Let A be a 2 times 2 matrix of the form A = b | vedclass

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(C) The matrix $A$ is given by $A = \begin{bmatrix} a & b \ 1 & 1 \end{bmatrix}$.For $A^{-1}$ to exist,the determinant $|A| = a - b$ must be non-zero

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$202$

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(C) The matrix $A$ is given by $A = \begin{bmatrix} a & b \ 1 & 1 \end{bmatrix}$.For $A^{-1}$ to exist,the determinant $|A| = a - b$ must be non-zero,i.e.,$a 
eq b$.The inverse of $A$ is given by $A^{-1} = \frac{1}{a-b} \begin{bmatrix} 1 & -b \ -1 & a \end{bmatrix} = \begin{bmatrix} \frac{1}{a-b} & -\frac{b}{a-b} \ -\frac{1}{a-b} & \frac{a}{a-b} \end{bmatrix}$.For $A^{-1}$ to contain only integer entries,each entry must be an integer.This requires $(a-b)$ to be a divisor of $1$,$-b$,$-1$,and $a$.Specifically,$(a-b)$ must divide $1$,which implies $a-b = 1$ or $a-b = -1$.Case $1$: $a-b = 1 \implies a = b+1$.Since $-50 \leq b \leq 50$,there are $101$ possible values for $b$,and for each $b$,$a$ is uniquely determined.Case $2$: $a-b = -1 \implies a = b-1$.Similarly,for $-50 \leq b \leq 50$,there are $101$ possible values for $b$,and for each $b$,$a$ is uniquely determined.Thus,the total number of such matrices is $101 + 101 = 202$.

Let $A$ be a $2 \times 2$ matrix of the form $A = \begin{bmatrix} a & b \\ 1 & 1 \end{bmatrix}$,where $a, b$ are integers and $-50 \leq b \leq 50$. The number of such matrices $A$ such that $A^{-1}$,the inverse of $A$,exists and $A^{-1}$ contains only integer entries is

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