Question

Let $A = \left[ {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right]$ and \[B = \left[ {\begin{array}{*{20}{c}}
  \alpha \\
  \beta
\end{array}} \right] \ne \left[ {\begin{array}{*{20}{c}}
  0 \\
  0
\end{array}} \right]\] such that \[AB = B\] and \[a + d = 2021\], then the value of ad-bcis equal to

Answer 1

Hint: To get the value of\[ad - bc\], multiply the matrix A and matrix B and equate it to matrix B. The equations obtained are solved in such a way that $\alpha $ and \[\beta \] are eliminated. Substitute the value of \[a + d\] to get the value of \[ad - bc\].

Complete step by step Solution:
Consider the given condition \[AB = B\]
$\left[ {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  \alpha \\
  \beta
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  \alpha \\
  \beta
\end{array}} \right]$
Multiply the Matrix A and Matrix B and equate it to matrix B.
\[\left[ {\begin{array}{*{20}{c}}
  {a\alpha + b\beta } \\
  {c\alpha + d\beta }
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  \alpha \\
  \beta
\end{array}} \right]\]
When two matrices have equal number of rows and columns and they are equal to each other then each row of the matrices is equal to the row of the equal matrices so we can use this in the question to find the values.
now we will equate Left hand side equations to the right hand side equations
\[a\alpha + b\beta = \alpha \]---- (equation 1)
\[c\alpha + d\beta = \beta \]--- (equation 2)
by equating we get the above equations so now we will solve them
Solving equation (1)
 \[a\alpha - \alpha = - b\beta \]
Take \[\alpha \]as common from above equation
 \[\alpha (a - 1) = - b\beta \]
\[\dfrac{\alpha }{\beta } = \dfrac{{ - b}}{{a - 1}}\]--- (equation 3)
Solving equation (2)
\[c\alpha = \beta - d\beta \]
Take \[\beta \] as common from above equation
\[c\alpha = \beta (1 - d)\]
\[\dfrac{\alpha }{\beta } = \dfrac{{(1 - d)}}{c}\] --- (equation 4)
Equating equation (3) to equation (4)
\[\dfrac{-b}{a-1}\] = \[\dfrac{(1-d)}{c}\]
Cross multiply the above equation
\[ - bc = (a - 1)(1 - d)\]
\[ - bc = a - ad - 1 + d\]
Solve for \[ad - bc\]
\[ad - bc = a + d - 1\]--- equation (5)
Substitute the given value of \[a + d\] in equation (5) i.e. \[a + d = 2021\]
Therefore \[ad - bc = 2021 - 1\]
\[ad - bc = 2020\]

Note:As \[\alpha \]and \[\beta \] have non zero values, solve in a such a way that \[\alpha \]and \[\beta \] gets eliminated to get the value of \[ad - bc\].Many students make mistakes while multiplying A and B so should take care of that. Also, while equating product equals to B then equating equation students can make mistakes. We should do that carefully to avoid any errors in the solution.