AB) = 4$ the find value of the $\det(BA)$

My attempt: I took A = $$\begin{bmatrix} 2 & 0 &0\\ 0 & 0 &2\\ \end{bmatrix}$$
B= $$\begin{bmatrix} 1 & 0 \\ 0 & 0 \\ 0 & 1 \\ \end{bmatrix}$$

It satisfies given condition and I get $\det(BA)=0$
But I have not proved it
How do I prove that it is always zero
(background)I am 12th grader and I know about adjoint,inverse,determinant,rank of a matrix and the other basics. However I do NOT know about eigenvalues and eigenvectors.

• Are you familiar with the concept of rank of a matrix? – Guy May 12 '16 at 16:51
• yes I know basics of rank – Gopalkrishna Nayak May 12 '16 at 16:52
• What is the rank of $BA$? Regarding your example, what can we say about $BA$ if we know that $\det\left(BA\right)=0$? – Guy May 12 '16 at 16:53
• In my example rank is less than 3 but is it always less than 3? – Gopalkrishna Nayak May 12 '16 at 16:57
• Yes, when you multiply a 3 by 2 matrix with a 2 by 3 matrix, you arrive at a 3 by 3 matrix whose rank cannot be 3 hence it's singular. – imranfat May 12 '16 at 17:06

You will always get $\det(BA)=0$. The reason for this is very simple : $BA$ is a $3\times 3$ matrix with rank at most $2$; thus it is not invertible and thus it has $0$ determinant.

One possible definition of the rank is that it is the dimension of the subspace generated by the columns or the lines of the matrices. The lines of $BA$ are obtained by taking linear combinations of the lines of $A$ : for example, if $$B=\begin{pmatrix}b_{11} & b_{12} \\ b_{21} & b_{22} \\ b_{31}& b_{32} \end{pmatrix},\ A= \begin{pmatrix}a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23} \end{pmatrix},$$ then the first line of $BA$ is $$b_{11}\cdot \begin{pmatrix}a_{11} & a_{12} & a_{13} \end{pmatrix} + b_{12}\cdot \begin{pmatrix} a_{21} & a_{22} & a_{23} \end{pmatrix}.$$

Since the lines of $BA$ are generated by only two lines (of $A$), the dimension of the generated subspace cannot be more than $2$.

This also means that the lines of $BA$ cannot be linearly independent; thus there must be some linear relation between them, which means that when you convert it to echelon matrix you will certainly get a line of $0$s.

• Why exactly is it rank 2 I know only basics of rank could you explain it to me? – Gopalkrishna Nayak May 12 '16 at 16:56
• What definition of the rank are you used to? – Arnaud D. May 12 '16 at 16:58
• The definition I use is that I convert it into echelon matrix and calculate non zero rows – Gopalkrishna Nayak May 12 '16 at 17:01
• Ah yes, it is not really obvious then. I'll edit my answer to take that into account. – Arnaud D. May 12 '16 at 17:03
• @GopalkrishnaNayak I've tried to make it clearer, let me know if it is not clear. – Arnaud D. May 12 '16 at 17:18

Let $T = BA$, then $T : \mathbb{R}^3 \to \mathbb{R}^3$ is a linear transformation with nontrivial kernel, as $A : \mathbb{R}^3 \to \mathbb{R}^2$ cannot be an injection. Thus, it is not invertible, and $\det(T) = 0$.

• The kernel of a linear transformation is its null space, i.e the set of all vectors that are mapped to zero by the transformation. Written formally, this means that for $T : V \to W$, we have $\ker T = \{ v \in V : T(v) = 0_W \}$. – Starfall May 12 '16 at 17:09
• If you are not familiar with vector spaces and linear maps, then simply take $v = (0, 1, 0)^T$ and consider $(BA)v = 0$. What does the existence of a nontrivial solution to the homogeneous system tell you about $BA$? – Starfall May 12 '16 at 17:15
• if $(BA)v=0$ the middle row elements of $BA$ must be 0 – Gopalkrishna Nayak May 13 '16 at 6:06
• I am not even sure what you meant by that, but it should tell you that $BA$ is not full rank. – Starfall May 13 '16 at 9:21
• What I meant was $BA$ is <br> $$\begin{bmatrix} a1 & 0 &a3\\ b1 & 0 &b3\\ c1 & 0 &c3\\ \end{bmatrix}$$ – Gopalkrishna Nayak May 13 '16 at 11:25

I think what you are looking for is an answer like this

$B=\begin{pmatrix}b_{11} & b_{12} \\ b_{21} & b_{22} \\ b_{31}& b_{32} \end{pmatrix}$, $A= \begin{pmatrix}a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23} \end{pmatrix}$,

Then \begin{align}BA & =\begin{pmatrix}b_{11}a_{11}+b_{12}a_{21} & b_{11}a_{12}+b_{12}a_{22} & b_{11}a_{13}+b_{12}a_{23}\\ b_{21}a_{11}+b_{22}a_{21} & b_{21}a_{12}+b_{22}a_{22} &b_{21}a_{13}+b_{22}a_{23} \\b_{31}a_{11}+b_{32}a_{21} & b_{31}a_{12}+b_{32}a_{22} & b_{31}a_{13}+b_{32}a_{23} \end{pmatrix} \\ & =\begin{pmatrix}b_{11}a_{11}+b_{12}a_{21}+0.0 & b_{11}a_{12}+b_{12}a_{22}+0.0 & b_{11}a_{13}+b_{12}a_{23}+0.0\\ b_{21}a_{11}+b_{22}a_{21}+0.0 & b_{21}a_{12}+b_{22}a_{22}+0.0 &b_{21}a_{13}+b_{22}a_{23}+0.0 \\b_{31}a_{11}+b_{32}a_{21}+0.0 &b_{31}a_{12}+b_{32}a_{22}+0.0 & b_{31}a_{13}+b_{32}a_{23} +0.0 \end{pmatrix}\\ & =\begin{pmatrix}b_{11} & b_{12} & 0\\ b_{21} & b_{22} &0 \\b_{31} & b_{32} & 0 \end{pmatrix} \begin{pmatrix}a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} &a_{23} \\0 & 0 & 0 \end{pmatrix}.\end{align}

Let $C=\begin{pmatrix}b_{11} & b_{12} & 0\\ b_{21} & b_{22} &0 \\b_{31} & b_{32} & 0 \end{pmatrix}$ and $D= \begin{pmatrix}a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} &a_{23} \\0 & 0 & 0 \end{pmatrix}$.

Now, $BA=CD$. Therefore, $\det(BA)=\det(CD)=\det(C)\cdot \det(D)=0\cdot 0=0$.