If a, b, c are real then value of determinant left| {begin{array}{*{20}{c}} {{a^2} + 1}&{a

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3 and 4 .Determinants and Matrices

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If $a, b, c$ are real then the value of determinant $\left| {\begin{array}{*{20}{c}} {{a^2} + 1}&{ab}&{ac}\\{ab}&{{b^2} + 1}&{bc}\\{ac}&{bc}&{{c^2} + 1}\end{array}}\right|$ $= 1$ if

$a + b + c = 0$

$a + b + c = 1$

$a + b + c = -1$

$a = b = c = 0$

Solution

Multiply $R_1$ by $a, R_2$ by $b$ and $R_3$ by $c$ and divide the determinant by $abc$. Now take $a, b$ and $c$ common from $c_1, c_2$ and $c_3$. Now use $C_1 \rightarrow C_1 + C_2 + C_3$ to get
$(a^2 + b^2 + c^2 + 1)$ $\left| {\,\begin{array}{*{20}{c}}1&1&1\\{{b^2}}&{{b^2} + 1}&{{b^2}}\\{{c^2}}&{{c^2}}&{{c^2} + 1}\end{array}\,} \right|$ $= 1$. Now use $c_1 \rightarrow c_1 – c_2$ and $c_2 \rightarrow c_2 – c_3$
we get $1 + a^2 + b^2 + c^2 = 1$ ==> $a = b = c = 0$

Standard 12

Mathematics

Let $P=\left[\begin{array}{ccc}1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1\end{array}\right]$ and $I$ be the identity matrix of order $3$ . If $\left.Q=q_{i j}\right]$ is a matrix such that $P^{50}-Q=I$, then $\frac{q_{31}+q_{32}}{q_{21}}$ equals

hard

(IIT-2016)

By using properties of determinants, show that:

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

medium

By using properties of determinants, show that:

$\left|\begin{array}{ccc}1+a^{2}-b^{2} & 2 a b & -2 b \\ 2 a b & 1-a^{2}+b^{2} & 2 a \\ 2 b & -2 a & 1-a^{2}-b^{2}\end{array}\right|=\left(1+a^{2}+b^{2}\right)^{3}$

hard

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

$\left| {\,\begin{array}{*{20}{c}}{{a^2} + {x^2}}&{ab}&{ca}\\{ab}&{{b^2} + {x^2}}&{bc}\\{ca}&{bc}&{{c^2} + {x^2}}\end{array}\,} \right|$ is divisor of

hard

If $a, b, c$ are real then the value of determinant $\left| {\begin{array}{*{20}{c}} {{a^2} + 1}&{ab}&{ac}\\{ab}&{{b^2} + 1}&{bc}\\{ac}&{bc}&{{c^2} + 1}\end{array}}\right|$ $= 1$ if

Solution

Similar Questions

Let $P=\left[\begin{array}{ccc}1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1\end{array}\right]$ and $I$ be the identity matrix of order $3$ . If $\left.Q=q_{i j}\right]$ is a matrix such that $P^{50}-Q=I$, then $\frac{q_{31}+q_{32}}{q_{21}}$ equals

By using properties of determinants, show that: $\left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|=(a+b)(b-c)(c-a)$

By using properties of determinants, show that: $\left|\begin{array}{ccc}1+a^{2}-b^{2} & 2 a b & -2 b \\ 2 a b & 1-a^{2}+b^{2} & 2 a \\ 2 b & -2 a & 1-a^{2}-b^{2}\end{array}\right|=\left(1+a^{2}+b^{2}\right)^{3}$

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

$\left| {\,\begin{array}{*{20}{c}}{{a^2} + {x^2}}&{ab}&{ca}\\{ab}&{{b^2} + {x^2}}&{bc}\\{ca}&{bc}&{{c^2} + {x^2}}\end{array}\,} \right|$ is divisor of

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By using properties of determinants, show that:

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

By using properties of determinants, show that:

$\left|\begin{array}{ccc}1+a^{2}-b^{2} & 2 a b & -2 b \\ 2 a b & 1-a^{2}+b^{2} & 2 a \\ 2 b & -2 a & 1-a^{2}-b^{2}\end{array}\right|=\left(1+a^{2}+b^{2}\right)^{3}$

By using properties of determinants, show that:

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