You have a table of a shopkeeper's prices arranged as a matrix. Suddenly every price must be doubled for a festival, or cut to 90% in a sale. You don't want to touch each number one by one — you want a single instruction: multiply the whole matrix by a number. That number is called a scalar, and the operation is scalar multiplication.
The Idea
To multiply a matrix A by a scalar k, you multiply every entry of A by k. Nothing else changes — the order (size) of the matrix stays exactly the same.
If A=[aij]m×n and k is a real number, then
kA=[kaij]m×n
An Example
A=[20−14],3A=[3⋅23⋅03⋅(−1)3⋅4]=[60−312]
A negative scalar flips every sign. In particular −A=(−1)A, which is exactly the matrix you use to subtract: A−B=A+(−1)B.
Properties (all inherited from ordinary numbers)
For scalars k,l and matrices A,B of the same order:
k(A+B)=kA+kB (distributes over matrix addition)
(k+l)A=kA+lA (distributes over scalar addition)
k(lA)=(kl)A
1⋅A=A and 0⋅A=O (the zero matrix)
Tip
Scalar multiplication and matrix addition together let you write neat combinations like 2A−3B: scale each matrix first, then add. This is the building block behind linear combinations of matrices.
Bottom line: scaling a matrix by k just scales each entry by k — the shape is untouched, and it behaves with all the friendly rules of number multiplication.
Scalar multiplication of a matrix, where every entry is multiplied by the same constant, is covered in the CBSE Class 12 Matrices chapter alongside matrix addition, and "scalar multiplication of matrix properties" is a commonly searched topic for quick revision. This operation, combined with addition, is what allows students to simplify expressions like 2A − 3B in board exam and JEE Main matrix questions.
Concept: Scalar Multiplication — multiplying a matrix by a constant multiplies every entry.
We treat the matrix equation 2A+3X=5B like a scalar equation — isolate X by subtracting 2A and dividing by 3. The result is X=31(5B−2A), which we compute entrywise to get X=−24−331−310314−37.
The core idea here is scalar multiplication — matrices obey the same algebraic rules as numbers when it comes to addition and multiplication by a constant. So you can solve for an unknown matrix X exactly as you would solve 2a+3x=5b for a number x. The only difference is that the operations are performed entry by entry.
Let’s walk through it.
Isolate the term containing X.
Start with 2A+3X=5B. Subtract 2A from both sides:
3X=5B−2A
Divide both sides by 3.
Since scalar multiplication is just multiplying every entry, dividing by 3 means multiplying by 31:
X=31(5B−2A)
Compute 5B and 2A separately.
Multiply each entry of B by 5:
5B=5×24−5−221=1020−25−10105
Multiply each entry of A by 2:
2A=2×8430−26=16860−412
Subtract 2A from 5B.
Subtract corresponding entries:
Multiply by 31 to get X.
Divide every entry by 3:
X=31−612−31−1014−7=−24−331−310314−37
Watch out
A common mistake is to forget that division by a scalar applies to every entry — not just the first row or first column. Also, be careful with signs when subtracting: 10−(−4)=14, not 6.
Tip
You can check your answer by plugging X back into 2A+3X and verifying you get 5B. It’s a quick sanity check that catches arithmetic errors.
✓Final answer
The required matrix is X=−24−331−310314−37.
Method: Solving a matrix equation for an unknown matrix using scalar algebra
Use this for equations like 2A+3X=5B where X is an unknown matrix.
Steps
Step 1: Isolate the term containing X as in scalar algebra.
Treat A,B,X like numbers for +, −, and scalar multiples: 3X=5B−2A.
Step 2: Compute the scalar multiples and combine entry-wise.
Form 5B and 2A by scaling every entry, then subtract corresponding entries.
Step 3: Divide by the scalar to get X.
Multiply the result by 31 (every entry), leaving fractional entries in exact form.
Common Mistakes
Mistake 1: Dividing only some entries by the scalar.
Why it's wrong: 31(5B−2A) divides every entry by 3; a partial division gives a wrong matrix. Correct approach: apply the 31 to all entries.
Mistake 2: Sign slips in the subtraction.
Why it's wrong: e.g. 10−(−4)=14, not 6. Correct approach: subtract each entry of 2A from the matching entry of 5B, keeping signs.
Mistake 3: Leaving improper or rounded fractions.
Why it's wrong: entries like −310 should stay exact, not be rounded. Correct approach: keep fractions in exact form.
