If A =[[1,2],[a,b]] A² =[[0,0],[0,0]] then the values of a and b are:

Given, \( A = \begin{bmatrix} 1 & 2 \\ a & b \end{bmatrix} \) and \( A^2 = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \).

We have, \( A^2 = A \cdot A = \begin{bmatrix} 1 & 2 \\ a & b \end{bmatrix} \begin{bmatrix} 1 & 2 \\ a & b \end{bmatrix} = \begin{bmatrix} 1+2a & 2+2b \\ a+ba & 2a+b^2 \end{bmatrix} \)

Since \( A^2 = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \), we can equate the elements:

\( 1 + 2a = 0 \) ...(1)
\( 2 + 2b = 0 \) ...(2)
\( a + ba = 0 \) ...(3)
\( 2a + b^2 = 0 \) ...(4)

From (1), \( 2a = -1 \), so \( a = -\frac{1}{2} \). 😮

From (2), \( 2b = -2 \), so \( b = -1 \). 😎

Now, let's check if these values satisfy equations (3) and (4).

From (3), \( a + ba = 0 \) => \( a(1+b) = 0 \). Substituting the values \( a = -\frac{1}{2} \) and \( b = -1 \), we get \( -\frac{1}{2}(1 - 1) = 0 \), which is true.

From (4), \( 2a + b^2 = 0 \). Substituting the values \( a = -\frac{1}{2} \) and \( b = -1 \), we get \( 2(-\frac{1}{2}) + (-1)^2 = -1 + 1 = 0 \), which is also true.

Therefore, the values of \( a \) and \( b \) are \( -\frac{1}{2} \) and \( -1 \) respectively. 🎉