BCA COURSE / BCS-012

 To show that \(A^2 - 4A + 5I_2 = 0\), where \(I_2\) is the 2x2 identity matrix, we simply substitute the matrix \(A\) into the expression and perform the operations:

Given:

\[ A = \begin{pmatrix} 3 & -1 \\ 2 & 1 \end{pmatrix} \]

First, calculate \(A^2\):

\[ A^2 = A \times A = \begin{pmatrix} 3 & -1 \\ 2 & 1 \end{pmatrix} \times \begin{pmatrix} 3 & -1 \\ 2 & 1 \end{pmatrix} \]

\[ = \begin{pmatrix} (3 \times 3 + (-1) \times 2) & (3 \times -1 + (-1) \times 1) \\ (2 \times 3 + 1 \times 2) & (2 \times -1 + 1 \times 1) \end{pmatrix} \]

\[ = \begin{pmatrix} 7 & -4 \\ 8 & -3 \end{pmatrix} \]

Now, calculate \(4A\):

\[ 4A = 4 \times \begin{pmatrix} 3 & -1 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} 12 & -4 \\ 8 & 4 \end{pmatrix} \]

Finally, add \(5I_2\):

\[ 5I_2 = 5 \times \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} \]

Now, let's subtract \(4A\) from \(A^2\) and add \(5I_2\):

\[ A^2 - 4A + 5I_2 = \begin{pmatrix} 7 & -4 \\ 8 & -3 \end{pmatrix} - \begin{pmatrix} 12 & -4 \\ 8 & 4 \end{pmatrix} + \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} \]

\[ = \begin{pmatrix} 7 & -4 \\ 8 & -3 \end{pmatrix} - \begin{pmatrix} 12 & -4 \\ 8 & 4 \end{pmatrix} + \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} \]

\[ = \begin{pmatrix} 7 & -4 \\ 8 & -3 \end{pmatrix} - \begin{pmatrix} 12 & -4 \\ 8 & 4 \end{pmatrix} + \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} \]

\[ = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \]

Thus, \(A^2 - 4A + 5I_2 = 0\).

Next, to find \(A^4\), we can simply square \(A^2\), as we already have \(A^2\) calculated:

\[ (A^2)^2 = \begin{pmatrix} 7 & -4 \\ 8 & -3 \end{pmatrix} \times \begin{pmatrix} 7 & -4 \\ 8 & -3 \end{pmatrix} \]

\[ = \begin{pmatrix} (7 \times 7 + (-4) \times 8) & (7 \times -4 + (-4) \times -3) \\ (8 \times 7 + (-3) \times 8) & (8 \times -4 + (-3) \times -3) \end{pmatrix} \]

\[ = \begin{pmatrix} 33 & 28 \\ 28 & 33 \end{pmatrix} \]

So, \(A^4 = \begin{pmatrix} 33 & 28 \\ 28 & 33 \end{pmatrix}\).