If A, B are two square matrices such that AB = A and BA = B, then prove that B² = B

If the matrices A = $\left[\begin{array}{ccc}2& 1& 3\\ 4& 1& 0\end{array}\right]$ and $\mathrm{B} = \left[\begin{array}{cc}1& - 1\\ 0& 2\\ 5& 0\end{array}\right]$, then AB will be

• $\left[\begin{array}{cc}17& 0\\ 4& - 2\end{array}\right]$

• $\left[\begin{array}{cc}4& 0\\ 0& 4\end{array}\right]$

• $\left[\begin{array}{cc}17& 4\\ 0& - 2\end{array}\right]$

• $\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

If A is a square matrix, then

• A + A^T is symmetric

• A A^T is skew-symmetric

• A^T + A is skew-symmetric

• A^TA is skew-symmetric

If A² - A + I = 0, then the inverse of the matrix A is

• A - I

• I - A

• A + I

• A

If A and B are square matrices of the same order and AB = 3I, then A^{- 1} is equal to

• 3B

• $\frac{1}{3}\mathrm{B}$

• 3B^{- 1}

• $\frac{1}{3}{\mathrm{B}}^{- 1}$

The values of x for which the given matrix

$\left[\begin{array}{ccc}- \mathrm{x}& \mathrm{x}& 2\\ 2& \mathrm{x}& - \mathrm{x}\\ \mathrm{x}& - 2& - \mathrm{x}\end{array}\right]$ will be non-singular, are

• $- 2 \le \mathrm{x} \le 2$

• for all x other than 2 and - 2

• $\mathrm{x} \ge 2$

• $\mathrm{x} \le - 2$

If the matrix $\left[\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\end{array}\right]$ is commutative with the matrix  $\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$, then

• a = 0, b = c

• b = 0, c = d

• c = 0, d = a

• d = 0, a = b

If 1, w, w² are the cube roots of unity, then

$\left|\begin{array}{ccc}1& {\mathrm{w}}^{\mathrm{n}}& {\mathrm{w}}^{2\mathrm{n}}\\ {\mathrm{w}}^{2\mathrm{n}}& 1& {\mathrm{w}}^{\mathrm{n}}\\ {\mathrm{w}}^{\mathrm{n}}& {\mathrm{w}}^{2\mathrm{n}}& 1\end{array}\right|$ has value

• 0

• w

• w²

• w + w²

If A is a square matrix such that A² = A and B = I - A, then AB + BA + I - (I - A)² is equal to

• A

• 2A

• - A

• I - A

If A = $\frac{1}{3}\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& - 2\\ \mathrm{a}& 2& \mathrm{b}\end{array}\right]$ is an orthagonal matrix, then

• a = 2, b = 1

• a = - 2, b = - 1

• a = 2, b = - 1

• a = - 2, b = 1