Skip to content
Exercise 12.1 · Q9

Q.(i) Let M={(xxxx):x∈R−{0}}M=\left\{\begin{pmatrix}x&x\\x&x\end{pmatrix} : x\in\mathbb{R}-\{0\}\right\} and let ∗* be the matrix multiplication. Determine whether MM is closed under ∗*. If so, examine the commutative and associative properties satisfied by ∗* on MM.

(ii) Let M={(xxxx):x∈R−{0}}M=\left\{\begin{pmatrix}x&x\\x&x\end{pmatrix} : x\in\mathbb{R}-\{0\}\right\} and let ∗* be the matrix multiplication. Determine whether MM is closed under ∗*. If so, examine the existence of identity, existence of inverse properties for the operation ∗* on MM.
Tamil Nadu DgeTextbookSubjectiveImportance★★★★★
11% · 9/84 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 →

Every element of MM has the special form Xx=(xxxx)X_x=\begin{pmatrix}x&x\\x&x\end{pmatrix} for some nonzero real xx; multiplying two such matrices collapses neatly, letting us track the whole operation through the single scalar xx.

Step 1. Multiply two general elements of MM. For Xx=(xxxx)X_x=\begin{pmatrix}x&x\\x&x\end{pmatrix} and Xy=(yyyy)X_y=\begin{pmatrix}y&y\\y&y\end{pmatrix}:

XxXy=(xxxx)(yyyy)=(xy+xyxy+xyxy+xyxy+xy)=(2xy2xy2xy2xy)=X2xy.X_x X_y=\begin{pmatrix}x&x\\x&x\end{pmatrix}\begin{pmatrix}y&y\\y&y\end{pmatrix}=\begin{pmatrix}xy+xy&xy+xy\\xy+xy&xy+xy\end{pmatrix}=\begin{pmatrix}2xy&2xy\\2xy&2xy\end{pmatrix}=X_{2xy}.

Step 2. Check closure. Since x,y≠0x,y\ne0, 2xy≠02xy\ne0, so X2xy∈MX_{2xy}\in M. MM is closed under ∗*.

Step 3. Commutative. Xx∗Xy=X2xyX_x*X_y=X_{2xy} and Xy∗Xx=X2yxX_y*X_x=X_{2yx}; since ordinary real multiplication is commutative, 2xy=2yx2xy=2yx, so Xx∗Xy=Xy∗XxX_x*X_y=X_y*X_x. Commutative.

Step 4. Associative. Matrix multiplication is associative in general, so ∗* is associative on MM too. (Directly: (Xx∗Xy)∗Xz=X2(2xy)z=X4xyz(X_x*X_y)*X_z=X_{2(2xy)z}=X_{4xyz} and Xx∗(Xy∗Xz)=X2x(2yz)=X4xyzX_x*(X_y*X_z)=X_{2x(2yz)}=X_{4xyz} -- equal.) Associative. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.