Q.The inverse of $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ is (A) $\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$ (B) $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ (C) $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ (D) None of these

Computing the inverse directly gives $\alpha=1$.

Let $A=\begin{bmatrix}2&-6\\1&-2\end{bmatrix}$.

Step 1 — determinant:

$\det A = (2)(-2)-(-6)(1) = -4+6 = 2$

Step 2 — adjoint: for a $2\times2$ matrix $\begin{bmatrix}a&b\\c&d\end{bmatrix}$, the adjoint is $\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$:

$\operatorname{adj}A=\begin{bmatrix}-2&6\\-1&2\end{bmatrix}$

Step 3 — inverse:

$A^{-1}=\frac{1}{\det A}\operatorname{adj}A=\frac{1}{2}\begin{bmatrix}-2&6\\-1&2\end{bmatrix}=\begin{bmatrix}-1&3\\-\frac12&1\end{bmatrix}$

Comparing with the given inverse $\begin{bmatrix}-1&3\\-\frac12&\alpha\end{bmatrix}$, we get $\alpha=1$.

The uniqueness of the matrix inverse is a standard theorem provable by contradiction.

Suppose $A$ has two inverses $B$ and $C$, so $AB=BA=I$ and $AC=CA=I$. Then:

$B = BI = B(AC) = (BA)C = IC = C$

So $B=C$, proving the inverse (when it exists) is unique.

A non-singular (invertible) square matrix has one and only one inverse — uniqueness of the matrix inverse is a standard theorem.

If $A$ is a non-singular square matrix, $A^{-1}$ exists and is unique: if $B$ and $C$ both satisfy $AB=BA=I$ and $AC=CA=I$, then $B=BI=B(AC)=(BA)C=IC=C$.

So the number of multiplicative inverses of a non-singular square matrix is exactly $1$. None of "$0$", "$\frac12$" or "infinite" is correct — this looks like a gap in the supplied choice list (the correct choice "$1$" appears to be missing from what was extracted), so the honestly correct fact is given directly.

$A$ is an orthogonal (rotation) matrix with $|A|=1$, so $A^{-1}$ equals its adjoint, which equals $A^T$.

$|A| = \cos\theta\cdot\cos\theta - \sin\theta\cdot(-\sin\theta) = \cos^2\theta+\sin^2\theta = 1$

Cofactors: $A_{11}=\cos\theta,\ A_{12}=\sin\theta,\ A_{21}=-\sin\theta,\ A_{22}=\cos\theta$

$adj(A) = \begin{bmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{bmatrix}$

$A^{-1} = \dfrac{1}{|A|}adj(A) = \begin{bmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{bmatrix}$

Compute $|A|$, then the adjoint, then $A^{-1} = \frac{1}{|A|}\,\text{adj}(A)$.

$A = \begin{bmatrix} 3 & 4 \\ 1 & 2 \end{bmatrix}$

$|A| = 3\times 2 - 4\times 1 = 6-4 = 2$.

The adjoint matrix (swap diagonal entries, negate off-diagonal entries) is:

$\text{adj}(A) = \begin{bmatrix} 2 & -4 \\ -1 & 3 \end{bmatrix}$

So:

$A^{-1} = \frac{1}{2}\begin{bmatrix} 2 & -4 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ -\frac{1}{2} & \frac{3}{2} \end{bmatrix}$