Q.A manufacturer produces three products x,y,z which he sells in two markets. Annual sales are indicated below:
Market I — x: 10,000, y: 2,000, z: 18,000
Market II — x: 6,000, y: 20,000, z: 8,000
(a) If unit sale prices of x,y and z are ₹ 2.50, ₹ 1.50 and ₹ 1.00, respectively, find the total revenue in each market with the help of matrix algebra.
(b) If the unit costs of the above three commodities are ₹ 2.00, ₹ 1.00 and 50 paise respectively. Find the gross profit.
Matrix Multiplication Compatibility: The Inner-Dimensions Rule
You cannot multiply just any two matrices. Multiplication is defined only when their sizes line up in a specific way, and this compatibility check is always the very first step of any product.
The Idea: A Row Meets a Column
When you multiply A by B, you take each row of A and pair it against each column of B, multiply corresponding entries, and add. For that pairing to work, a row of A must have exactly as many entries as a column of B.
Note
Think of a handshake: each finger of one hand must meet a finger of the other. If one hand has 4 fingers and the other has 3, the handshake fails.
The Precise Statement
Let A be m×n and B be p×q.
A×B is defined if and only if n=p — the number of columns of A equals the number of rows of B. The product C=AB then has order m×q.
Writing the sizes side by side, (m×n)(p×q), the inner numbers (n,p) must match; the outer numbers (m,q) give the result's shape.
Why the Rule Exists
Each entry of the product is
cij=∑k=1naikbkj.
Here k runs over the columns of A (up to n) and the rows of B (up to p). If n=p, the sum runs out of matching terms and is meaningless — that is exactly why compatibility demands n=p.
Watch out
Even when both AB and BA are defined, they usually differ. For A of order 2×3 and B of order 3×2, AB is 2×2 but BA is 3×3 — different sizes entirely. Matrix multiplication is not commutative.
Treating the sales table as a 2×3 matrix and the prices/costs as 3×1 columns: revenue is SP=[4600053000], and the gross profit (total revenue minus total cost) is ₹32,000.
Multiplying a row of quantities by a column of prices sums each product's quantity times its price, so stacking both markets into one matrix gives all the totals in a single product.
Method: Modelling revenue / cost / profit with a quantity–price matrix product
Use this for word problems where quantities and unit prices (or costs) must be combined "with the help of matrix algebra".
Steps
Step 1: Arrange quantities as a matrix and prices as a column.
Put the sales quantities in a matrix S (one row per market/seller, one column per product) and the unit prices in a column vector P (one entry per product). The inner dimensions must match so SP is defined.
Mistake 1: Multiplying in an order whose shapes don't line up.
Why it's wrong: to get a per-market total you need (markets × products)(products × 1); reversing the factors is undefined or gives the wrong meaning. Correct approach: check that the number of products (columns of S) equals the rows of the price column.
Mistake 2: Not converting paise to rupees consistently. …
The expression A2+B(A+B) simplifies to (A+B)2−AB by using the distributive property of matrix multiplication. We calculate (A+B)2 and then subtract AB to find the result 431646623.
The core idea here is to simplify the given expression A2+B(A+B) using the properties of matrix algebra, specifically the distributive property of matrix multiplication over addition. We are given A+B and AB, so we should try to express the target expression in terms of these known quantities.
First, let's expand B(A+B):
B(A+B)=BA+B2
So, the expression we need to evaluate becomes:
A2+B(A+B)=A2+BA+B2
Now, let's recall the expansion of (A+B)2 for matrices:
For matrices A and B, (A+B)2=(A+B)(A+B)=A(A+B)+B(A+B)=A2+AB+BA+B2.
Watch out
It is crucial to remember that matrix multiplication is generally not commutative, meaning AB=BA. Therefore, (A+B)2 is not equal to A2+2AB+B2 unless AB=BA.
Comparing the expression we need, A2+BA+B2, with the expansion of (A+B)2, which is A2+AB+BA+B2, we can see a direct relationship.
If we subtract AB from (A+B)2, we get:
(A+B)2−AB=(A2+AB+BA+B2)−AB
=A2+(AB−AB)+BA+B2
=A2+0+BA+B2
=A2+BA+B2
This shows that A2+B(A+B)=(A+B)2−AB. This simplification is key because we are given A+B and AB directly.