Verify Property 2 for Delta=left|begin{array}{ccc}2 & -3 & 5 6 & 0 & 4 1 & 5 &am

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3 and 4 .Determinants and Matrices

easy

Verify Property $2$ for $\Delta=\left|\begin{array}{ccc}2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7\end{array}\right|$

Option A

Option B

Option C

Option D

Solution

$\Delta = \left| {\begin{array}{*{20}{r}}
2&{ – 3}&5 \\
6&0&4 \\
1&5&{ – 7}
\end{array}} \right| = – 28{\text{ }}$

Interchanging rows $\mathrm{R}_{2}$ and $\mathrm{R}_{3}$ i.e., $\mathrm{R}_{2} \leftrightarrow \mathrm{R}_{3},$ we have

${\Delta _1} = \left| {\begin{array}{*{20}{r}}
2&{ – 3}&5 \\
1&5&{ – 7} \\
6&0&4
\end{array}} \right|$

Expanding the determinant $\Delta_{1}$ along first row, we have

${\Delta _1} = 2\left| {\begin{array}{*{20}{c}}
5&{ – 7} \\
0&4
\end{array}} \right| – ( – 3)\left| {\begin{array}{*{20}{c}}
1&{ – 7} \\
6&4
\end{array}} \right| + 5\left| {\begin{array}{*{20}{c}}
1&5 \\
6&0
\end{array}} \right|$

$ = 2(20 – 0) + 3(4 + 42) + 5(0 – 30)$

$ = 40 + 138 – 150 = 28$

Clearly $\quad \Delta_{1}=-\Delta$

Hence, Property $2$ is verified.

Standard 12

Mathematics

Let $P$ be a matrix of order $3 \times 3$ such that all the entries in $P$ are from the set $\{-1,0,1\}$. Then, the maximum possible value of the determinant of $P$ is. . . . . . .

normal

(IIT-2018)

Evaluate $\left|\begin{array}{ccc}x & y & x+y \\ y & x+y & x \\ x+y & x & y\end{array}\right|$

medium

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

hard

(IIT-1980)

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

hard

$$f(x)=\left| {\begin{array}{*{20}{c}} {{{\sin }^2}x}&{ – 2 + {{\cos }^2}x}&{\cos 2x} \\ {2 + {{\sin }^2}x}&{{{\cos }^2}x}&{\cos 2x} \\ {{{\sin }^2}x}&{{{\cos }^2}x}&{1 + \cos 2x} \end{array}} \right| ,x \in[0, \pi]$$

Then the maximum value of $f(x)$ is equal to $…..$

hard

(JEE MAIN-2021)

Verify Property $2$ for $\Delta=\left|\begin{array}{ccc}2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7\end{array}\right|$

Solution

Similar Questions

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Evaluate $\left|\begin{array}{ccc}x & y & x+y \\ y & x+y & x \\ x+y & x & y\end{array}\right|$

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

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

$$f(x)=\left| {\begin{array}{*{20}{c}} {{{\sin }^2}x}&{ – 2 + {{\cos }^2}x}&{\cos 2x} \\ {2 + {{\sin }^2}x}&{{{\cos }^2}x}&{\cos 2x} \\ {{{\sin }^2}x}&{{{\cos }^2}x}&{1 + \cos 2x} \end{array}} \right| ,x \in[0, \pi]$$ Then the maximum value of $f(x)$ is equal to $…..$

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$$f(x)=\left| {\begin{array}{*{20}{c}} {{{\sin }^2}x}&{ – 2 + {{\cos }^2}x}&{\cos 2x} \\ {2 + {{\sin }^2}x}&{{{\cos }^2}x}&{\cos 2x} \\ {{{\sin }^2}x}&{{{\cos }^2}x}&{1 + \cos 2x} \end{array}} \right| ,x \in[0, \pi]$$

Then the maximum value of $f(x)$ is equal to $…..$