If A = left[ {begin{array}{*{20}{c}}0&10&0end{array}} right],I is unit matrix of order 2 and

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3 and 4 .Determinants and Matrices

easy

If $A = \left[ {\begin{array}{*{20}{c}}0&1\\0&0\end{array}} \right],I$ is the unit matrix of order $2$ and $a, b$ are arbitrary constants, then ${(aI + bA)^2}$ is equal to

${a^2}I + abA$

${a^2}I + 2abA$

${a^2}I + {b^2}A$

None of these

Solution

(b) ${(aI + bA)^2} = \left[ {\begin{array}{*{20}{c}}a&b\\0&a\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}a&b\\0&a\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{a^2}}&{2ab}\\0&{{a^2}}\end{array}} \right] = {a^2}I + 2abA$.

Standard 12

Mathematics

The set of all $2 \times 2$ matrices over the real numbers is not a group under matrix multiplication because

normal

If $A = \left[ {\begin{array}{*{20}{c}}2&{ – 3}\\{ – 4}&1\end{array}} \right],$ then $adj\;\left( {3{A^2} + 12A} \right) = $ . . . .

medium

(JEE MAIN-2017)

$A$ and $B$ are two square matrix such that $A^2B = BA$ and If $(AB)^{10} = A^K B^{10}$ then $k$ is

normal

Let $A = \left( {\begin{array}{{20}{c}}
{\cos \,\alpha }&{ – \sin \,\alpha }\\
{\sin \,\alpha }&{\cos \,\alpha }
\end{array}} \right)$, $\left( {\alpha \in R} \right)$ such that ${A^{32}} = \left( {\begin{array}{{20}{c}}
0&{ – 1}\\
1&0
\end{array}} \right)$. Then a value of $\alpha $ is

hard

(JEE MAIN-2019)

Let $A=\left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3\end{array}\right)$. Then the sum of the diagonal elements of the matrix $( A + I )^{11}$ is equal to:

hard

(JEE MAIN-2023)