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
KCET 2021Set A-11 markMCQ
Q.If A and B are matrices of order 3 and ∣A∣=5, ∣B∣=3 then ∣3AB∣ is
(A) 425
(B) 405
(C) 565
(D) 585
›Reveal solutionSolution
Use the two determinant laws ∣kA∣=kn∣A∣ (order n) and ∣AB∣=∣A∣∣B∣.
Step 1 — The scalar-multiple law, and why the power n appears.
Multiplying a matrix by a scalar k multiplies every one of its n rows by k. A determinant is linear in each row separately, so each row contributes one factor of k:
∣kA∣=kn∣A∣(A of order n).
Here n=3, so ∣3AB∣=33∣AB∣=27∣AB∣.
Step 2 — The product law.
∣AB∣=∣A∣⋅∣B∣=5×3=15.
Step 3 — Combine.
∣3AB∣=27×15=405.
Common trap: writing ∣3AB∣=3∣A∣∣B∣=45, i.e. forgetting to raise 3 to the power of the order. The presence of 405=27×15 among the options is exactly the check that the cube was applied.
✓Final answer
The correct option is (B) — 405.
ANSWER: B
KCET 2023Set A-21 markMCQ
Q.If a+2b+3c=0 and (a×b)+(b×c)+(c×a)=λ(b×c) then the value of λ is equal to
(A) 3
(B) 4
(C) 6
(D) 2
›Reveal solutionSolution
Eliminate a using the linear relation, then reduce every cross product to a multiple of b×c using v×v=0 and anti-commutativity.
The scalar triple product simplifies using linearity and the fact that any repeated vector makes the product zero. The final result is 3[a,b,c], which corresponds to option (D).
The scalar triple product [x,y,z] is defined as x⋅(y×z). It is linear in each argument, and it changes sign when two arguments are swapped. A key property: if any two vectors are the same (or linearly dependent), the triple product is zero. This problem is all about using these properties to expand a complicated-looking expression into simpler pieces.
We are given:
[a+2b−c,a−b,a−b−c]
Let’s denote the three vectors as:
u=a+2b−c,v=a−b,w=a−b−c
We need to compute [u,v,w].
Use linearity in the first argument.
The triple product is linear in each slot. So expand u:
[a+2b−c,v,w]=[a,v,w]+2[b,v,w]−[c,v,w]
Now expand each of these three terms using linearity in the second and third arguments.
Start with [a,v,w] where v=a−b and w=a−b−c:
[a,a−b,a−b−c]=[a,a,a−b−c]−[a,b,a−b−c]
The first term [a,a,…]=0 because two arguments are identical. So:
[a,v,w]=−[a,b,a−b−c]
Now expand the third argument:
−[a,b,a−b−c]=−[a,b,a]+[a,b,b]+[a,b,c]
The first two terms are zero (repeated vectors). So:
[a,v,w]=[a,b,c]
Next, compute [b,v,w].
[b,a−b,a−b−c]=[b,a,a−b−c]−[b,b,a−b−c]
The second term is zero. So:
[b,v,w]=[b,a,a−b−c]
Expand the third argument:
[b,a,a]−[b,a,b]−[b,a,c]
The first two terms are zero. So:
[b,v,w]=−[b,a,c]
Swapping two arguments changes sign: [b,a,c]=−[a,b,c]. Therefore:
[b,v,w]=−(−[a,b,c])=[a,b,c]
Finally, compute [c,v,w].
[c,a−b,a−b−c]=[c,a,a−b−c]−[c,b,a−b−c]
Expand each:
First term: [c,a,a]−[c,a,b]−[c,a,c]=0−[c,a,b]−0=−[c,a,b]
Second term: [c,b,a]−[c,b,b]−[c,b,c]=[c,b,a]−0−0=[c,b,a]
So:
[c,v,w]=−[c,a,b]−[c,b,a]
But [c,b,a]=−[c,a,b] (swap the last two). So:
[c,v,w]=−[c,a,b]−(−[c,a,b])=0
Put it all together.
From step 1:
[u,v,w]=[a,v,w]+2[b,v,w]−[c,v,w]
Substitute the results:
=[a,b,c]+2[a,b,c]−0=3[a,b,c]
Watch out
A common mistake is forgetting the sign change when swapping arguments. For example, [b,a,c]=−[a,b,c], not equal. Always track the sign carefully.
Tip
You can often skip full expansion by noticing patterns. Here, the first and third vectors differ by 2b and the second vector is a−b. The symmetry leads to a clean multiple of [a,b,c] without messy algebra.
✓Final answer
The value is 3[a,b,c], so the correct option is (D).