Let A = [ cases a & b c & d cases ] and B = [ array ll α β array ] ≠ [ array ll 0 0 array ] such that AB = B and a + d =2021, then the value of ad - bc is equal to ______ .

Correct answer: 0 Given A = [ cases a & b c & d cases ], B = [ array ll α β array ] And, AB = B ⇒ ( A - I ) B = O ⇒ |( A - I ) B| = 0 ⇒ |( A - I )| × |B| = 0 ⇒ | A - I | = 0, since B ≠ 0 ⇒ | cases ( a - 1 ) & b c & ( d - 1 ) cases | = 0 ⇒ (a - 1)(d - 1) - bc = 0 ⇒ ad - a - d + 1 - bc = 0 ⇒ ad - bc = a + d - 1 ⇒ ad - bc = 2021 - 1 = 2020 .

JEE Main 202117 MarchEvening Shift

Exam: JEE Main
Year: 2021
Date: 17 March
Shift: Evening Shift
Subject: Mathematics
Chapter: Matrices

JEE Main 2021 · 17 March · Evening Shift · Mathematics · Matrices

Let A = [ cases a & b c & d cases ] and B = [ array ll α β array ]...

Let

A = \left[

\begin{cases}a & b \\ c & d\end{cases}

\right]

and

B = \left[

\begin{array}{ll}\alpha \\ \beta\end{array}

\right] \ne \left[

\begin{array}{ll}0 \\ 0\end{array}

\right]

such that

AB = B

and

a + d =2021,

then the value of

ad - bc

is equal to ______ .

Correct answer

Step-by-step solution

Given $A = \left[$ \begin{cases}a & b \\ c & d\end{cases} $\right], B = \left[$ $\begin{array}{ll}\alpha \\ \beta\end{array}$ $\right]$

And, $AB = B$

$⇒ ( A - I ) B = O$

$⇒ \left|( A - I ) B\right| = 0$

$⇒ \left|( A - I )\right| \times \left|B\right| = 0$

$⇒ | A - I | = 0,$ since $B \ne 0$

$⇒ \left|$ \begin{cases}( a - 1 ) & b \\ c & ( d - 1 )\end{cases} $\right| = 0$

$⇒ \left(a - 1\right)\left(d - 1\right) - bc = 0$

$⇒ ad - a - d + 1 - bc = 0$

$⇒ ad - bc = a + d - 1$

$⇒ ad - bc = 2021 - 1 = 2020 .$

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