Question

Consider the following statements:
I. If any two rows or columns of a determinant are identical , then the value of the determinant is zero.
II. If the corresponding rows and columns of a determinant are interchanged , then the value of the determinant does not change.
III. If any two rows or columns of a determinant are interchanged , then the value of the determinant changes in sign. Which of these is/are correct?
$\left( 1 \right)$ I and II
$\left( 2 \right)$ II and III
$\left( 3 \right)$ I , II and III
$\left( 4 \right)$ I and III

Answer 1

Hint: This question is from properties of determinants. Determinant of a matrix is denoted by $\det {\text{A or }}\left| {\text{A}} \right|$ . For a $2 \times 2$ matrix ${\text{A}} = \left( {\begin{array}{*{20}{c}}
  a&c \\
  b&d
\end{array}} \right)$ , $\left| {\text{A}} \right| = {\text{ad}} - {\text{bc}}$ . For a $3 \times 3$ matrix ${\text{A = }}\left( {\begin{array}{*{20}{c}}
  a&d&g \\
  b&e&h \\
  c&f&i
\end{array}} \right)$ , $\left| {\text{A}} \right| = a\left| {\left( {\begin{array}{*{20}{c}}
  e&h \\
  f&i
\end{array}} \right)} \right| - d\left| {\left( {\begin{array}{*{20}{c}}
  b&h \\
  c&i
\end{array}} \right)} \right| + g\left| {\left( {\begin{array}{*{20}{c}}
  b&e \\
  c&f
\end{array}} \right)} \right|$ ( expanding along row $1$ )which is $\left| {\text{A}} \right| = a\left( {ei - fh} \right) - d\left( {bi - ch} \right) + g\left( {bf - ce} \right)$ .

Complete step-by-step answer:
Statement I : . If any two rows or columns of a determinant are identical or same , then the value of the determinant is zero.
Let $A = \left( {\begin{array}{*{20}{c}}
  a&b&c \\
  a&b&c \\
  x&y&z
\end{array}} \right)$
In the above matrix row $1$ and row $2$ are identical;
For example; $A = \left( {\begin{array}{*{20}{c}}
  3&2&3 \\
  3&2&3 \\
  2&2&3
\end{array}} \right)$
$ \Rightarrow \det {\text{A}} = \left| {\text{A}} \right| = \left| {\left( {\begin{array}{*{20}{c}}
  3&2&3 \\
  3&2&3 \\
  2&2&3
\end{array}} \right)} \right|$
Here, expanding along row $1$ ;
$ \Rightarrow \left| {\text{A}} \right| = 3\left[ {\left( {\begin{array}{*{20}{c}}
  2&3 \\
  2&3
\end{array}} \right)} \right] - 2\left| {\left( {\begin{array}{*{20}{c}}
  3&3 \\
  2&3
\end{array}} \right)} \right| + 3\left[ {\left( {\begin{array}{*{20}{c}}
  3&2 \\
  2&2
\end{array}} \right)} \right]$
$ \Rightarrow \left| {\text{A}} \right| = 3\left( {6 - 6} \right) - 2\left( {9 - 6} \right) + 3\left( {6 - 4} \right)$
$ \Rightarrow \left| {\text{A}} \right| = 0 - 6 + 6$
$ \Rightarrow \left| {\text{A}} \right| = 0$
Therefore, statement I is true .
Statement II. If the corresponding rows and columns of a determinant are interchanged , then the value of the determinant does not change.
$B = \left( {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\
  {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\
  {{a_{31}}}&{{a_{32}}}&{{a_{33}}}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{21}}}&{{a_{31}}} \\
  {{a_{12}}}&{{a_{22}}}&{{a_{32}}} \\
  {{a_{13}}}&{{a_{23}}}&{{a_{33}}}
\end{array}} \right)$ ( rows and columns are interchanged)
${\text{For example ; }}B = \left( {\begin{array}{*{20}{c}}
  1&1&{ - 2} \\
  2&1&{ - 3} \\
  5&3&{ - 9}
\end{array}} \right)$
Now, expanding along row $1$ ;
$ \Rightarrow \det {\text{B}} = \left| {\text{B}} \right| = 1\left[ {\left( {\begin{array}{*{20}{c}}
  1&{ - 3} \\
  3&{ - 9}
\end{array}} \right)} \right] - 1\left[ {\left( {\begin{array}{*{20}{c}}
  2&{ - 3} \\
  5&{ - 9}
\end{array}} \right)} \right] - 2\left[ {\left( {\begin{array}{*{20}{c}}
  2&1 \\
  5&3
\end{array}} \right)} \right]$
$ \Rightarrow \left| {\text{B}} \right| = 1\left( { - 9 + 9} \right) - 1\left( { - 18 + 15} \right) - 2\left( {6 - 5} \right)$
$ \Rightarrow \left| {\text{B}} \right| = 1$
$ \Rightarrow {B^T} = B' = \left( {\begin{array}{*{20}{c}}
  1&2&5 \\
  1&1&3 \\
  { - 2}&{ - 3}&{ - 9}
\end{array}} \right)$ ( This is transpose of the matrix $B$ )
$ \Rightarrow \det {{\text{B}}^T} = \left| {{{\text{B}}^T}} \right| = 1\left[ {\left( {\begin{array}{*{20}{c}}
  1&3 \\
  { - 3}&{ - 9}
\end{array}} \right)} \right] - 2\left[ {\left( {\begin{array}{*{20}{c}}
  1&3 \\
  { - 2}&{ - 9}
\end{array}} \right)} \right] + 5\left[ {\left( {\begin{array}{*{20}{c}}
  1&1 \\
  { - 2}&{ - 3}
\end{array}} \right)} \right]$
$ \Rightarrow \left| {{B^T}} \right| = 1\left( { - 9 + 9} \right) - 2\left( { - 9 + 6} \right) + 5\left( { - 3 + 2} \right)$
$ \Rightarrow \left| {{B^T}} \right| = 1$
Therefore, the value of the determinant does not change.
In other words, we can say that $\left| B \right| = \left| {{B^T}} \right|$ are equal .
Hence statement II is true.

