Show that $\left|\begin{array}{ccc}a & b & c \\ a+2x & b+2y & c+2z \\ x & y & z\end{array}\right|=0$.
(A) Given the determinant $\Delta = \left|\begin{array}{ccc}a & b & c \\ a+2x & b+2y & c+2z \\ x & y & z\end{array}\right|$.
Using the property of determinants,we can split the second row into two parts:
$\Delta = \left|\begin{array}{ccc}a & b & c \\ a & b & c \\ x & y & z\end{array}\right| + \left|\begin{array}{ccc}a & b & c \\ 2x & 2y & 2z \\ x & y & z\end{array}\right|$.
In the first determinant,the first row and second row are identical $(R_1 = R_2)$,so its value is $0$.
In the second determinant,we can factor out $2$ from the second row:
$\Delta = 0 + 2 \left|\begin{array}{ccc}a & b & c \\ x & y & z \\ x & y & z\end{array}\right|$.
Here,the second row and third row are identical $(R_2 = R_3)$,so its value is $0$.
Thus,$\Delta = 0 + 2(0) = 0$.
Using the property of determinants,we can split the second row into two parts:
$\Delta = \left|\begin{array}{ccc}a & b & c \\ a & b & c \\ x & y & z\end{array}\right| + \left|\begin{array}{ccc}a & b & c \\ 2x & 2y & 2z \\ x & y & z\end{array}\right|$.
In the first determinant,the first row and second row are identical $(R_1 = R_2)$,so its value is $0$.
In the second determinant,we can factor out $2$ from the second row:
$\Delta = 0 + 2 \left|\begin{array}{ccc}a & b & c \\ x & y & z \\ x & y & z\end{array}\right|$.
Here,the second row and third row are identical $(R_2 = R_3)$,so its value is $0$.
Thus,$\Delta = 0 + 2(0) = 0$.
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Similar Questions
If $A_1B_1C_1, A_2B_2C_2, A_3B_3C_3$ are three-digit numbers,each of which is divisible by $k$,and $\Delta = \begin{vmatrix} A_1 & B_1 & C_1 \\ A_2 & B_2 & C_2 \\ A_3 & B_3 & C_3 \end{vmatrix}$,then $\Delta$ is divisible by:
If $A = \begin{bmatrix} 2 & 5 \\ 3 & 7 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 3 \\ 4 & 1 \end{bmatrix}$,then which of the following properties holds true?
Easy
View SolutionConsider the following statements :
$(a)$ If any two rows or columns of a determinant are identical,then the value of the determinant is zero.
$(b)$ If the corresponding rows and columns of a determinant are interchanged,then the value of the determinant does not change.
$(c)$ If any two rows (or columns) of a determinant are interchanged,then the value of the determinant changes in sign.
Which of these are correct?
$(a)$ If any two rows or columns of a determinant are identical,then the value of the determinant is zero.
$(b)$ If the corresponding rows and columns of a determinant are interchanged,then the value of the determinant does not change.
$(c)$ If any two rows (or columns) of a determinant are interchanged,then the value of the determinant changes in sign.
Which of these are correct?
By using properties of determinants,show that:
$\left|\begin{array}{ccc}y+k & y & y \\ y & y+k & y \\ y & y & y+k\end{array}\right|=k^{2}(3y+k)$
$\left|\begin{array}{ccc}y+k & y & y \\ y & y+k & y \\ y & y & y+k\end{array}\right|=k^{2}(3y+k)$
Medium
View Solution$2\,\,\left| {\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ {a^2 - bc} & {b^2 - ac} & {c^2 - ab} \end{array}} \right| = $
Difficult
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