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 — when a matrix is multiplied by a scalar, every element is multiplied by that scalar. The determinant of the resulting matrix is the scalar raised to the order of the matrix times the original determinant.
Step 1: Compute ∣A∣.
∣A∣=(1)(2)−(2)(4)=2−8=−6.
Step 2: Compute 2A:
2A=[2844].
Step 3: Compute ∣2A∣:
∣2A∣=(2)(4)−(4)(8)=8−32=−24.
Step 4: Compare: 4∣A∣=4(−6)=−24, which equals ∣2A∣.
✓Final answer
Hence ∣2A∣=4∣A∣ is verified.
The property ∣kA∣=kn∣A∣ for an n×n matrix is verified here: ∣2A∣=4∣A∣ because n=2, so 22=4.
The key idea is scalar multiplication of a determinant. When you multiply a matrix by a scalar k, every single entry gets multiplied by k. But a determinant is a multilinear function — it scales linearly with each row (or column). So if you have an n×n matrix, multiplying the whole matrix by k multiplies each of the n rows by k, and the determinant gets multiplied by k a total of n times. That is, ∣kA∣=kn∣A∣.
Here n=2, so ∣2A∣=22∣A∣=4∣A∣. Let's verify this concretely.
Compute ∣A∣ first.
For A=[1422],
∣A∣=(1)(2)−(2)(4)=2−8=−6.
Form 2A by multiplying every entry by 2.
2A=[2844].
Compute ∣2A∣.
∣2A∣=(2)(4)−(4)(8)=8−32=−24.
Compare ∣2A∣ with 4∣A∣.
4∣A∣=4×(−6)=−24.
They match exactly: ∣2A∣=−24=4∣A∣.
Watch out
A common mistake is to think ∣2A∣=2∣A∣. That would be true only for a 1×1 matrix. For a 2×2 matrix, the factor is 22=4. Always check the dimension n: ∣kA∣=kn∣A∣.
Tip
This property works for any square matrix. If you ever forget, just test with a simple 2×2 identity matrix: ∣2I∣=4, while 2∣I∣=2 — the factor 4 is correct.
✓Final answer
We have shown that ∣2A∣=−24=4∣A∣, so the statement holds.
Method: Verifying the Scalar-Multiplication Property ∣kA∣=kn∣A∣
Use this whenever asked to verify (or apply) how a determinant scales when the whole matrix is multiplied by a scalar.
Steps
Step 1: State the general property and why it holds
A determinant is multilinear — it scales by k for every row that gets multiplied by k. Multiplying the whole n×n matrix by k scales every row by k, so the determinant picks up a factor of k once per row:
∣kA∣=kn∣A∣.
Step 2: Compute ∣A∣ directly
Evaluate the original determinant using the standard ad−bc (or cofactor) method.
Step 3: Form kA by scaling every entry, then compute ∣kA∣ directly
Multiply each entry of A by k (not just one row), then evaluate the new determinant the same way.
Step 4: Confirm ∣kA∣=kn∣A∣
Compare the two computed numbers against kn times the original determinant to confirm they match.
Common Mistakes
Mistake 1: Assuming ∣kA∣=k∣A∣ for any matrix
Why it's wrong: this treats the matrix as if only a single row were scaled, but multiplying the whole matrix by k scales every row by k — for an n×n matrix the determinant picks up a factor of k once per row, giving kn, not k. Correct approach: always state ∣kA∣=kn∣A∣ and identify n from the actual order of the matrix before applying it.
Mistake 2: Only scaling one row or column when forming kA
Why it's wrong: kA means multiplying every entry of the matrix by k, not just one row — partial scaling gives a wrong intermediate matrix and a wrong determinant. Correct approach: multiply every single entry of A by k before computing ∣kA∣ directly, as a check against the kn∣A∣ formula.
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.
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).
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.