Matrices Question 10
Question 10 - 30 January - Shift 1
Let $A=(\begin{matrix} p & q\end{matrix} ), d=|A| \neq 0 \quad|A-d(Adj A)|=0$. Then
(1) $(1+d)^{2}=(m+q)^{2}$
(2) $1+d^{2}=(m+q)^{2}$
(3) $(1+d)^{2}=m^{2}+q^{2}$
(4) $1+d^{2}=m^{2}+q^{2}$
Show Answer
Answer: (1)
Solution:
Formula: Properties of Adjoint of a Matrix, Determinant properties, Properties of Scalar Multiplication, Properties of Matrix Addition
$A= \begin{bmatrix} m & n \\ p & q\end{bmatrix} , \quad|A-d(adj A)|=0$
$\Rightarrow m-d(adj A)|=| \begin{bmatrix} m & n \\ p & q\end{bmatrix} -d \begin{bmatrix} q & -n \\ -p & m\end{bmatrix} \rvert,$
$= \begin{vmatrix} m-qd & n(1+d) \\ p(1+d) & q-md\end{vmatrix} =0$
$\Rightarrow \quad(m-qd)(q-md)-np(1+d)^{2}=0$
$\Rightarrow \quad m q-m^{2} d-q^{2} d+m q d^{2}-n p(1+d)^{2}=0$
$(m q-n p)+d^{2}(m q-n p)-d(m^{2}+q^{2}+2 n p)=0$
$\Rightarrow d+d^{3}-d((m+q)^{2}-2 d)=0$
$\Rightarrow 1+d^{2}=(m+q)^{2}-2 d$
$\Rightarrow \quad(1+d)^{2}=(m+q)^{2}$
$\therefore \quad$ Option (1) is correct.
