Accedmychevron_right11thchevron_rightmathchevron_rightMatrices And Determinantschevron_rightExercise 4.1

Solution

In ordinary algebra, we often expand using commutativity (i.e. ab=baab=ba). For matrices, in general:

ABBAAB\ne BA

So expansions that rely on combining ABAB and BABA are not valid.


(i) Why in general (A+B)2A2+2AB+B2(A+B)^2\ne A^2+2AB+B^2

Expand:

(A+B)2=(A+B)(A+B)=A2+AB+BA+B2(A+B)^2=(A+B)(A+B)=A^2+AB+BA+B^2

For it to equal A2+2AB+B2A^2+2AB+B^2, we would need AB+BA=2ABAB+BA=2AB, i.e. BA=ABBA=AB, which is not true in general.

Hence:

(A+B)2A2+2AB+B2 (in general)\boxed{(A+B)^2\ne A^2+2AB+B^2\ \text{(in general)}}

(ii) Why in general (AB)2A22AB+B2(A-B)^2\ne A^2-2AB+B^2

Expand:

(AB)2=(AB)(AB)=A2ABBA+B2(A-B)^2=(A-B)(A-B)=A^2-AB-BA+B^2

For it to equal A22AB+B2A^2-2AB+B^2, we would need ABBA=2AB-AB-BA=-2AB, i.e. BA=ABBA=AB, which is not true in general.

So:

(AB)2A22AB+B2 (in general)\boxed{(A-B)^2\ne A^2-2AB+B^2\ \text{(in general)}}

(iii) Why in general (A+B)(AB)A2B2(A+B)(A-B)\ne A^2-B^2

Expand:

(A+B)(AB)=A2AB+BAB2(A+B)(A-B)=A^2-AB+BA-B^2

For it to equal A2B2A^2-B^2, we would need AB+BA=0-AB+BA=0, i.e. BA=ABBA=AB, which is not true in general.

Thus:

(A+B)(AB)A2B2 (in general)\boxed{(A+B)(A-B)\ne A^2-B^2\ \text{(in general)}}