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 $\text{A}=\left(\begin{array}{cc}a& c\\ b& d\end{array}\right)$ , $\left|\text{A}\right|=\text{ad}-\text{bc}$ . For a $3\times 3$ matrix $\text{A = }\left(\begin{array}{ccc}a& d& g\\ b& e& h\\ c& f& i\end{array}\right)$ , $\left|\text{A}\right|=a\left|\left(\begin{array}{cc}e& h\\ f& i\end{array}\right)\right|-d\left|\left(\begin{array}{cc}b& h\\ c& i\end{array}\right)\right|+g\left|\left(\begin{array}{cc}b& e\\ c& f\end{array}\right)\right|$ ( 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 $A=\left(\begin{array}{ccc}a& b& c\\ a& b& c\\ x& y& z\end{array}\right)$
In the above matrix row $1$ and row $2$ are identical;
For example; $A=\left(\begin{array}{ccc}3& 2& 3\\ 3& 2& 3\\ 2& 2& 3\end{array}\right)$
$\Rightarrow det\text{A}=\left|\text{A}\right|=\left|\left(\begin{array}{ccc}3& 2& 3\\ 3& 2& 3\\ 2& 2& 3\end{array}\right)\right|$
Here, expanding along row $1$ ;
$\Rightarrow \left|\text{A}\right|=3\left[\left(\begin{array}{cc}2& 3\\ 2& 3\end{array}\right)\right]-2\left|\left(\begin{array}{cc}3& 3\\ 2& 3\end{array}\right)\right|+3\left[\left(\begin{array}{cc}3& 2\\ 2& 2\end{array}\right)\right]$
$\Rightarrow \left|\text{A}\right|=3\left(6-6\right)-2\left(9-6\right)+3\left(6-4\right)$
$\Rightarrow \left|\text{A}\right|=0-6+6$
$\Rightarrow \left|\text{A}\right|=0$
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.
$B=\left(\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right)=\left(\begin{array}{ccc}{a}_{11}& a_{21}& a_{31}\\ a_{12}& a_{22}& a_{32}\\ a_{13}& a_{23}& a_{33}\end{array}\right)$ ( rows and columns are interchanged)
$\text{For example ; }B=\left(\begin{array}{ccc}1& 1& -2\\ 2& 1& -3\\ 5& 3& -9\end{array}\right)$
Now, expanding along row $1$ ;
$\Rightarrow det\text{B}=\left|\text{B}\right|=1\left[\left(\begin{array}{cc}1& -3\\ 3& -9\end{array}\right)\right]-1\left[\left(\begin{array}{cc}2& -3\\ 5& -9\end{array}\right)\right]-2\left[\left(\begin{array}{cc}2& 1\\ 5& 3\end{array}\right)\right]$
$\Rightarrow \left|\text{B}\right|=1\left(-9+9\right)-1\left(-18+15\right)-2\left(6-5\right)$
$\Rightarrow \left|\text{B}\right|=1$
$\Rightarrow B^T=B^{\prime }=\left(\begin{array}{ccc}1& 2& 5\\ 1& 1& 3\\ -2& -3& -9\end{array}\right)$ ( This is transpose of the matrix $B$ )
$\Rightarrow det{\text{B}}^T=\left|{\text{B}}^T\right|=1\left[\left(\begin{array}{cc}1& 3\\ -3& -9\end{array}\right)\right]-2\left[\left(\begin{array}{cc}1& 3\\ -2& -9\end{array}\right)\right]+5\left[\left(\begin{array}{cc}1& 1\\ -2& -3\end{array}\right)\right]$
$\Rightarrow \left|B^T\right|=1\left(-9+9\right)-2\left(-9+6\right)+5\left(-3+2\right)$
$\Rightarrow \left|B^T\right|=1$
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 $C=\left(\begin{array}{ccc}2& 3& 1\\ 8& 1& 2\\ 3& 4& 2\end{array}\right)$
Here, expanding along row $1$ ;
$\Rightarrow \left|\text{C}\right|=2\left[\left(\begin{array}{cc}1& 2\\ 4& 2\end{array}\right)\right]-3\left[\left(\begin{array}{cc}8& 2\\ 3& 2\end{array}\right)\right]+1\left[\left(\begin{array}{cc}8& 1\\ 3& 4\end{array}\right)\right]$
$\Rightarrow \left|\text{C}\right|=2\left(-6\right)-3\left(10\right)+29$
$\Rightarrow \left|\text{C}\right|=-13$
$\Rightarrow R_1\leftrightarrow R_2$ (interchanging row $1$ and row $2$ )
$\text{Let , }D=\left(\begin{array}{ccc}8& 1& 2\\ 2& 3& 1\\ 3& 4& 2\end{array}\right)$
Here, expanding along row $1$ ;
$\Rightarrow \left|\text{D}\right|=8\left[\left(\begin{array}{cc}3& 1\\ 4& 2\end{array}\right)\right]-1\left[\left(\begin{array}{cc}2& 1\\ 3& 2\end{array}\right)\right]+2\left[\left(\begin{array}{cc}2& 3\\ 3& 4\end{array}\right)\right]$
$\Rightarrow \left|\text{D}\right|=8\left(2\right)-1\left(1\right)+2\left(-1\right)$
$\Rightarrow \left|\text{D}\right|=13$
$\therefore \left|\text{C}\right|=-\left|\text{D}\right|$
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;$$\mathrm{\Delta }=\left|\left(\begin{array}{ccc}p+q& q+r& r+p\\ r& p& q\\ 1& 1& 1\end{array}\right)\right|=0$$ . Let $R_1\to R_1+R_2\left(\text{here }k=1\right)$ then $$\mathrm{\Delta }=\left|\left(\begin{array}{ccc}p+q+r& p+q+r& p+q+r\\ r& p& q\\ 1& 1& 1\end{array}\right)\right|=p+q+r\left|\left(\begin{array}{ccc}1& 1& 1\\ r& p& q\\ 1& 1& 1\end{array}\right)\right|=0$$ (By statement I , two rows are identical, hence the determinant is zero ).