MCQ
Let $A =\left(\begin{array}{cc} m & n \\ p & q \end{array}\right), d =| A | \neq 0| A - d (\operatorname{Adj} A )|=0$. Then
- ✓$(1+ d )^2=( m + q )^2$
- B$1+ d ^2=( m + q )^2$
- C$(1+ d )^2= m ^2+ q ^2$
- D$1+ d ^2= m ^2+ q ^2$
✓
Answer
Correct option: A.
$(1+ d )^2=( m + q )^2$
a
$A =\left[\begin{array}{cc} m & n \\ p & q \end{array}\right], \quad| A - d (\operatorname{adj} A )|=0$
$A =\left[\begin{array}{cc} m & n \\ p & q \end{array}\right], \quad| A - d (\operatorname{adj} A )|=0$
$\Rightarrow \quad|A-d(\operatorname{adj} A)|=\left|\left[\begin{array}{cc} m & n \\ p & q \end{array}\right]- d \left[\begin{array}{cc} q & - n \\ - p & m \end{array}\right]\right|$
$=\left|\begin{array}{ll} m - qd & n (1+ d ) \\ p (1+ d ) & q - md \end{array}\right|=0$
$\Rightarrow \quad(m-q d)(q-m d)-n p(1+d)^2=0$
$\Rightarrow \quad m q-m^2 d-q^2 d+m q d^2-n p(1+d)^2=0$
$\Rightarrow \quad(m q-n p)+d^2(m q-n p)-d\left(m^2+q^2+2 n p\right)=0$
$\Rightarrow \quad d+d^3-d\left((m+q)^2-2 d\right)=0$
$\Rightarrow \quad 1+d^2=(m+q)^2-2 d$
$\Rightarrow \quad(1+d)^2=(m+q)^2$
$\therefore \quad$ Option $(1)$ is correct.
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