If A=left[begin{array}{rr}1 & 0 0 & -1end{array}right] and B=left[begin{array}{ll}0 & 1

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3 and 4 .Determinants and Matrices

medium

If $A=\left[\begin{array}{rr}1 & 0 \\ 0 & -1\end{array}\right]$ and $B=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right],$ then $A B=\left[\begin{array}{rr}0 & 1 \\ -1 & 0\end{array}\right]$. and $\mathrm{BA}=\left[\begin{array}{rr}0 & -1 \\ 1 & 0\end{array}\right] .$ Clearly $\mathrm{AB} \neq \mathrm{BA}$.

Thus matrix multiplication is not commutative.

Option A

Option B

Option C

Option D

Solution

This does not mean that $\mathrm{AB} \neq \mathrm{BA}$ for every pair of matrices $\mathrm{A}, \mathrm{B}$ for which $\mathrm{AB}$ and $\mathrm{BA}$, are defined. For instance,

If $A=\left[\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right], B=\left[\begin{array}{ll}3 & 0 \\ 0 & 4\end{array}\right],$ then $A B=B A=\left[\begin{array}{ll}3 & 0 \\ 0 & 8\end{array}\right]$

Observe that multiplication of diagonal matrices of same order will be commutative.

Standard 12

Mathematics

Construct a $2 \times 2$ matrix, $A=\left[a_{ij}\right]$, whose elements are given by : $a_{i j}=\frac{(i+j)^{2}}{2}$

medium

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

easy

Let $A$ be a matrix such that $A.\,\left[ {\begin{array}{*{20}{c}}
1&2\\
0&3
\end{array}} \right]$ is a scalar matrix and $\left| {3A} \right| = 108$ . Then $A^2$ equals

hard

(JEE MAIN-2018)

If $A,B,C$ are three square matrices such that $AB = AC$ implies $B = C$, then the matrix $A$ is always a/an

normal

If $A$ is square matrix such that $A^{2}=A$, then $(1+A)^{3}-7 A$ is equal to

medium

Solution

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Let $A$ be a matrix such that $A.\,\left[ {\begin{array}{*{20}{c}}
1&2\\
0&3
\end{array}} \right]$ is a scalar matrix and $\left| {3A} \right| = 108$ . Then $A^2$ equals