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 See how quickly you looked at the hint, solution, and answer. This is important for making sure you will finish the FE Exam in time.
  • Hint: Not clicked
  • Solution: Not clicked
  • Answer: Not clicked

Similar Problems from FE Sub Section: Inverse
278. Matrix Inverse
511. Inverse Matrix

Similar Problems from FE Section: Matrices
277. Matrix Transpose
278. Matrix Inverse
279. Matrix Multiplication
507. Transpose Matrix
509. Transpose Size
511. Inverse Matrix
518. Matrix Multiply