Prove that $\Delta = \left| \begin{array}{ccc} a+bx & c+dx & p+qx \\ ax+b & cx+d & px+q \\ u & v & w \end{array} \right| = (1-x^2) \left| \begin{array}{ccc} a & c & p \\ b & d & q \\ u & v & w \end{array} \right|$
(A) Applying the row operation $R_1 \rightarrow R_1 - x R_2$ to the determinant $\Delta$,we get:
$\Delta = \left| \begin{array}{ccc} a+bx - x(ax+b) & c+dx - x(cx+d) & p+qx - x(px+q) \\ ax+b & cx+d & px+q \\ u & v & w \end{array} \right|$
$= \left| \begin{array}{ccc} a(1-x^2) & c(1-x^2) & p(1-x^2) \\ ax+b & cx+d & px+q \\ u & v & w \end{array} \right|$
Taking $(1-x^2)$ common from $R_1$:
$= (1-x^2) \left| \begin{array}{ccc} a & c & p \\ ax+b & cx+d & px+q \\ u & v & w \end{array} \right|$
Now,applying the row operation $R_2 \rightarrow R_2 - x R_1$ to the resulting determinant:
$= (1-x^2) \left| \begin{array}{ccc} a & c & p \\ (ax+b) - x(a) & (cx+d) - x(c) & (px+q) - x(p) \\ u & v & w \end{array} \right|$
$= (1-x^2) \left| \begin{array}{ccc} a & c & p \\ b & d & q \\ u & v & w \end{array} \right|$
Hence,the identity is proved.
$\Delta = \left| \begin{array}{ccc} a+bx - x(ax+b) & c+dx - x(cx+d) & p+qx - x(px+q) \\ ax+b & cx+d & px+q \\ u & v & w \end{array} \right|$
$= \left| \begin{array}{ccc} a(1-x^2) & c(1-x^2) & p(1-x^2) \\ ax+b & cx+d & px+q \\ u & v & w \end{array} \right|$
Taking $(1-x^2)$ common from $R_1$:
$= (1-x^2) \left| \begin{array}{ccc} a & c & p \\ ax+b & cx+d & px+q \\ u & v & w \end{array} \right|$
Now,applying the row operation $R_2 \rightarrow R_2 - x R_1$ to the resulting determinant:
$= (1-x^2) \left| \begin{array}{ccc} a & c & p \\ (ax+b) - x(a) & (cx+d) - x(c) & (px+q) - x(p) \\ u & v & w \end{array} \right|$
$= (1-x^2) \left| \begin{array}{ccc} a & c & p \\ b & d & q \\ u & v & w \end{array} \right|$
Hence,the identity is proved.
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Similar Questions
Consider 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?
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View SolutionIf $A$ is any square matrix of order $3 \times 3$,then $|3A|$ is equal to:
For non-zero $a, b, c$,if $\Delta = \begin{vmatrix} 1 + a & 1 & 1 \\ 1 & 1 + b & 1 \\ 1 & 1 & 1 + c \end{vmatrix} = 0$,then the value of $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = $
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View SolutionIf $x, y, z$ are distinct and $\Delta=\left|\begin{array}{lll}x & x^{2} & 1+x^{3} \\ y & y^{2} & 1+y^{3} \\ z & z^{2} & 1+z^{3}\end{array}\right|=0,$ then show that $1+x y z=0$.
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