Row Matrix

A row Matrix is just a mathematical way of representing a list of numbers in horizontal form. We may encounter different sets of data which can be represented in a list or an array form, such as the price of different stationary items or marks got by a student in different subjects and many more. Writing this data in the form of a Row Matrix makes it simple to read and understand the data.

What is Row Matrix?

A Row Matrix is a matrix having a single row. Recall that a Matrix is a rectangular array of numbers or symbols which is used to represent some physical object, having numbers arranged in rows and columns. In mathematical form, A Matrix having m rows and n columns is represented by

\[A = {\left[ {{a_{ij}}} \right]_{m \times n}}\]

So, a Row Matrix will have only one row and n columns. Hence, a Row Matrix is represented by \[{[{a_{1j}}]_{1 \times n}}\]where \[{a_{1j}}\]is the (1,j)th entry of the matrix.

Representation and Size

A Row Matrix is written in the form:

\[{[{a_{1j}}]_{1 \times n}}\]where \[{a_{1j}}\]is the (1,j)th entry of the matrix. The number of entries of the matrix \[\left[a_{1j}\right]_{1 \times n}\] is n. The number of entities in a row matrix depends on the number of columns. In other words, the number of entities of a row matrix is equal to the number of columns. A Row Matrix always has only one row. It can have any number of columns. We say that a row Matrix has n columns. So, a Row Matrix is n-dimensional.

Examples of Row Matrix

• A Row Matrix of order 1×2 is \[\left[ {\begin{array}{*{20}{c}}2&5\end{array}} \right]\].

• A Row Matrix of size 5 is \[\left[ {\begin{array}{*{20}{c}}{2.5}&{3.6}&{5.1}&{ - 2.0}&{6.3}\end{array}} \right]\].

• Row Matrix containing 3 elements is \[\left[ {\begin{array}{*{20}{c}}{60}&{40}&{50}\end{array}} \right]\].

• Row Matrix of order 1×4 with alternative entries 1 and 0 is \[\left[ {\begin{array}{*{20}{c}}1&0&1&0\end{array}} \right]\].

Practical Uses of Row Matrix

• While sending a message over a network, the sender device encodes it into a binary format using an array of 0's and 1's in Row Matrix Form. This is known as Cryptography.

• We can store data such as the Family's total monthly cost using a one-dimensional horizontal matrix, i.e., a Row Matrix.

Properties of Row Matrix

• Commutativity

• \[A + B = B + A\]

• Associativity

• \[A + \left( {B + C} \right) = \left( {A + B} \right) + C\]

Operations on Row Matrix

1. Addition of Row Matrices

We can add two row matrices by simply adding their corresponding entries.

This is depicted as:

Given two matrices \[A = \left[ {\begin{array}{*{20}{c}}1&2&3\end{array}} \right]\] and \[B = \left[ {\begin{array}{*{20}{c}}2&3&4\end{array}} \right]\], We need to find \[C = A + B\]. Implies

\[\begin{array}{l}C = \left[ {\begin{array}{*{20}{c}}1&2&3\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}2&3&4\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{1 + 2}&{2 + 3}&{3 + 4}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}3&5&7\end{array}} \right]\end{array}\]

So, the addition of two matrices is \[\left[ {\begin{array}{*{20}{c}}3&5&7\end{array}} \right]\].

Similarly, we can add three matrices together by adding their corresponding entries as:

Given \[P = \left[ {\begin{array}{*{20}{c}}2&{2.5}&3&{1.5}\end{array}} \right]\], \[Q = \left[ {\begin{array}{*{20}{c}}{1.5}&{ - 1}&{ - 4.25}&5\end{array}} \right]\], \[R = \left[ {\begin{array}{*{20}{c}}0&{ - 1.5}&1&{ - 2.5}\end{array}} \right]\]. We need to find \[S = P + Q + R\]. Putting values of P, Q, and R, we get

\[S = \left[ {\begin{array}{*{20}{c}}2&{2.5}&3&{1.5}\end{array}\left] + \right[\begin{array}{*{20}{c}}{1.5\;\,}&{ - 1}&{ - 4.25}&5\end{array}\left] + \right[\begin{array}{*{20}{c}}{0\;}&{ - 1.5}&1&{ - 2.5}\end{array}} \right]\]

\[ = \left[ {\begin{array}{*{20}{c}}{2 + 1.5 + 0\;}&{2.5 + \left( { - 1} \right) + \left( { - 1.5} \right)}&{3 + \left( { - 4.25} \right) + 1\;}&{1.5 + 5 + \left( { - 2.5} \right)}\end{array}\;} \right]\]

\[ = {\rm{ }}\left[ {\begin{array}{*{20}{c}}{3.5}&0&{ - 0.25}&{4.0}\end{array}} \right]\]

