Question

Let \[A\] be a \[3 \times 3\] matrix and determinant of the matrix is 4. Let \[{R_i}\] denote the \[{i^{th}}\] row of \[A\]. If matrix \[B\] is obtained by performing the row operation \[{R_2} \to 2{R_2} + 5{R_3}\] on a matrix \[2A\]. Then find the determinant of matrix \[B\].
A. 64
B. 16
C. 80
D. 128

Answer 1

Hint In the given question, a \(3 \times 3\) matrix \(A\) and its determinant are given. By performing the given operations, we will find the matrix \(2A\) and matrix \(B\). After that we will perform a row operation \[{R_2} \to 2{R_2} + 5{R_3}\]. Then we will apply the row operation \[{R_2} \to {R_2} - 5{R_3}\]. Then by using the determinant formula of a \(3 \times 3\) matrix, we will get the determinant of matrix \(B\).

Formula used
Determinant of a \[3 \times 3\] matrix \[M = \left( {\begin{array}{*{20}{c}}{m{a_{11}}}&{m{a_{12}}}&{m{a_{13}}}\\{n{a_{21}}}&{n{a_{22}}}&{n{a_{23}}}\\{p{a_{31}}}&{p{a_{32}}}&{p{a_{33}}}\end{array}} \right)\] is,
\[\left| M \right| = m \times n \times p \times \left| {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\{{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\{{a_{31}}}&{{a_{32}}}&{{a_{33}}}\end{array}} \right|\]

Complete step by step solution:
It is given that the order of matrix \[A\] is \[3 \times 3\] and value of the determinant is 4.
Let consider
\[A = \left( {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\{{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\{{a_{31}}}&{{a_{32}}}&{{a_{33}}}\end{array}} \right)\]
Then the matrix \[2A\] is,
\[2A = \left( {\begin{array}{*{20}{c}}{2{a_{11}}}&{2{a_{12}}}&{2{a_{13}}}\\{2{a_{21}}}&{2{a_{22}}}&{2{a_{23}}}\\{2{a_{31}}}&{2{a_{32}}}&{2{a_{33}}}\end{array}} \right)\]
Now perform the given row operation \[{R_2} \to 2{R_2} + 5{R_3}\] on matrix \[2A\] and obtain the matrix \[B\].
\[B = \left( {\begin{array}{*{20}{c}}{2{a_{11}}}&{2{a_{12}}}&{2{a_{13}}}\\{4{a_{21}} + 10{a_{31}}}&{4{a_{22}} + 10{a_{32}}}&{4{a_{23}} + 10{a_{33}}}\\{2{a_{31}}}&{2{a_{32}}}&{2{a_{33}}}\end{array}} \right)\]
If a multiple of a row is subtracted from a row, then the value of determinant is remaining the same.
Perform the row operation \[{R_2} \to {R_2} - 5{R_3}\]
\[B = \left( {\begin{array}{*{20}{c}}{2{a_{11}}}&{2{a_{12}}}&{2{a_{13}}}\\{4{a_{21}}}&{4{a_{22}}}&{4{a_{23}}}\\{2{a_{31}}}&{2{a_{32}}}&{2{a_{33}}}\end{array}} \right)\]
Now calculate the determinant of the matrix \[B\].
\[\left| B \right| = \left| {\begin{array}{*{20}{c}}{2{a_{11}}}&{2{a_{12}}}&{2{a_{13}}}\\{4{a_{21}}}&{4{a_{22}}}&{4{a_{23}}}\\{2{a_{31}}}&{2{a_{32}}}&{2{a_{33}}}\end{array}} \right|\]
Factor out the common numbers from each row.
\[\left| B \right| = 2 \times 4 \times 2 \times \left| {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\{{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\{{a_{31}}}&{{a_{32}}}&{{a_{33}}}\end{array}} \right|\]
\[ \Rightarrow \]\[\left| B \right| = 16\left| {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\{{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\{{a_{31}}}&{{a_{32}}}&{{a_{33}}}\end{array}} \right|\]
\[ \Rightarrow \]\[\left| B \right| = 16 \times 4\] [Since \[\left| A \right| = 4\]]
\[ \Rightarrow \]\[\left| B \right| = 64\]
Hence the correct option is A.
Note: Students often do a common mistake that they calculate the value of determinant \[\left| B \right| = \left| {\begin{array}{*{20}{c}}{2{a_{11}}}&{2{a_{12}}}&{2{a_{13}}}\\{4{a_{21}} + 10{a_{31}}}&{4{a_{22}} + 10{a_{32}}}&{4{a_{23}} + 10{a_{33}}}\\{2{a_{31}}}&{2{a_{32}}}&{2{a_{33}}}\end{array}} \right|\] by expanding method. They don’t perform the row operation \[{R_2} \to {R_2} - 5{R_3}\]. Don’t reach the required solution.