Let $left| {begin{array}{*{20}{c}} {{{left( {a - x} right)}^2}}&{{{left( {a

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3 and 4 .Determinants and Matrices

normal

Let $\left| {\begin{array}{*{20}{c}}
{{{\left( {a - x} \right)}^2}}&{{{\left( {a - y} \right)}^2}}&{{{\left( {a - z} \right)}^2}} \\
{{{\left( {b - x} \right)}^2}}&{{{\left( {b - y} \right)}^2}}&{{{\left( {b - z} \right)}^2}} \\
{{{\left( {c - x} \right)}^2}}&{{{\left( {c - y} \right)}^2}}&{{{\left( {c - z} \right)}^2}}
\end{array}} \right| = \frac{{ - 351}}{8}$ . If $x, y , z$ are the roots of the equation $8t^3 - 62t^2 + 43t -7 = 0$ and satisfy the determinant above, and $a, b, c$ are distinct number then value of $|(a - b) (b - c) (c - a)|$ is

$2$

$4$

$10$

$14$

Solution

Roots of given cubic is $7, \frac{1}{2} \cdot \frac{1}{4}$

Given determinant is

$\left| {\begin{array}{*{20}{c}}
{{a^2}}&{ – 2a}&1\\
{{b^2}}&{2b}&1\\
{{c^2}}&{ – 2b}&1
\end{array}} \right| \times \left| {\begin{array}{*{20}{c}}
1&1&1\\
x&y&z\\
{{x^2}}&{{y^2}}&{{z^2}}
\end{array}} \right|$

$2(a-b)(b-c)(c-a)(x-y)(y-z)(z-x)$

Hence $|(a-b)(b-c)(c-a)|=2$

Standard 12

Mathematics

If $A = \left| {\,\begin{array}{*{20}{c}}5&6&3\\{ – 4}&3&2\\{ – 4}&{ – 7}&3\end{array}\,} \right|\,, $ then cofactors of the elements of $2^{nd}$ row are

easy

$\left| {\,\begin{array}{{20}{c}}{{{\log }_3}512}&{{{\log }_4}3}\\{{{\log }_3}8}&{{{\log }_4}9}\end{array}\,} \right| \times \left| {\,\begin{array}{{20}{c}}{{{\log }_2}3}&{{{\log }_8}3}\\{{{\log }_3}4}&{{{\log }_3}4}\end{array}\,} \right|=$

medium

Let $A = {[{a_{ij}}]_{n \times n}}$ be a square matrix and let ${c_{ij}}$ be cofactor of ${a_{ij}}$ in $A$. If $C = [{c_{ij}}]$ ,then

normal

The cofactor of the element $'4'$ in the determinant $\left| {\,\begin{array}{*{20}{c}}1&3&5&1\\2&3&4&2\\8&0&1&1\\0&2&1&1\end{array}\,} \right|$ is

easy

If $\left| {\begin{array}{{20}{c}}
1&1&1\\
a&b&c\\
{{a^2}}&{{b^2}}&{{c^2}}
\end{array}} \right| = 5$ , then $\left| {\begin{array}{{20}{c}}
{b{c^2} – {b^2}c}&{{a^2}c – a{c^2}}&{a{b^2} – b{a^2}}\\
{{b^2} – {c^2}}&{{c^2} – {a^2}}&{{a^2} – {b^2}}\\
{c – b}&{a – c}&{b – a}
\end{array}} \right|$ is equal to

hard

Solution

Similar Questions

If $A = \left| {\,\begin{array}{*{20}{c}}5&6&3\\{ – 4}&3&2\\{ – 4}&{ – 7}&3\end{array}\,} \right|\,, $ then cofactors of the elements of $2^{nd}$ row are

$\left| {\,\begin{array}{*{20}{c}}{{{\log }_3}512}&{{{\log }_4}3}\\{{{\log }_3}8}&{{{\log }_4}9}\end{array}\,} \right| \times \left| {\,\begin{array}{*{20}{c}}{{{\log }_2}3}&{{{\log }_8}3}\\{{{\log }_3}4}&{{{\log }_3}4}\end{array}\,} \right|=$

Let $A = {[{a_{ij}}]_{n \times n}}$ be a square matrix and let ${c_{ij}}$ be cofactor of ${a_{ij}}$ in $A$. If $C = [{c_{ij}}]$ ,then

The cofactor of the element $'4'$ in the determinant $\left| {\,\begin{array}{*{20}{c}}1&3&5&1\\2&3&4&2\\8&0&1&1\\0&2&1&1\end{array}\,} \right|$ is

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$\left| {\,\begin{array}{{20}{c}}{{{\log }_3}512}&{{{\log }_4}3}\\{{{\log }_3}8}&{{{\log }_4}9}\end{array}\,} \right| \times \left| {\,\begin{array}{{20}{c}}{{{\log }_2}3}&{{{\log }_8}3}\\{{{\log }_3}4}&{{{\log }_3}4}\end{array}\,} \right|=$