Prove that left|begin{array}{ccc}a & a+b & a+b+c 2 a & 3 a+2 b & 4 a+3 b+2 c 3 a &am

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3 and 4 .Determinants and Matrices

medium

Prove that $\left|\begin{array}{ccc}a & a+b & a+b+c \\ 2 a & 3 a+2 b & 4 a+3 b+2 c \\ 3 a & 6 a+3 b & 10 a+6 b+3 c\end{array}\right|=a^{3}$

Option A

Option B

Option C

Option D

Solution

Solution Applying operations $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-2 \mathrm{R}_{1}$ and $\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-3 \mathrm{R}_{1}$ to the given determinant $\Delta$, we have

$\Delta = \left| {\begin{array}{*{20}{c}}
a&{a + b}&{a + b + c} \\
0&a&{2a + b} \\
0&{3a}&{7a + 3b}
\end{array}} \right|$

Now applying $\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-3 \mathrm{R}_{2},$ we get

$\Delta=\left|\begin{array}{ccc}
a & a+b & a+b+c \\
0 & a & 2 a+b \\
0 & 0 & a
\end{array}\right|$

Expanding along $C_{1},$ we obtain

$\Delta = a\left| {\begin{array}{*{20}{c}}
a&{2a + b} \\
0&a
\end{array}} \right| + 0 + 0$

$ = a\left( {{a^2} – 0} \right) = a\left( {{a^2}} \right) = {a^3}$

Standard 12

Mathematics

If $\mathrm{a, b, c}$ are in $\mathrm{A.P}$, find value of

$\left|\begin{array}{ccc}
2 y+4 & 5 y+7 & 8 y+a \\
3 y+5 & 6 y+8 & 9 y+b \\
4 y+6 & 7 y+9 & 10 y+c
\end{array}\right|$

medium

If $D =$ $\left| {\,\begin{array}{*{20}{c}}{\frac{1}{z}}&{\frac{1}{z}}&{ – \frac{{(x + y)}}{{{z^2}}}}\\{ – \frac{{(y + z)}}{{{x^2}}}}&{\frac{1}{x}}&{\frac{1}{x}}\\{ – \frac{{y(y + z)}}{{{x^2}z}}}&{\frac{{x + 2y + z}}{{xz}}}&{ – \frac{{y(x +y)}}{{x{z^2}}}}\end{array}\,} \right|$ then, the incorrect statement is

normal

By using properties of determinants, show that:

$\left|\begin{array}{lll}x & x^{2} & y z \\ y & y^{2} & z x \\ z & z^{2} & x y\end{array}\right|=(x-y)(y-z)(z-x)(x y+y z+z x)$

hard

The value of the determinant $\left| {\,\begin{array}{*{20}{c}}1&1&1\\{b + c}&{c + a}&{a + b}\\{b + c – a}&{c + a – b}&{a + b – c}\end{array}\,} \right|$ is

easy

Let $a, b, c, d$ be in arithmetic progression with common difference $\lambda$. If

$\left|\begin{array}{lll} x+a-c & x+b & x+a \\ x-1 & x+c & x+b \\ x-b+d & x+d & x+c \end{array}\right|=2$

then value of $\lambda^{2}$ is equal to $…..$

medium

(JEE MAIN-2021)

Prove that $\left|\begin{array}{ccc}a & a+b & a+b+c \\ 2 a & 3 a+2 b & 4 a+3 b+2 c \\ 3 a & 6 a+3 b & 10 a+6 b+3 c\end{array}\right|=a^{3}$

Solution

Similar Questions

If $\mathrm{a, b, c}$ are in $\mathrm{A.P}$, find value of $\left|\begin{array}{ccc} 2 y+4 & 5 y+7 & 8 y+a \\ 3 y+5 & 6 y+8 & 9 y+b \\ 4 y+6 & 7 y+9 & 10 y+c \end{array}\right|$

By using properties of determinants, show that: $\left|\begin{array}{lll}x & x^{2} & y z \\ y & y^{2} & z x \\ z & z^{2} & x y\end{array}\right|=(x-y)(y-z)(z-x)(x y+y z+z x)$

The value of the determinant $\left| {\,\begin{array}{*{20}{c}}1&1&1\\{b + c}&{c + a}&{a + b}\\{b + c – a}&{c + a – b}&{a + b – c}\end{array}\,} \right|$ is

Let $a, b, c, d$ be in arithmetic progression with common difference $\lambda$. If $\left|\begin{array}{lll} x+a-c & x+b & x+a \\ x-1 & x+c & x+b \\ x-b+d & x+d & x+c \end{array}\right|=2$ then value of $\lambda^{2}$ is equal to $…..$

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If $\mathrm{a, b, c}$ are in $\mathrm{A.P}$, find value of

$\left|\begin{array}{ccc}
2 y+4 & 5 y+7 & 8 y+a \\
3 y+5 & 6 y+8 & 9 y+b \\
4 y+6 & 7 y+9 & 10 y+c
\end{array}\right|$

By using properties of determinants, show that:

$\left|\begin{array}{lll}x & x^{2} & y z \\ y & y^{2} & z x \\ z & z^{2} & x y\end{array}\right|=(x-y)(y-z)(z-x)(x y+y z+z x)$

Let $a, b, c, d$ be in arithmetic progression with common difference $\lambda$. If

$\left|\begin{array}{lll} x+a-c & x+b & x+a \\ x-1 & x+c & x+b \\ x-b+d & x+d & x+c \end{array}\right|=2$

then value of $\lambda^{2}$ is equal to $…..$