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Step-by-Step Solution
Step 1: Write Down the Given Matrices and Conditions
We have
A =
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
and
B =
\begin{bmatrix}
\alpha \\
\beta
\end{bmatrix}
\neq
\begin{bmatrix}
0 \\
0
\end{bmatrix}.
The problem states that
AB = B
and
a + d = 2021.
We want to find
ad - bc.
Step 2: Express the Equation AB = B
From
AB = B,
we get
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
\begin{bmatrix}
\alpha \\
\beta
\end{bmatrix}
=
\begin{bmatrix}
\alpha \\
\beta
\end{bmatrix}.
Carrying out the matrix multiplication on the left-hand side gives
\begin{bmatrix}
a\alpha + b\beta \\
c\alpha + d\beta
\end{bmatrix}
=
\begin{bmatrix}
\alpha \\
\beta
\end{bmatrix}.
Therefore, we have two equations:
a\alpha + b\beta = \alpha \quad (1)
and
c\alpha + d\beta = \beta \quad (2).
Step 3: Rearrange Each Equation
From equation (1):
a\alpha + b\beta = \alpha
\ \Rightarrow\
a\alpha - \alpha = -\,b\beta
\ \Rightarrow\
\alpha(a - 1) = -\,b\beta.
Hence,
\frac{\alpha}{\beta} = \frac{-\,b}{a - 1}
(assuming
\beta \neq 0).
From equation (2):
c\alpha + d\beta = \beta
\ \Rightarrow\
d\beta - \beta = -\,c\alpha
\ \Rightarrow\
\beta(d - 1) = -\,c\alpha.
Rearranging gives
c\alpha = \beta(1 - d),
so
\frac{\alpha}{\beta} = \frac{1 - d}{c}.
Step 4: Equate the Two Expressions for α/β
Since both expressions represent
\frac{\alpha}{\beta},
we set
\frac{-\,b}{a - 1} = \frac{1 - d}{c}.
Cross-multiplying:
-\,b\,c = (a - 1)(1 - d).
Step 5: Expand the Right-Hand Side and Simplify
Expand
(a - 1)(1 - d):
(a - 1)(1 - d) = a - ad - 1 + d.
Thus,
-\,bc = a - ad - 1 + d.
Rearrange to isolate
ad - bc:
ad - bc = a + d - 1.
Step 6: Use the Given Sum of Diagonal Elements
We know
a + d = 2021.
Substitute this into
ad - bc = a + d - 1:
ad - bc = 2021 - 1 = 2020.
Final Answer
ad - bc = 2020.