Statement III : If any two rows or columns of a determinant are interchanged , then the value of the determinant changes in sign.
Let $C = \left( {\begin{array}{*{20}{c}}
  2&3&1 \\
  8&1&2 \\
  3&4&2
\end{array}} \right)$
Here, expanding along row $1$ ;
$ \Rightarrow \left| {\text{C}} \right| = 2\left[ {\left( {\begin{array}{*{20}{c}}
  1&2 \\
  4&2
\end{array}} \right)} \right] - 3\left[ {\left( {\begin{array}{*{20}{c}}
  8&2 \\
  3&2
\end{array}} \right)} \right] + 1\left[ {\left( {\begin{array}{*{20}{c}}
  8&1 \\
  3&4
\end{array}} \right)} \right]$
$ \Rightarrow \left| {\text{C}} \right| = 2\left( { - 6} \right) - 3\left( {10} \right) + 29$
$ \Rightarrow \left| {\text{C}} \right| = - 13$
$ \Rightarrow {R_1} \leftrightarrow {R_2}$ (interchanging row $1$ and row $2$ )
${\text{Let , }}D = \left( {\begin{array}{*{20}{c}}
  8&1&2 \\
  2&3&1 \\
  3&4&2
\end{array}} \right)$
Here, expanding along row $1$ ;
$ \Rightarrow \left| {\text{D}} \right| = 8\left[ {\left( {\begin{array}{*{20}{c}}
  3&1 \\
  4&2
\end{array}} \right)} \right] - 1\left[ {\left( {\begin{array}{*{20}{c}}
  2&1 \\
  3&2
\end{array}} \right)} \right] + 2\left[ {\left( {\begin{array}{*{20}{c}}
  2&3 \\
  3&4
\end{array}} \right)} \right]$
$ \Rightarrow \left| {\text{D}} \right| = 8\left( 2 \right) - 1\left( 1 \right) + 2\left( { - 1} \right)$
$ \Rightarrow \left| {\text{D}} \right| = 13$
$\therefore \left| {\text{C}} \right| = - \left| {\text{D}} \right|$
Therefore, we can notice here that by interchanging two rows the sign of the determinant has changed.
Hence statement III is also true.
So, in conclusion, statements I , II , III all are true.
Therefore, the correct answer for this question is option $\left( 3 \right)$ .
So, the correct answer is “Option 3”.

Note: Some of the other important properties of a determinant are listed as under: $\left( 1 \right)$ If every element of a row or a column of a matrix is multiplied by a constant $k$ , then it’s determinant value is also multiplied by the same constant $k$ . $\left( 2 \right)$ If some or all elements of a row or a column of a matrix are expressed as a sum of two or more terms, then it’s determinant can also be expressed as sum of two or more determinants. $\left( 3 \right)$ The value of the determinant remains same or unchanged if ${R_i} = {R_i} + k{R_j}$ or ${C_i} = {C_i} + k{C_j}$ . Example: Without expanding show that;\[\Delta = \left| {\left( {\begin{array}{*{20}{c}}
  {p + q}&{q + r}&{r + p} \\
  r&p&q \\
  1&1&1
\end{array}} \right)} \right| = 0\] . Let ${R_1} \to {R_1} + {R_2}{\text{ }}\left( {{\text{here }}k = 1} \right)$ then \[\Delta = \left| {\left( {\begin{array}{*{20}{c}}
  {p + q + r}&{p + q + r}&{p + q + r} \\
  r&p&q \\
  1&1&1
\end{array}} \right)} \right| = p + q + r\left| {\left( {\begin{array}{*{20}{c}}
  1&1&1 \\
  r&p&q \\
  1&1&1
\end{array}} \right)} \right| = 0\] (By statement I , two rows are identical, hence the determinant is zero ).