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:
The scalar that, when multiplied by any matrix, yields the null matrix is the number zero (0). This follows directly from the definition of scalar multiplication of matrices: every entry of the matrix is multiplied by the scalar, and only multiplying by zero makes every entry zero.
The core idea. Scalar multiplication of a matrix is entry-wise. If you have a matrix A=[aij] and a scalar k, then kA=[k⋅aij]. The result is the null matrix (every entry is 0) if and only if k⋅aij=0 for every single entry aij in the matrix.
Why it must be zero. For this to be true for any matrix A, the scalar k must work regardless of what the entries aij are. Consider a matrix that contains a non-zero entry, say A=[1]. For kA to be the null matrix [0], we need k⋅1=0. The only number that satisfies this is k=0.
Checking the other direction. If k=0, then for any matrix A, every entry of 0⋅A is 0⋅aij=0. So the result is always the null matrix. …
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:
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: