PPA-9

Question

Accept a square matrix \(A\) and an integer \(s\) as input and print the matrix \(s \cdot A\) as output. Multiplying a matrix by an integer \(s\) is equivalent to multiplying each element of the matrix by \(s\). For example: \[ 2 \cdot \begin{bmatrix} 1 & 2\\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 2 & 4\\ 6 & 8 \end{bmatrix} \]

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The first line of input is a positive integer, \(n\), that denotes the dimension of the matrix \(A\). Each of the next \(n\) lines contains a sequence of space-separated integers. The last line of the input contains the integer \(s\).

Print the matrix \(s \cdot A\) as output. Each row of the matrix must be printed as a sequence of space separated integers, one row on each line. There should not be any space after the last number on each line.

Hint

One way to do it is to accept the matrix \(A\) as input and then create a new matrix \(B\), such that: \[ B[i][j] = s \cdot A[i][j] \]

B = [ ]
n = len(A)
for i in range(n):
    row = [ ]
    for j in range(n):
        row.append(s * A[i][j])
    B.append(row)

Though this forms a part of a valid solution, is there another way of solving this problem?

Solution

n = int(input())
matrix = [ ]
for i in range(n):
    row = [ ]
    for x in input().split(' '):
        row.append(int(x))
    matrix.append(row)
s = int(input())

for i in range(n):
    for j in range(n):
        matrix[i][j] *= s

for i in range(n):
    for j in range(n):
        if j != n - 1:
            print(matrix[i][j], end = ' ')
        else:
            print(matrix[i][j])