Mathematics Class 12 — Part…
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3.5.3 Matrix Multiplication

🎓 Class 12 Mathematics CBSE Theory Ch 3 — Matrices ⏱ ~15 min
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This MCQ module is based on: 3.5.3 Matrix Multiplication

This mathematics assessment will be based on: 3.5.3 Matrix Multiplication
Targeting Class 12 level in Matrices, with Advanced difficulty.

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3.5.3 Matrix Multiplication

Matrix product
For \(A=[a_{ij}]\) of order \(m\times p\) and \(B=[b_{jk}]\) of order \(p\times n\) (the inner dimensions match), the product \(AB\) is an \(m\times n\) matrix with \[\boxed{\;(AB)_{ik}=\sum_{j=1}^{p}a_{ij}\,b_{jk}\;}\] That is, the \((i,k)\) entry of \(AB\) is the dot product of the \(i\)-th row of \(A\) with the \(k\)-th column of \(B\). If the inner dimensions don't match, \(AB\) is undefined.

Example: \(A=\begin{bmatrix}1&2\\3&4\end{bmatrix}\) (2×2), \(B=\begin{bmatrix}5&6\\7&8\end{bmatrix}\) (2×2). Inner dimension: 2=2. Product is 2×2: \[AB=\begin{bmatrix}1\cdot 5+2\cdot 7 & 1\cdot 6+2\cdot 8\\ 3\cdot 5+4\cdot 7 & 3\cdot 6+4\cdot 8\end{bmatrix}=\begin{bmatrix}19 & 22\\ 43 & 50\end{bmatrix}.\]

Properties of multiplication

Five key properties
  1. Associative: \((AB)C=A(BC)\) (when shapes are compatible).
  2. Distributive: \(A(B+C)=AB+AC\) and \((A+B)C=AC+BC\).
  3. Identity: \(I_m A=A I_n=A\) for an \(m\times n\) matrix \(A\).
  4. Zero product: \(O\,A=O\) and \(A\,O=O\) (when shapes match).
  5. NOT commutative: in general \(AB\ne BA\). Sometimes one is defined and the other isn't.
Critical pitfalls
  • \(AB=O\) does NOT imply \(A=O\) or \(B=O\). Matrices have "zero divisors": e.g. \(\begin{bmatrix}1&0\\0&0\end{bmatrix}\begin{bmatrix}0&0\\0&1\end{bmatrix}=O\).
  • Cancellation: \(AB=AC\) does NOT imply \(B=C\) (even when \(A\ne O\)).
  • You must respect the order: \((AB)^k\ne A^kB^k\) in general.

3.6 Transpose of a Matrix

Transpose
The transpose of an \(m\times n\) matrix \(A=[a_{ij}]\) is the \(n\times m\) matrix \(A'\) (or \(A^T\)) with entries \[(A')_{ij}=a_{ji}.\] Equivalently: rows become columns and vice versa.

Example: \(A=\begin{bmatrix}1 & 2 & 3\\ 4 & 5 & 6\end{bmatrix}\Rightarrow A'=\begin{bmatrix}1 & 4\\ 2 & 5\\ 3 & 6\end{bmatrix}\).

Properties of transpose
  1. \((A')'=A\) (taking transpose twice returns the original).
  2. \((kA)'=kA'\) for scalar \(k\).
  3. \((A+B)'=A'+B'\).
  4. \((AB)'=B'\,A'\) — note the order REVERSAL.

3.6.1 Symmetric and Skew-Symmetric Matrices

Symmetric / Skew-symmetric
A square matrix \(A\) is
  • symmetric if \(A'=A\), i.e. \(a_{ij}=a_{ji}\) for all \(i,j\). Mirror across the main diagonal.
  • skew-symmetric if \(A'=-A\), i.e. \(a_{ij}=-a_{ji}\). Setting \(i=j\): \(a_{ii}=-a_{ii}\Rightarrow a_{ii}=0\). So diagonal of a skew-symmetric matrix is all zero.
Decomposition theorem
Every square matrix \(A\) can be uniquely written as the sum of a symmetric matrix \(P\) and a skew-symmetric matrix \(Q\): \[A=P+Q,\quad P=\dfrac{1}{2}(A+A'),\quad Q=\dfrac{1}{2}(A-A').\] Verify: \(P'=P\) and \(Q'=-Q\).

3.7 Invertible Matrices

Inverse
A square matrix \(A\) of order \(n\) is invertible if there exists a square matrix \(B\) of the same order satisfying \[AB=BA=I_n.\] The matrix \(B\) is unique (when it exists) and is called the inverse of \(A\), denoted \(A^{-1}\).
Properties of inverse
  1. \((A^{-1})^{-1}=A\).
  2. \((AB)^{-1}=B^{-1}A^{-1}\) — note order reversal, like transpose.
  3. \((A')^{-1}=(A^{-1})'\) — transpose and inverse commute.
  4. The identity is its own inverse: \(I_n^{-1}=I_n\).

Not every matrix has an inverse. Necessary and sufficient: \(\det A\ne 0\) (next chapter, Determinants).

Worked Examples

Example 6. \(A=\begin{bmatrix}1 & 2 & 3\\ 4 & 5 & 6\end{bmatrix},\ B=\begin{bmatrix}1\\2\\3\end{bmatrix}\). Find \(AB\). Is \(BA\) defined?
\(A\) is 2×3, \(B\) is 3×1, inner dim 3=3. Product is 2×1.
\(AB=\begin{bmatrix}1+4+9\\ 4+10+18\end{bmatrix}=\begin{bmatrix}14\\ 32\end{bmatrix}\). \(BA\): 3×1 times 2×3, inner 1≠2 — NOT defined.
Example 7. Show that \(AB\) and \(BA\) are not equal in general. Take \(A=\begin{bmatrix}1&0\\0&-1\end{bmatrix}\), \(B=\begin{bmatrix}0&1\\1&0\end{bmatrix}\).
\(AB=\begin{bmatrix}0&1\\-1&0\end{bmatrix}\), \(BA=\begin{bmatrix}0&-1\\1&0\end{bmatrix}\). They differ — not commutative.
Example 8. Find the transpose of \(A=\begin{bmatrix}3 & \sqrt 3 & 2\\ 4 & 2 & 0\end{bmatrix}\) and verify \((A')'=A\).
\(A'=\begin{bmatrix}3 & 4\\ \sqrt 3 & 2\\ 2 & 0\end{bmatrix}\). Transpose again gives back the original \(A\). ✓
Example 9. If \(A=\begin{bmatrix}1 & -1\\ 2 & 3\end{bmatrix}\) and \(B=\begin{bmatrix}3 & 1\\ -1 & 2\end{bmatrix}\), verify \((AB)'=B'A'\).
\(AB=\begin{bmatrix}4 & -1\\ 3 & 8\end{bmatrix}\Rightarrow (AB)'=\begin{bmatrix}4 & 3\\ -1 & 8\end{bmatrix}\). \(B'A'=\begin{bmatrix}3 & -1\\ 1 & 2\end{bmatrix}\begin{bmatrix}1 & 2\\ -1 & 3\end{bmatrix}=\begin{bmatrix}4 & 3\\ -1 & 8\end{bmatrix}\). Match. ✓
Example 10. Express \(A=\begin{bmatrix}2 & 4\\ -1 & 3\end{bmatrix}\) as the sum of a symmetric and a skew-symmetric matrix.
\(A'=\begin{bmatrix}2 & -1\\ 4 & 3\end{bmatrix}\). \(P=\frac{1}{2}(A+A')=\begin{bmatrix}2 & 3/2\\ 3/2 & 3\end{bmatrix}\) (symmetric ✓). \(Q=\frac{1}{2}(A-A')=\begin{bmatrix}0 & 5/2\\ -5/2 & 0\end{bmatrix}\) (skew ✓). Verify \(P+Q=A\). ✓
Example 11. Show that \(B=\begin{bmatrix}2 & -3\\ -1 & 2\end{bmatrix}\) is the inverse of \(A=\begin{bmatrix}2 & 3\\ 1 & 2\end{bmatrix}\).
\(AB=\begin{bmatrix}4-3 & 6-6\\ 2-2 & 3-4\end{bmatrix}=\begin{bmatrix}1 & 0\\ 0 & -1\end{bmatrix}\). Hmm, that's NOT \(I_2\). Let me re-check: \(B=\begin{bmatrix}2 & -3\\ -1 & 2\end{bmatrix}\), \(AB_{11}=2\cdot 2+3\cdot(-1)=1\); \(AB_{12}=2\cdot(-3)+3\cdot 2=0\); \(AB_{21}=1\cdot 2+2\cdot(-1)=0\); \(AB_{22}=1\cdot(-3)+2\cdot 2=1\). So \(AB=I_2\). Similarly \(BA=I_2\). Hence \(B=A^{-1}\).
Activity: Verify Non-Commutativity
L3 Apply
Materials: Pen, paper.
Predict: For 2×2 matrices, are there ANY pairs (A, B) where AB = BA? Will it always fail?
  1. Compute AB and BA for several pairs. Try \(A=\begin{bmatrix}1&1\\0&1\end{bmatrix}\) and \(B=\begin{bmatrix}1&0\\1&1\end{bmatrix}\) — not equal.
  2. Try \(A=I_2\) and any B: \(IB=BI=B\) — they DO commute (identity is the universal commuter).
  3. Try \(A=\begin{bmatrix}2&0\\0&3\end{bmatrix}\) (diagonal) and \(B=\begin{bmatrix}5&0\\0&7\end{bmatrix}\) — both diagonal, so AB=BA (entrywise multiplication on diagonal).
  4. Conclude: commutativity is the exception, not the rule. Diagonal pairs and identity-pairs commute; most others don't.
Two matrices commute iff they are simultaneously diagonalisable (have a common eigenbasis). Most pairs don't share that — so non-commutativity is generic. This is why "matrix algebra ≠ scalar algebra": you cannot freely re-order products.

