Question

If A and B are square matrices of the same order, explain, why in general
(i) (A + B)² ≠ A² + 2AB + B²
(ii) (A − B)² ≠ A² − 2AB + B²
(iii) (A + B) (A − B) ≠ A² − B².

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{LHS}={\left(A+B\right)}^2 \\ =\left(A+B\right)\left(A+B\right) \\ =A\left(A+B\right)+B\left(A+B\right) \\ =A^2+AB+BA+B^2 \end{gathered}$$

We know that a matrix does not have commutative property. So,
AB ≠ BA
Thus,
${\left(A+B\right)}^2$ ≠ $A^2+2AB+B^2$

$$\begin{gathered} \left(ii\right)\mathrm{LHS}={\left(A-B\right)}^2 \\ =\left(A-B\right)\left(A-B\right) \\ =A\left(A-B\right)-B\left(A-B\right) \\ =A^2-AB-BA+B^2 \end{gathered}$$

We know that a matrix does not have commutative property. So,
AB ≠ BA
Thus,
${\left(A-B\right)}^2$ ≠ $A^2-2AB+B^2$

$$\begin{gathered} \left(iii\right)\mathrm{LHS}=\left(A+B\right)\left(A-B\right) \\ =A\left(A-B\right)+B\left(A-B\right) \\ =A^2-AB+BA-B^2 \end{gathered}$$

We know that a matrix does not have commutative property. So,
AB ≠ BA
Thus,
$\left(A+B\right)\left(A-B\right)$ ≠ $A^2-B^2$