Let A = begin{bmatrix} m & n p & q end{bmatr | vedclass

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(A) Given $A = \begin{bmatrix} m & n \ p & q \end{bmatrix}$,where $d = |A| = mq - np 
eq 0$.The adjoint of $A$ is $\operatorname{adj} A = \begin{bma

Vedclass · Accepted Answer

$(1+d)^2 = (m+q)^2$

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(A) Given $A = \begin{bmatrix} m & n \ p & q \end{bmatrix}$,where $d = |A| = mq - np 
eq 0$.The adjoint of $A$ is $\operatorname{adj} A = \begin{bmatrix} q & -n \ -p & m \end{bmatrix}$.We are given $|A - d(\operatorname{adj} A)| = 0$.Substituting the matrices:$|\begin{bmatrix} m & n \ p & q \end{bmatrix} - d \begin{bmatrix} q & -n \ -p & m \end{bmatrix}| = 0$$|\begin{bmatrix} m - qd & n + nd \ p + pd & q - md \end{bmatrix}| = 0$$|\begin{bmatrix} m - qd & n(1+d) \ p(1+d) & q - md \end{bmatrix}| = 0$$(m - qd)(q - md) - np(1+d)^2 = 0$$mq - m^2d - q^2d + mqd^2 - np(1+d)^2 = 0$$(mq - np) + d^2(mq - np) - d(m^2 + q^2 + 2np) = 0$Since $d = mq - np$,we have:$d + d^3 - d(m^2 + q^2 + 2np) = 0$Divide by $d$ (as $d 
eq 0$):$1 + d^2 - (m^2 + q^2 + 2np) = 0$$1 + d^2 = m^2 + q^2 + 2np$Since $(m+q)^2 = m^2 + q^2 + 2mq$,we can write $m^2 + q^2 = (m+q)^2 - 2mq$.$1 + d^2 = (m+q)^2 - 2mq + 2np$$1 + d^2 = (m+q)^2 - 2(mq - np)$$1 + d^2 = (m+q)^2 - 2d$$1 + 2d + d^2 = (m+q)^2$$(1+d)^2 = (m+q)^2$.

Let $A = \begin{bmatrix} m & n \\ p & q \end{bmatrix}$,$d = |A| \neq 0$ and $|A - d(\operatorname{Adj} A)| = 0$. Then:

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