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The matrix, A=\[\left( {\begin{array}{*{20}{c}}
  0&0&4 \\
  0&4&0 \\
  4&0&0
\end{array}} \right)\] is a
A. Square matrix
B. Diagonal matrix
C. Unit matrix
D. None of these

seo-qna
Last updated date: 19th Jun 2024
Total views: 403.5k
Views today: 9.03k
Answer
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Hint: In this question, we will first see the definition of square matrix. Then we will see the definition of diagonal matrix and unit matrix. The definition of these types of matrices helps us to get the answer for the question.

Complete step-by-step answer:
The given matrix is A=\[\left( {\begin{array}{*{20}{c}}
  0&0&4 \\
  0&4&0 \\
  4&0&0
\end{array}} \right)\] .
Now first let’s talk about a square matrix.
Square matrix-
If the number of rows of a matrix is equal to the number of columns then it is called a square matrix.
It is denoted as a matrix of order$n \times n$, where ‘n’ denotes the number of rows or number of columns.
In the given matrix A, the number of rows = the number of columns = 3.
So, it is a square matrix of order 3.
Now let’s talk about a diagonal matrix.
Diagonal matrix-
 If a matrix is having only diagonal elements and rest all elements are zero then it is said to be a diagonal matrix. Thus in diagonal matrix ${a_{ij}} = 0{\text{ ; i}} \ne {\text{j}}$ and ${a_{ij}} \ne 0{\text{; i = j}}$.
In the given matrix, non- diagonal elements are non-zero. So, the given matrix is not a diagonal matrix.

Now let’s talk about a unit matrix.
Unit matrix-
It is a diagonal matrix whose each of diagonal element is 1 and non-diagonal elements are zero or we can say that ${a_{ij}} = 0{\text{ ; i}} \ne {\text{j}}$ and ${a_{ij}}{\text{ = 1; i = j}}$.
Since the given matrix is not a diagonal matrix. So, it cannot be a unit matrix.

So, the correct answer is “Option A”.

Note: While solving these types of problems it’s better to have the knowledge about different types of matrix like scalar, unit, square, upper triangular, lower triangular etc. Direct applications of their definitions are quite common in exams. Minimum number of zeros in a diagonal matrix = n (n – 1).