If A = left[ {begin{array}{*{20}{c}}{ - 1}&0&00&{ - 1}&00&0&{

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3 and 4 .Determinants and Matrices

easy

If $A = \left[ {\begin{array}{*{20}{c}}{ - 1}&0&0\\0&{ - 1}&0\\0&0&{ - 1}\end{array}} \right]$, then ${A^2}$is

Null matrix

Unit matrix

$A$

$2A$

Solution

(b) ${A^2} = \left[ {\begin{array}{*{20}{c}}{ – 1}&0&0\\0&{ – 1}&0\\0&0&{ – 1}\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}{ – 1}&0&0\\0&{ – 1}&0\\0&0&{ – 1}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right] = I$.

Standard 12

Mathematics

If $A = \left[ {\begin{array}{*{20}{c}}x&1\\1&0\end{array}} \right]$ and ${A^2}$ is the identity matrix, then $x =$

easy

If $A$ is a square matrix of order $n$ and $A = kB$, where $k$ is a scalar, then $|A|=$

easy

Order of $\left[ {x\,y\,z} \right]\,\left[ {\begin{array}{{20}{c}}
a&h&g\\
h&b&f\\
g&f&c
\end{array}} \right]\,\left[ {\begin{array}{{20}{c}}
x\\
y\\
z
\end{array}} \right]$ is

medium

If $A = \left[ {\begin{array}{{20}{c}}\alpha &0\\1&1\end{array}} \right]$ and $B = \left[ {\begin{array}{{20}{c}}1&0\\5&1\end{array}} \right]$, then value of $\alpha $for which ${A^2} = B$, is

easy

(IIT-2003)

Let be a $3 \times 3$ matrix such that $X ^{ T } AX = O$ for all nonzero $3 \times 1$ matrices $X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$. If $A \left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=\left[\begin{array}{l}1 \\ 4 \\ -5\end{array}\right], A \left[\begin{array}{l}1 \\ 2 \\ 1\end{array}\right]=\left[\begin{array}{l}0 \\ 4 \\ -8\end{array}\right]$, and $\operatorname{det}(\operatorname{adj}(2(A+ I )))=2^\alpha 3^\beta 5^\gamma, \alpha, \beta, \gamma, \in N$, then $\alpha^2+\beta^2+\gamma^2$ is

normal

(JEE MAIN-2025)

If $A = \left[ {\begin{array}{*{20}{c}}{ - 1}&0&0\\0&{ - 1}&0\\0&0&{ - 1}\end{array}} \right]$, then ${A^2}$is

Solution

Similar Questions

If $A = \left[ {\begin{array}{*{20}{c}}x&1\\1&0\end{array}} \right]$ and ${A^2}$ is the identity matrix, then $x =$

If $A$ is a square matrix of order $n$ and $A = kB$, where $k$ is a scalar, then $|A|=$

Order of $\left[ {x\,y\,z} \right]\,\left[ {\begin{array}{*{20}{c}} a&h&g\\ h&b&f\\ g&f&c \end{array}} \right]\,\left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right]$ is

If $A = \left[ {\begin{array}{*{20}{c}}\alpha &0\\1&1\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}1&0\\5&1\end{array}} \right]$, then value of $\alpha $for which ${A^2} = B$, is

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Order of $\left[ {x\,y\,z} \right]\,\left[ {\begin{array}{{20}{c}}
a&h&g\\
h&b&f\\
g&f&c
\end{array}} \right]\,\left[ {\begin{array}{{20}{c}}
x\\
y\\
z
\end{array}} \right]$ is

If $A = \left[ {\begin{array}{{20}{c}}\alpha &0\\1&1\end{array}} \right]$ and $B = \left[ {\begin{array}{{20}{c}}1&0\\5&1\end{array}} \right]$, then value of $\alpha $for which ${A^2} = B$, is