If $A=\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}$ and $B=\begin{bmatrix}0 & -1\\1 & 0\end{bmatrix}$ show that $(A+B)(A-B)\neq A^2-B^2$.

If A= 0 1; 1 0 and B= 0 -1; 1 0 show that (A+B)(A-B)≠ A²-B².

Miscellaneous Exercise 4(B) · Q201

A=\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★est

95% · 201/212 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

$A=\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}$ , $B=\begin{bmatrix}0 & -1\\1 & 0\end{bmatrix}$ .

$A+B=\begin{bmatrix}0 & 0\\2 & 0\end{bmatrix}$ , $A-B=\begin{bmatrix}0 & 2\\0 & 0\end{bmatrix}$ .

$(A+B)(A-B)=\begin{bmatrix}0 & 0\\2 & 0\end{bmatrix}\begin{bmatrix}0 & 2\\0 & 0\end{bmatrix}=\begin{bmatrix}0 & 0\\0 & 4\end{bmatrix}$ .

$A^2=\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}=I$ ; $B^2=\begin{bmatrix}0 & -1\\1 & 0\end{bmatrix}\begin{bmatrix}0 & -1\\1 & 0\end{bmatrix}=\begin{bmatrix}-1 & 0\\0 & -1\end{bmatrix}=-I$ .

$A^2-B^2=I-(-I)=2I=\begin{bmatrix}2 & 0\\0 & 2\end{bmatrix}$ . …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ200 — If A= -3 2; 2 -4 , B= 1 x; y 0 , and (A+B)(A-B)=A²-B², find x and y.→Next questionQ202 — If A= 2 -1; 3 -2 , find A³.

Q.If $A=\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}$ and $B=\begin{bmatrix}0 & -1\\1 & 0\end{bmatrix}$ show that $(A+B)(A-B)\neq A^2-B^2$ .

Concept understanding — Multiplication of Matrices

Q.If A=[0110]A=\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}A=[01​10​] and B=[0−110]B=\begin{bmatrix}0 & -1\\1 & 0\end{bmatrix}B=[01​−10​] show that (A+B)(A−B)≠A2−B2(A+B)(A-B)\neq A^2-B^2(A+B)(A−B)=A2−B2.

🎓Unlock everything free for 14 days

Q.If $A=\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}$ and $B=\begin{bmatrix}0 & -1\\1 & 0\end{bmatrix}$ show that $(A+B)(A-B)\neq A^2-B^2$ .

Unlock everything free for 14 days