Inverse of Matrix
What is the inverse of the shown matrix?
Expand Hint
The inverse of a square
$$n\times n$$
matrix is:
$$$A^{-1}=\frac{adj(A)}{|A|}$$$
where
$$|A|$$
is the determinant of matrix A, and
$$adj(A)$$
is the adjoint of matrix A, which is obtained by replacing
$$A^T$$
elements with their cofactors.
Hint 2
$$$\begin{bmatrix}a & b\\ c & d\end{bmatrix}^{-1}=\frac{1}{ad-bc}\begin{bmatrix}d & -b\\ -c & a\end{bmatrix}$$$
The inverse of a square
$$n\times n$$
matrix is:
$$$A^{-1}=\frac{adj(A)}{|A|}$$$
where
$$|A|$$
is the determinant of matrix A, and
$$adj(A)$$
is the adjoint of matrix A, which is obtained by replacing
$$A^T$$
elements with their cofactors. Thus, for a 2 x 2 matrix:
$$$\begin{bmatrix}a & b\\ c & d\end{bmatrix}^{-1}=\frac{1}{ad-bc}\begin{bmatrix}d & -b\\ -c & a\end{bmatrix}$$$
Therefore,
$$$\begin{bmatrix}1 & 0\\ -1 & 4\end{bmatrix}^{-1}=\frac{1}{(1)(4)-(0)(-1)}\begin{bmatrix}4 & -0\\ 1 & 1\end{bmatrix}$$$
$$$=\frac{1}{4-0}\begin{bmatrix}4 & 0\\ 1 & 1\end{bmatrix}=\frac{1}{4}\begin{bmatrix}4 & 0\\ 1 & 1\end{bmatrix}=\begin{bmatrix}1 & 0\\ 0.25 & 0.25\end{bmatrix}$$$
The solution can be verified by multiplying the original matrix with its inverse. The resultant is the identity matrix (
$$n\times n$$
matrix with 1’s on the diagonal and 0’s everywhere else).
$$$[A][A]^{-1}=[A]^{-1}[A]=[I]$$$
where
$$[I]$$
is the identity matrix,
$$[A]$$
is the original matrix, and
$$[A]^{-1}$$
is the inverse matrix.
$$$\begin{bmatrix}1 & 0\\ 0.25 & 0.25\end{bmatrix}$$$
Time Analysis
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