2. Subtraction of Row Matrices

We can subtract one matrix from another by subtracting their corresponding ith entries. This is depicted as:

Given two matrices \[A = \left[ {\begin{array}{*{20}{c}}1&2&3\end{array}} \right]\] and\[B = \left[ {\begin{array}{*{20}{c}}2&3&4\end{array}} \right]\]. We need to find \[D = A - B\]. Implies

\[\begin{array}{l}D = \left[ {\begin{array}{*{20}{c}}1&2&3\end{array}} \right]{\rm{ }} - \left[ {\begin{array}{*{20}{c}}2&3&4\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{1 - 2}&{2 - 3}&{3 - 4}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - 1}&{ - 1}&{ - 1}\end{array}} \right]\end{array}\]

So, the subtraction of second matrix from first is \[\left[ {\begin{array}{*{20}{c}}{ - 1}&{ - 1}&{ - 1}\end{array}} \right]\].

Similarly, we can subtract two matrices from a given matrix together as:

Given \[P = \left[ {\begin{array}{*{20}{c}}2&{2.5}&3&{1.5}\end{array}} \right]\], \[Q = \left[ {\begin{array}{*{20}{c}}{1.5}&{ - 1}&{ - 4.25}&5\end{array}} \right]\], \[R = \left[ {\begin{array}{*{20}{c}}0&{ - 1.5}&1&{ - 2.5}\end{array}} \right]\]. We need \[T = P - Q - R\]. Putting values of P, Q, and R, we get

\[\begin{array}{l}S = \left[ {\begin{array}{*{20}{c}}2&{2.5}&3&{1.5}\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{1.5}&{ - 1}&{ - 4.25}&5\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}0&{ - 1.5}&1&{ - 2.5}\end{array}} \right]{\rm{ }}\\ = \left[ {\begin{array}{*{20}{c}}{2 - 1.5 - 0\;}&{2.5 - \left( { - 1} \right) - \left( { - 1.5} \right)}&{3 - \left( { - 4.25} \right) - 1\;}&{1.5 - 5 - \left( { - 2.5} \right)}\end{array}\;} \right]\\ = \left[ {\begin{array}{*{20}{c}}{0.5}&{5.0}&{6.25}&{ - 1}\end{array}{\rm{.0}}} \right]\end{array}\].

This can also be done as \[T = P - \left( {Q + R} \right)\].

Put \[Q + R = M\], we get

\[\begin{array}{*{20}{l}}{M = \left[ {\begin{array}{*{20}{c}}{1.5 + 0}&{ - 1 + \left( { - 1.5} \right)}&{\left( { - 4.25} \right) + 1}&{{\rm{ }}5 + \left( { - 2.5} \right)}\end{array}} \right]}\\{ = \left[ {\begin{array}{*{20}{c}}{1.5}&{ - 2.5}&{ - 3.25}&{2.5}\end{array}} \right]}\end{array}\]

Now,

\[\begin{array}{*{20}{l}}{S = P - M}\\\begin{array}{l} = \left[ {\begin{array}{*{20}{c}}2&{2.5}&3&{1.5}\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{1.5}&{ - 2.5}&{ - 3.25}&{2.5}\end{array}} \right]{\rm{ }}\\ = \left[ {\begin{array}{*{20}{c}}{0.5}&{5.0}&{6.25}&{ - 1.0}\end{array}} \right]\end{array}\end{array}\]

3. Multiplication of Row Matrix by a Scalar

We can multiply a matrix by a scalar as: Given \[B = [\begin{array}{*{20}{c}}{{b_{11}}}&{{b_{12}}}&{{b_{13}}}& \ldots &{{b_{1n}}}\end{array}]\]. Multiplying B by a scalar k, we get \[kB = k[\begin{array}{*{20}{c}}{{b_{11}}}&{{b_{12}}}&{{b_{13}}}& \ldots &{{b_{1n}}}\end{array}]\]

\[kB = [\begin{array}{*{20}{c}}{k{b_{11}}}&{k{b_{12}}}&{k{b_{13}}}& \ldots &{k{b_{1n}}}\end{array}]\].

