Given that matrix ` B = [{:( a,b),(c,d) :}]` find `lambda `so that `B^(2) - ( a+d) B =lambda l.` where I is a unit matrix of order 2.

Answer

To solve the problem, we need to find the value of \( \lambda \) such that

\[ B^2 - (a + d)B = \lambda I \]

where \( B = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) and \( I \) is the identity matrix of order 2, which is

\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}. \]

### Step 1: Calculate \( B^2 \)

To find \( B^2 \), we multiply matrix \( B \) by itself:

\[ B^2 = B \cdot B = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix}. \]

Calculating this gives:

\[ B^2 = \begin{pmatrix} a^2 + bc & ab + bd \\ ac + cd & bc + d^2 \end{pmatrix}. \]

### Step 2: Calculate \( (a + d)B \)

Next, we calculate \( (a + d)B \):

\[ (a + d)B = (a + d) \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} (a + d)a & (a + d)b \\ (a + d)c & (a + d)d \end{pmatrix}. \]

This simplifies to:

\[ (a + d)B = \begin{pmatrix} a^2 + da & ab + db \\ ac + dc & (a + d)d \end{pmatrix}. \]

### Step 3: Subtract \( (a + d)B \) from \( B^2 \)

Now we need to subtract \( (a + d)B \) from \( B^2 \):

\[ B^2 - (a + d)B = \begin{pmatrix} a^2 + bc & ab + bd \\ ac + cd & bc + d^2 \end{pmatrix} - \begin{pmatrix} a^2 + da & ab + db \\ ac + dc & (a + d)d \end{pmatrix}. \]

Calculating this gives:

\[ B^2 - (a + d)B = \begin{pmatrix} (bc - da) & (bd - db) \\ (cd - dc) & (bc + d^2 - (a + d)d) \end{pmatrix}. \]

This simplifies to:

\[ B^2 - (a + d)B = \begin{pmatrix} bc - ad & 0 \\ 0 & bc - ad \end{pmatrix}. \]

### Step 4: Set the result equal to \( \lambda I \)

We set the result equal to \( \lambda I \):

\[ \begin{pmatrix} bc - ad & 0 \\ 0 & bc - ad \end{pmatrix} = \begin{pmatrix} \lambda & 0 \\ 0 & \lambda \end{pmatrix}. \]

### Step 5: Solve for \( \lambda \)

From the above equation, we can see that:

\[ bc - ad = \lambda. \]

Thus, we find:

\[ \lambda = bc - ad. \]

### Final Answer

The value of \( \lambda \) is:

\[ \lambda = bc - ad. \]