Solveeit Logo

Question

Question: Two rectangular matrices of order \(n\times m\) and \(m\times k\) are multiplied in the same order. ...

Two rectangular matrices of order n×mn\times m and m×km\times k are multiplied in the same order. The resulting matrix formed is a
This question has multiple correct options
A. Rectangular Matrix of order n×kn\times k
B. Square matrix of order m
C. Square matrix of order n if n = k
D. Rectangular Matrix of order k×nk\times n

Explanation

Solution

Hint: If there are m rows and n columns in the matrix, then the matrix is called a rectangular matrix of order m×nm\times n and a matrix having equal number of rows and columns is called square matrix of order n.

Complete step by step answer:
Let us consider the example,
Let A be the matrix of order 3×23\times 2 and B be another matrix of order 2×12\times 1, where A=[1 3 1  2 0 1 ]3×2A={{\left[ \begin{matrix} 1 \\\ 3 \\\ 1 \\\ \end{matrix}\text{ }\begin{matrix} 2 \\\ 0 \\\ 1 \\\ \end{matrix} \right]}_{3\times 2}} and B=[2 1 ]2×1B={{\left[ \begin{matrix} 2 \\\ 1 \\\ \end{matrix} \right]}_{2\times 1}}
Now, AB=[1 3 1  2 0 1 ]3×2[2 1 ]2×1=[2+2 3+0 2+1 ]3×1=[4 3 3 ]3×1AB={{\left[ \begin{matrix} 1 \\\ 3 \\\ 1 \\\ \end{matrix}\text{ }\begin{matrix} 2 \\\ 0 \\\ 1 \\\ \end{matrix} \right]}_{3\times 2}}{{\left[ \begin{matrix} 2 \\\ 1 \\\ \end{matrix} \right]}_{2\times 1}}={{\left[ \begin{matrix} 2+2 \\\ 3+0 \\\ 2+1 \\\ \end{matrix} \right]}_{3\times 1}}={{\left[ \begin{matrix} 4 \\\ 3 \\\ 3 \\\ \end{matrix} \right]}_{3\times 1}}
Two rectangular matrices of order 3×23\times 2 and 2×12\times 1 are multiplied, then the resulting matrix is a rectangular matrix of the order 3×13\times 1.

Hence multiplying two rectangular matrices of order n×mn\times m and m×km\times k is a rectangular matrix of order n×kn\times k.

Therefore, the correct option is option (a).

Also, consider another example,
Let C be the matrix of order 2×32\times 3 and D be another matrix of order 3×23\times 2, where C=[2 1  01 12 ]2×3C={{\left[ \begin{matrix} 2 \\\ 1 \\\ \end{matrix}\text{ }\begin{matrix} 0 & 1 \\\ 1 & 2 \\\ \end{matrix} \right]}_{2\times 3}} and D=[2 1 1  0 3 1 ]3×2D={{\left[ \begin{matrix} 2 \\\ 1 \\\ 1 \\\ \end{matrix}\text{ }\begin{matrix} 0 \\\ 3 \\\ 1 \\\ \end{matrix} \right]}_{3\times 2}}

Now, CD=[2 1  01 12 ]2×3[2 1 1  0 3 1 ]3×2=[4+0+1 2+1+2  0+0+1 0+3+2 ]2×2=[5 5  1 5 ]2×2CD={{\left[ \begin{matrix} 2 \\\ 1 \\\ \end{matrix}\text{ }\begin{matrix} 0 & 1 \\\ 1 & 2 \\\ \end{matrix} \right]}_{2\times 3}}{{\left[ \begin{matrix} 2 \\\ 1 \\\ 1 \\\ \end{matrix}\text{ }\begin{matrix} 0 \\\ 3 \\\ 1 \\\ \end{matrix} \right]}_{3\times 2}}={{\left[ \begin{matrix} 4+0+1 \\\ 2+1+2 \\\ \end{matrix}\text{ }\begin{matrix} 0+0+1 \\\ 0+3+2 \\\ \end{matrix} \right]}_{2\times 2}}={{\left[ \begin{matrix} 5 \\\ 5 \\\ \end{matrix}\text{ }\begin{matrix} 1 \\\ 5 \\\ \end{matrix} \right]}_{2\times 2}}
Two rectangular matrices of order 2×32\times 3 and 3×23\times 2 are multiplied, then the resulting matrix is a square matrix of the order 2×22\times 2.

Hence multiplying two rectangular matrices of order n×mn\times m and m×km\times k is a square matrix of order n if n = k.

Therefore, the correct option is option (c).

Hence correct options for the given question are option (A) and option (C).

Note: The possibility for the mistake is that you might get confused with the concept that two matrices A and B are said to be conformable for the product AB if the number of columns in A is equal to the number of rows in B.