If A = left[ {begin{array}{*{20}{c}}a&bc&dend{array}} right] (where bc ne 0 ) satisfie

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3 and 4 .Determinants and Matrices

normal

If $A =$ $\left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]$ (where $bc \ne 0$) satisfies the equations $x^2 + k = 0$, then

$a + d = 0$

$k = -|A|$

$k = |A|$

both $(A)$ and $(C)$

Solution

We have $A^2 =$ $\left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]$ $\left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]$ $=$ $\left[{\begin{array}{*{20}{c}}{{a^2} + bc}&{ab + db}\\{ac + cd}&{bc + {d^2}} \end{array}} \right]$ $= 0$

As $A$ satisfies, $x^2 + k = 0, A^2 + kI = O$

==>$\left[ {\begin{array}{*{20}{c}}{{a^2} + bc + k}&{(a + d)b}\\{(a + d)c}&{bc + {d^2} + k}\end{array}} \right]$

==>$a^2 + bc + k = 0 = bc + d^2 + k = 0$ and $(a + d)b = (a + d) c = 0$

As $bc \ne 0, b \ne 0, c \ne 0$ ==> $a + d = 0$ ==> $a = -d$

Also, $k = -(a^2 + bc)$ $= -(d^2 + bc)$ $= – ( (-ad) + bc ) = |A|$

Standard 12

Mathematics

If $a, b, c$ are all different from zero and $\left| {\begin{array}{*{20}{c}} {1 + a}&1&1\\ 1&{1 + b}&1\\ 1&1&{1 + c} \end{array}} \right| = 0$ , then the value of $a^{-1} + b^{-1} + c^{-1}$ is

normal

The total number of distinct $x \in \mathbb{R}$ for which $\left|\begin{array}{ccc}x & x^2 & 1+x^3 \\ 2 x & 4 x^2 & 1+8 x^3 \\ 3 x & 9 x^2 & 1+27 x^3\end{array}\right|=10$ is

normal

(IIT-2016)

If $p, q, r, s$ are in $A.P.$ and $f (x) =$ $\left| {\,\begin{array}{*{20}{c}} {p\,\, + \,\,\sin \,x}&{q\,\, + \,\,\sin \,x}&{p\,\, – \,\,r\,\, + \,\,\sin \,x}\\ {q\,\, + \,\,\sin \,x}&{r\,\, + \,\,\sin \,x}&{ – \,1\,\, + \,\,\sin \,x}\\ {r\,\, + \,\,\sin \,x}&{s\,\, + \,\,\sin \,x}&{s\,\, – \,\,q\,\, + \,\,\sin \,x} \end{array}\,} \right|$ such that $f (x)dx = – 4$ then the common difference of the $A.P.$ can be :

normal

For a non – zero, real $a, b$ and $c$ $\left| {\begin{array}{*{20}{c}}{\frac{{{a^2} + {b^2}}}{c}}&c&c\\a&{\frac{{{b^2} + {c^2}}}{a}}&a\\b&b&{\frac{{{c^2} + {a^2}}}{b}} \end{array}} \right|$ $= \alpha \, abc$, then the values of $\alpha$ is

normal

Using the property of determinants and without expanding, prove that:

$\left|\begin{array}{lll}2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86\end{array}\right|=0$

easy

If $A =$ $\left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]$ (where $bc \ne 0$) satisfies the equations $x^2 + k = 0$, then

Solution

Similar Questions

If $a, b, c$ are all different from zero and $\left| {\begin{array}{*{20}{c}} {1 + a}&1&1\\ 1&{1 + b}&1\\ 1&1&{1 + c} \end{array}} \right| = 0$ , then the value of $a^{-1} + b^{-1} + c^{-1}$ is

The total number of distinct $x \in \mathbb{R}$ for which $\left|\begin{array}{ccc}x & x^2 & 1+x^3 \\ 2 x & 4 x^2 & 1+8 x^3 \\ 3 x & 9 x^2 & 1+27 x^3\end{array}\right|=10$ is

For a non – zero, real $a, b$ and $c$ $\left| {\begin{array}{*{20}{c}}{\frac{{{a^2} + {b^2}}}{c}}&c&c\\a&{\frac{{{b^2} + {c^2}}}{a}}&a\\b&b&{\frac{{{c^2} + {a^2}}}{b}} \end{array}} \right|$ $= \alpha \, abc$, then the values of $\alpha$ is

Using the property of determinants and without expanding, prove that: $\left|\begin{array}{lll}2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86\end{array}\right|=0$

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Using the property of determinants and without expanding, prove that:

$\left|\begin{array}{lll}2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86\end{array}\right|=0$