Competency-Based Questions

Scenario: Two states A, B have item costs (pencil ₹5, notebook ₹50). State A bought 3 pencils, 4 notebooks. State B bought 2 pencils, 2 notebooks. Encode as matrices and use multiplication to compute total bills.
Q1. The total cost matrix for both states is best computed by multiplying:
L3 Apply
Answer: Quantity matrix \(Q=\begin{bmatrix}3&4\\ 2&2\end{bmatrix}\) (state × item) times price column \(P=\begin{bmatrix}5\\ 50\end{bmatrix}\) gives \(QP=\begin{bmatrix}215\\ 110\end{bmatrix}\) — the totals for state A and state B.
Q2. (T/F) "If \(A^2=I\), then \(A=I\) or \(A=-I\)." Justify with a counter-example or proof.
L5 Evaluate
False. \(A=\begin{bmatrix}1 & 0\\ 0 & -1\end{bmatrix}\) has \(A^2=I\) but is neither \(I\) nor \(-I\). Many "involutions" (matrices with A²=I) exist.
Q3. If A is a symmetric matrix and B is a skew-symmetric matrix of the same order, prove AB+BA is skew-symmetric.
L4 Analyse
Solution: Use \((AB+BA)'=B'A'+A'B'=(-B)A+A(-B)=-BA-AB=-(AB+BA)\). Hence \((AB+BA)'=-(AB+BA)\), i.e. AB+BA is skew-symmetric. \(\square\)
Q4. Compute \(\begin{bmatrix}\cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha\end{bmatrix}\begin{bmatrix}\cos\beta & -\sin\beta\\ \sin\beta & \cos\beta\end{bmatrix}\) and identify the result.
L4 Analyse
Solution: Using cos(α+β)=cosα·cosβ−sinα·sinβ and sin(α+β)=sinα·cosβ+cosα·sinβ, the product equals \(\begin{bmatrix}\cos(\alpha+\beta) & -\sin(\alpha+\beta)\\ \sin(\alpha+\beta) & \cos(\alpha+\beta)\end{bmatrix}\). Rotation by \(\alpha\) followed by rotation by \(\beta\) is rotation by \(\alpha+\beta\) — exactly what matrix multiplication encodes for 2-D rotations.
Q5. Design a 3×3 symmetric matrix and a 3×3 skew-symmetric matrix whose sum is \(A=\begin{bmatrix}1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9\end{bmatrix}\).
L6 Create
Solution: \(A'=\begin{bmatrix}1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9\end{bmatrix}\). \(P=\frac{1}{2}(A+A')=\begin{bmatrix}1 & 3 & 5\\ 3 & 5 & 7\\ 5 & 7 & 9\end{bmatrix}\) (symmetric). \(Q=\frac{1}{2}(A-A')=\begin{bmatrix}0 & -1 & -2\\ 1 & 0 & -1\\ 2 & 1 & 0\end{bmatrix}\) (skew). \(P+Q=A\) ✓.

Assertion–Reason Questions

Assertion (A): Matrix multiplication is not commutative.
Reason (R): If A is m × p and B is p × n, then AB is m × n; whereas BA requires n = m and is generally a different shape.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a). Even when both products exist (m=n=p), they need not be equal — but R also captures the more basic shape obstruction.
Assertion (A): The diagonal entries of any skew-symmetric matrix are zero.
Reason (R): \(a_{ii}=-a_{ii}\) implies \(a_{ii}=0\).
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a). Setting \(i=j\) in the skew condition gives R; R says A.
Assertion (A): \((AB)^{-1}=B^{-1}A^{-1}\).
Reason (R): The inverse "reverses" the order of multiplication.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a). Verify: \((AB)(B^{-1}A^{-1})=A(BB^{-1})A^{-1}=AIA^{-1}=AA^{-1}=I\). The order reversal is essential.

Frequently Asked Questions

How do you multiply two matrices?
For A of order m × p and B of order p × n (inner dimensions match), the product AB has order m × n. Each (i,j) entry is the dot product of row i of A with column j of B.
Is matrix multiplication commutative?
NO. In general AB ≠ BA. Sometimes one product is even undefined when the other is. Matrix multiplication is associative and distributive, but NOT commutative.
What is the transpose of a matrix?
The transpose A' of an m × n matrix A is the n × m matrix obtained by interchanging rows and columns: (A')_ij = a_ji.
What is a symmetric matrix?
A square matrix A is symmetric if A' = A, i.e. a_ij = a_ji for all i, j.
What is a skew-symmetric matrix?
A square matrix A is skew-symmetric if A' = -A. Diagonal entries must be zero.
What is an invertible matrix?
A square matrix A is invertible if there exists B with AB = BA = I. The unique B is denoted A⁻¹.
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