#### Need solution for RD Sharma maths class 12 chapter Algebra of matrices exercise 4.3 question 6

Hence proved $A^2=B^2=C^2=I_2$

Hint: matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.

Given: $A=\begin{bmatrix} 1 &0 \\ 0 &1 \end{bmatrix}, B =\begin{bmatrix} 1 &0 \\ 0&-1 \end{bmatrix}$ and $C=\begin{bmatrix} 0 &1 \\ 1 &0 \end{bmatrix}$

To Prove: $A^2=B^2=C^2=I_2$

We know that $A^2=AA$

$A^{2}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 \times 1+0 \times 0 & 1 \times 0+0 \times 1 \\ 0 \times 1+1 \times 0 & 0 \times 0+1 \times 1 \end{array}\right]=\left[\begin{array}{ll} 1+0 & 0+0 \\ 0+0 & 0+1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$                        ...(i)

Again we know that,

$B^2=BB$

$B^{2}=\left[\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right]\left[\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right]=\left[\begin{array}{cc} 1 \times 1+0 \times 0 & 1 \times 0+0 \times(-1) \\ 0 \times 1+(-1) \times 0 & 0 \times 0+(-1) \times(-1) \end{array}\right]$

$B^{2}=\left[\begin{array}{ll} 1+0 & 0-0 \\ 0-0 & 0+1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$                             ...(ii)

Now consider,

$C^2=CC$

$C^{2}=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]\left[\begin{array}{ll} 1 & 0 \\ 1 & 0 \end{array}\right]=\left[\begin{array}{ll} 0 \times 0+1 \times 1 & 0 \times 1+1 \times 0 \\ 1 \times 0+0 \times 1 & 1 \times 1+0 \times 0 \end{array}\right]$

$C^{2}=\begin{bmatrix} 0+1 & 0+0\\ 0+0 &1+0 \end{bmatrix}=\begin{bmatrix} 1 &0 \\ 0 &1 \end{bmatrix}$                           ...(iii)

And $I_2=\begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}$  where $I_2$ refers to an identity matrix having order 2x2 ( or matrix with two rows and two columns) and 1’s in the main diagonal.

$I_2=\begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}$                                                        ...(iv)

Now, from equation (i), (ii), (iii) and (iv), it is clear that $A^2=B^2=C^2=I_2$