Cos θ left[ {begin{array}{*{20}{c}}{cos θ }&{sin θ }{

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3 and 4 .Determinants and Matrices

easy

$\cos \theta \left[ {\begin{array}{{20}{c}}{\cos \theta }&{\sin \theta }\\{ - \sin \theta }&{\cos \theta }\end{array}} \right] + \sin \theta \left[ {\begin{array}{{20}{c}}{\sin \theta }&{ - \cos \theta }\\{\cos \theta }&{\sin \theta }\end{array}} \right] = $

$\left[ {\begin{array}{*{20}{c}}0&0\\0&0\end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}}1&0\\0&0\end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}}0&1\\1&0\end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]$

Solution

(d) $\cos \theta \left[ {\begin{array}{*{20}{c}}{\cos \theta }&{\sin \theta }\\{ – \sin \theta }&{\cos \theta }\end{array}} \right] + \sin \theta \left[ {\begin{array}{*{20}{c}}{\sin \theta }&{ – \cos \theta }\\{\cos \theta }&{\sin \theta }\end{array}} \right]$

=$\left[ {\begin{array}{*{20}{c}}{{{\cos }^2}\theta + {{\sin }^2}\theta }&0\\0&{{{\cos }^2}\theta + {{\sin }^2}\theta }\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]$.

Standard 12

Mathematics

If $A =$ $\left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right)$ satisfies the equation $x^2 – (a + d) x + k = 0$, then

medium

A trust fund has Rs. $30,000$ that must be invested in two different types of bonds. The first bond pays $5 \%$ interest per year, and the second bond pays $7 \%$ interest per year. Using matrix multiplication, determine how to divide Rs. $30,000$ among the two types of bonds. If the trust fund must obtain an annual total interest of Rs. $1800$.

hard

Let $A=\left[a_{i j}\right]_{2 \times 2}$ where $a_{i j} \neq 0$ for all $i, j$ and $A^2=I$. Let a be the sum of all diagonal elements of $A$ and $b =| A |$, then $3 a ^2+4 b ^2$ is equal to

hard

(JEE MAIN-2023)

Square matrix ${[{a_{ij}}]_{n \times n}}$ will be an upper triangular matrix, if

easy

Let $\mathrm{A}=\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0\end{array}\right) .$ Then $\mathrm{A}^{2025}-\mathrm{A}^{2020}$ is equal to :

medium

(JEE MAIN-2021)

$\cos \theta \left[ {\begin{array}{*{20}{c}}{\cos \theta }&{\sin \theta }\\{ - \sin \theta }&{\cos \theta }\end{array}} \right] + \sin \theta \left[ {\begin{array}{*{20}{c}}{\sin \theta }&{ - \cos \theta }\\{\cos \theta }&{\sin \theta }\end{array}} \right] = $