For example:

\[\begin{array}{*{20}{l}}\begin{array}{l}A = {\left[ {\begin{array}{*{20}{c}}2&{ - 1}&3\end{array}} \right]_{1 \times 3}},k = 2,{\rm{ }}\\kA = 2A = 2\left[ {\begin{array}{*{20}{c}}2&{ - 1}&3\end{array}} \right]\end{array}\\\begin{array}{l} = \left[ {\begin{array}{*{20}{c}}{2(2)}&{2( - 1)}&{2(3)}\end{array}} \right]{\rm{ }}\\ = \left[ {\begin{array}{*{20}{c}}4&{ - 2}&6\end{array}} \right]\end{array}\end{array}\]

Solved Questions

1. Find the value of\[{\bf{A}} - {\bf{B}} + {\bf{2C}}\], where \[{\bf{A}} = \left[ {\begin{array}{*{20}{c}}1&2\end{array}} \right],{\rm{ }}{\bf{B}} = \left[ {\begin{array}{*{20}{c}}3&4\end{array}} \right],{\rm{ }}{\bf{C}} = \left[ {\begin{array}{*{20}{c}}{ - 1}&{ - 2}\end{array}} \right]\].

Answer:

\[\begin{array}{*{20}{l}}{Let {\rm{ }}S = A - B + 2C}\\{ = \left[ {\begin{array}{*{20}{c}}1&2\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}3&4\end{array}} \right] + 2\left[ {\begin{array}{*{20}{c}}{ - 1}&{ - 2}\end{array}} \right]}\\\begin{array}{l} = \left[ {\begin{array}{*{20}{c}}1&2\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}3&4\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 2}&{ - 4}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{1 - 3 - 2}&{2 - 4 - 4}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - 4}&{ - 6}\end{array}} \right]\end{array}\end{array}\]

2. Given\[{\bf{A}} = \left[ {\begin{array}{*{20}{c}}2&3&5\end{array}} \right],{\rm{ }}{\bf{B}} = \left[ {\begin{array}{*{20}{c}}{ - 1}&5&6\end{array}} \right],{\rm{ }}{\bf{M}} = \left[ {\begin{array}{*{20}{c}}{ - 4}&2&{ - 5}\end{array}} \right]\]. Find Matrix C for\[{\bf{M}} = {\bf{C}}+{\bf{3A}} + {\bf{2B}} \].

Answer:

Given \[{\bf{M}} = {\bf{3A}} + {\bf{2B}} +{\bf{C}}\]. Solving for C, we get \[C = M - 3A - 2B\]. Putting values of A, B, and M, we get

\[\begin{array}{*{20}{l}}{C = \left[ {\begin{array}{*{20}{c}}{ - 4}&2&{ - 5}\end{array}} \right] - 3\left[ {\begin{array}{*{20}{c}}2&3&5\end{array}} \right] - 2\left[ {\begin{array}{*{20}{c}}{ - 1}&5&6\end{array}} \right]}\\\begin{array}{l} = \left[ {\begin{array}{*{20}{c}}{ - 4}&2&{ - 5}\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}6&9&{15}\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{ - 2}&{10}&{12}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - 4 - 6 + 2}&{2 - 9 - 10}&{ - 5 - 15 - 12}\end{array}} \right]{\rm{ }}\\ = \left[ {\begin{array}{*{20}{c}}{ - 8}&{ - 17}&{ - 32}\end{array}} \right]\end{array}\end{array}\]

Practice Questions

1. If \[A = \left[ {\begin{array}{*{20}{c}}2&3&x\end{array}} \right]\], \[B = \left[ {\begin{array}{*{20}{c}}y&3&5\end{array}} \right]\] and A = B, then find the value of x and y.

Answer: x = 5, y = 2

2. If A and B two row matrices and AB exist, then find the number of columns of A.

Answer: The number of columns of A is 1.

3. Why does your Word document need to be formatted?

Conclusion

A Row Matrix is a horizontal matrix. It can also be called an array. Two or more Row Matrices can be added or subtracted if the order of both matrices is the same. We can multiply a scalar with a row matrix. Row matrix follows associativity property.