Prove that $Delta=left|begin{array}{ccc} a+b x & c+d x & p+q x a x+b & c x+d & p x+q

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3 and 4 .Determinants and Matrices

medium

Prove that

$\Delta=\left|\begin{array}{ccc}
a+b x & c+d x & p+q x \\
a x+b & c x+d & p x+q \\
u & v & w
\end{array}\right|=\left(1-x^{2}\right)\left|\begin{array}{lll}
a & c & p \\
b & d & q \\
u & v & m
\end{array}\right|$

Option A

Option B

Option C

Option D

Solution

Applying $\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-x \mathrm{R}_{2}$ to $\Delta,$ we get

${\Delta = \left| {\begin{array}{*{20}{c}}
{a\left( {1 – {x^2}} \right)}&{c\left( {1 – {x^2}} \right)}&{p\left( {1 – {x^2}} \right)} \\
{ax + b}&{cx + d}&{px + q} \\
u&v&w
\end{array}} \right|}$

${ = \left( {1 – {x^2}} \right)\left| {\begin{array}{*{20}{c}}
a&c&p \\
{ax + b}&{cx + d}&{px + q} \\
u&v&w
\end{array}} \right|}$

Applying $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-x \mathrm{R}_{1},$ we get

$\Delta=\left(1-x^{2}\right)\left|\begin{array}{lll}
a & c & p \\
b & d & q \\
u & v & w
\end{array}\right|$

Standard 12

Mathematics

If $A$, $B$ and $C$ are the angles of a triangle then the value of the determinant

$\left| {\begin{array}{*{20}{c}}
{ – 1 + \cos B}&{\cos C + \cos B}&{\cos B} \\
{\cos C + \cos A}&{ – 1 + \cos A}&{\cos A} \\
{ – 1 + \cos B}&{ – 1 + \cos A}&{ – 1}
\end{array}} \right|$

normal

Which of the following matrices can $NOT$ be obtained from the matrix $\left[\begin{array}{cc}-1 & 2 \\ 1 & -1\end{array}\right]$ by a single elementary row operation?

easy

(JEE MAIN-2022)

By using properties of determinants, show that:

$\left|\begin{array}{ccc}-a^{2} & a b & a c \\ b a & -b^{2} & b c \\ c a & c b & -c^{2}\end{array}\right|=4 a^{2} b^{2} c^{2}$

medium

Let $D_1 =$ $\left| {\,\begin{array}{{20}{c}}a&b&{a + b}\\c&d&{c + d}\\a&b&{a – b}\end{array}\,} \right|$ and $D_2 =$ $\left| {\,\begin{array}{{20}{c}}a&c&{a + c}\\b&d&{b + d}\\a&c&{a + b + c}\end{array}\,} \right|$ then the value of $\frac{{{D_1}}}{{{D_2}}}$ where $b \ne 0$ and $ad \ne bc$, is

normal

The value of the determinant $\left| {\,\begin{array}{*{20}{c}}{31}&{37}&{92}\\{31}&{58}&{71}\\{31}&{105}&{24}\end{array}\,} \right|$ is

easy

Prove that $\Delta=\left|\begin{array}{ccc} a+b x & c+d x & p+q x \\ a x+b & c x+d & p x+q \\ u & v & w \end{array}\right|=\left(1-x^{2}\right)\left|\begin{array}{lll} a & c & p \\ b & d & q \\ u & v & m \end{array}\right|$

Solution

Similar Questions

If $A$, $B$ and $C$ are the angles of a triangle then the value of the determinant $\left| {\begin{array}{*{20}{c}} { – 1 + \cos B}&{\cos C + \cos B}&{\cos B} \\ {\cos C + \cos A}&{ – 1 + \cos A}&{\cos A} \\ { – 1 + \cos B}&{ – 1 + \cos A}&{ – 1} \end{array}} \right|$

Which of the following matrices can $NOT$ be obtained from the matrix $\left[\begin{array}{cc}-1 & 2 \\ 1 & -1\end{array}\right]$ by a single elementary row operation?

By using properties of determinants, show that: $\left|\begin{array}{ccc}-a^{2} & a b & a c \\ b a & -b^{2} & b c \\ c a & c b & -c^{2}\end{array}\right|=4 a^{2} b^{2} c^{2}$

Let $D_1 =$ $\left| {\,\begin{array}{*{20}{c}}a&b&{a + b}\\c&d&{c + d}\\a&b&{a – b}\end{array}\,} \right|$ and $D_2 =$ $\left| {\,\begin{array}{*{20}{c}}a&c&{a + c}\\b&d&{b + d}\\a&c&{a + b + c}\end{array}\,} \right|$ then the value of $\frac{{{D_1}}}{{{D_2}}}$ where $b \ne 0$ and $ad \ne bc$, is

The value of the determinant $\left| {\,\begin{array}{*{20}{c}}{31}&{37}&{92}\\{31}&{58}&{71}\\{31}&{105}&{24}\end{array}\,} \right|$ is

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Prove that

$\Delta=\left|\begin{array}{ccc}
a+b x & c+d x & p+q x \\
a x+b & c x+d & p x+q \\
u & v & w
\end{array}\right|=\left(1-x^{2}\right)\left|\begin{array}{lll}
a & c & p \\
b & d & q \\
u & v & m
\end{array}\right|$

If $A$, $B$ and $C$ are the angles of a triangle then the value of the determinant

$\left| {\begin{array}{*{20}{c}}
{ – 1 + \cos B}&{\cos C + \cos B}&{\cos B} \\
{\cos C + \cos A}&{ – 1 + \cos A}&{\cos A} \\
{ – 1 + \cos B}&{ – 1 + \cos A}&{ – 1}
\end{array}} \right|$

By using properties of determinants, show that:

$\left|\begin{array}{ccc}-a^{2} & a b & a c \\ b a & -b^{2} & b c \\ c a & c b & -c^{2}\end{array}\right|=4 a^{2} b^{2} c^{2}$

Let $D_1 =$ $\left| {\,\begin{array}{{20}{c}}a&b&{a + b}\\c&d&{c + d}\\a&b&{a – b}\end{array}\,} \right|$ and $D_2 =$ $\left| {\,\begin{array}{{20}{c}}a&c&{a + c}\\b&d&{b + d}\\a&c&{a + b + c}\end{array}\,} \right|$ then the value of $\frac{{{D_1}}}{{{D_2}}}$ where $b \ne 0$ and $ad \ne bc$, is