Given A = left[ {begin{array}{*{20}{c}}1&32&2end{array}} right] ; I = left[ {begin{array}{*{

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3 and 4 .Determinants and Matrices

normal

Given $A =$$\left[ {\begin{array}{{20}{c}}1&3\\2&2\end{array}} \right]$ ; $I =$$\left[ {\begin{array}{{20}{c}}1&0\\0&1\end{array}} \right]$ . $If A - \lambda I$ is a singular matrix then

$\lambda \in \phi$

$\lambda ^2 - 3\lambda - 4 = 0$

$\lambda ^2 + 3\lambda + 4 = 0$

$\lambda ^2 - 3\lambda - 6 = 0$

Solution

$A – \lambda I$

$=$ $\left[ {\begin{array}{*{20}{c}}1&3\\2&2\end{array}} \right]$ – $\left[ {\begin{array}{*{20}{c}}\lambda &0\\0&\lambda\end{array}} \right]$ $=$ $\left[ {\begin{array}{*{20}{c}}{1 – \lambda }&3\\2&{2 – \lambda }\end{array}} \right]$

$= (1 – \lambda ) (2 – \lambda ) = \lambda ^2 – 3\lambda + 2 = 0$

i.e. for $A – \lambda I$ to be singular $\lambda ^ 2 – 3\lambda + 2 = 0$

since $A – \lambda I$ is singular ==> det. $(A – \lambda I)$ $= 0$
.hence $\left[ {\begin{array}{*{20}{c}}{1 – \lambda }&3\\2&{2 – \lambda }\end{array}} \right]$ $= 0$

$==> 2 – \lambda – 2\lambda + \lambda ^2 – 6 = 0$ or $\lambda ^2 – 3\lambda – 4 = 0$

Standard 12

Mathematics

For the matrix $A=\left[\begin{array}{ll}1 & 5 \\ 6 & 7\end{array}\right],$ verify that $\left(A+A^{\prime}\right)$ is a symmetric matrix

easy

Let $A=\left\{X=(x, y, z)^{T}: P X=0\right.$ and $\left.\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}=1\right\}$ where $\mathrm{P}=\left[\begin{array}{ccc}1 & 2 & 1 \\ -2 & 3 & -4 \\ 1 & 9 & -1\end{array}\right]$ then the set $\mathrm{A}$

hard

(JEE MAIN-2020)

An orthogonal matrix is

easy

Let $A=\left[\begin{array}{cc}0 & -2 \\ 2 & 0\end{array}\right]$. If $M$ and $N$ are two matrices given by $M =\sum \limits_{ k =1}^{10} A ^{2 k }$ and $N =\sum \limits_{ k =1}^{10} A ^{2 k -1}$ then $MN ^{2}$ is

medium

(JEE MAIN-2022)

If $A=\left(\begin{array}{cc}\frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \\ \frac{-2}{\sqrt{5}} & \frac{1}{\sqrt{5}}\end{array}\right), B=\left(\begin{array}{ll}1 & 0 \\ i & 1\end{array}\right), i=\sqrt{-1}$, and

$\mathrm{Q}=\mathrm{A}^{\mathrm{T}} \mathrm{BA}$, then the inverse of the matrix $\mathrm{A} \mathrm{Q}^{2021} \mathrm{~A}^{\mathrm{T}}$ is equal to :

hard

(JEE MAIN-2021)

Given $A =$$\left[ {\begin{array}{*{20}{c}}1&3\\2&2\end{array}} \right]$ ; $I =$$\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]$ . $If A - \lambda I$ is a singular matrix then