Answer
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Hint: At first, we will find the matrix of order 3 which is commutative with every square matrix of order 3.
Then, by using the trace of the matrix we will find the elements of the matrix.
Then we will find the determinant of \[A.\]
Complete step-by-step answer:
It is given that, \[A\] is a diagonal matrix of order \[3 \times 3\]. And it is commutative with every square matrix of order \[3 \times 3\] under multiplication.
Also, trace \[A = 12\]
For a diagonal matrix of any order, other than the diagonal elements all are zero.
Let us consider the diagonal matrix of order \[3 \times 3\] is commutative with every square matrix of order \[3 \times 3\] under multiplication as,
\[A = \left( {\begin{array}{*{20}{c}}
x&0&0 \\
0&x&0 \\
0&0&x
\end{array}} \right)\]
It is given that trace\[A = 12\].
That means the sum of the diagonal elements are \[ = 12\]
So, we can write as,
\[x + x + x = 12\]
On simplifying the above equation we get,
\[x = 4\]
So, the matrix is \[A = \left( {\begin{array}{*{20}{c}}
4&0&0 \\
0&4&0 \\
0&0&4
\end{array}} \right)\]
Now, we will find the determinant.
\[\det A = \left| {\begin{array}{*{20}{c}}
4&0&0 \\
0&4&0 \\
0&0&4
\end{array}} \right|\]
Let us solve the det A, then we get,
\[\det A = 4(4 \times 4 - 0) - 0 + 0 = 64\]
Hence, \[\det A = 64\]
The correct answer is option A, \[\left| A \right| = 64\].
Note:
A square matrix in which every element except the principal diagonal elements is zero, it is called a Diagonal Matrix. The general form of a diagonal matrix of order three is, \[A = \left( {\begin{array}{*{20}{c}}
{{a_{11}}}&0&0 \\
0&{{a_{22}}}&0 \\
0&0&{{a_{33}}}
\end{array}} \right)\]
If A is a square matrix of order 3, then the Trace of A denoted tr(A) is the sum of all of the entries in the main diagonal. It can be written as,
tr(A) \[ = {a_{11}} + {a_{22}} + {a_{33}}\]
Let us consider two matrices \[A,B\] which are square matrices, and are called commutative under multiplication operation, we have \[AB = BA\].
Then, by using the trace of the matrix we will find the elements of the matrix.
Then we will find the determinant of \[A.\]
Complete step-by-step answer:
It is given that, \[A\] is a diagonal matrix of order \[3 \times 3\]. And it is commutative with every square matrix of order \[3 \times 3\] under multiplication.
Also, trace \[A = 12\]
For a diagonal matrix of any order, other than the diagonal elements all are zero.
Let us consider the diagonal matrix of order \[3 \times 3\] is commutative with every square matrix of order \[3 \times 3\] under multiplication as,
\[A = \left( {\begin{array}{*{20}{c}}
x&0&0 \\
0&x&0 \\
0&0&x
\end{array}} \right)\]
It is given that trace\[A = 12\].
That means the sum of the diagonal elements are \[ = 12\]
So, we can write as,
\[x + x + x = 12\]
On simplifying the above equation we get,
\[x = 4\]
So, the matrix is \[A = \left( {\begin{array}{*{20}{c}}
4&0&0 \\
0&4&0 \\
0&0&4
\end{array}} \right)\]
Now, we will find the determinant.
\[\det A = \left| {\begin{array}{*{20}{c}}
4&0&0 \\
0&4&0 \\
0&0&4
\end{array}} \right|\]
Let us solve the det A, then we get,
\[\det A = 4(4 \times 4 - 0) - 0 + 0 = 64\]
Hence, \[\det A = 64\]
The correct answer is option A, \[\left| A \right| = 64\].
Note:
A square matrix in which every element except the principal diagonal elements is zero, it is called a Diagonal Matrix. The general form of a diagonal matrix of order three is, \[A = \left( {\begin{array}{*{20}{c}}
{{a_{11}}}&0&0 \\
0&{{a_{22}}}&0 \\
0&0&{{a_{33}}}
\end{array}} \right)\]
If A is a square matrix of order 3, then the Trace of A denoted tr(A) is the sum of all of the entries in the main diagonal. It can be written as,
tr(A) \[ = {a_{11}} + {a_{22}} + {a_{33}}\]
Let us consider two matrices \[A,B\] which are square matrices, and are called commutative under multiplication operation, we have \[AB = BA\].
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