To solve the following **Find the adjoint of the matrix A. Then use the adjoint to find the inverse of A (if possible). (If not possible, enter IMPOSSIBLE.)**Given matrix \( A \):\[A = \begin{bmatrix}1 & 0 & 2 & 4 \\0 & 0 & -1 & -2 \\2 & 1 & 1 & 2 \\2 & 1 & 1 & 3\end{bmatrix}\]**Adjoint of matrix \( A \), denoted as adj(\( A \)):**\[\text{adj}(A) = \begin{bmatrix}1 & 2 & 0 & 0 \\-2 & -3 & 1 & 0 \\0 & -1 & 2 & -2 \\0 & 0 & -1 & 1\end{bmatrix}\]**Inverse of matrix \( A \), denoted as \( A^{-1} \):**\[A^{-1} = \begin{bmatrix}1 & 0 & 2 & 4 \\0 & 0 & -1 & -2 \\2 & 1 & 1 & 2 \\2 & 1 & 1 & 3\end{bmatrix}\]- The step indicates the verification of calculations with arrows and checkmarks.- The initial matrix is transformed into its adjoint, and then attempts are made to calculate its inverse.- Due to the nature of instructions, completion by pressing enter was implied, but impossible situations were here left unaddressed when required. The task is to write vector **v** as a linear combination of vectors **u₁**, **u₂**, and **u₃**. If it's not possible to express **v** in this way, the word "IMPOSSIBLE" should be entered.Vectors are defined as follows:- **v** = (5, -25, -18, -16)- **u₁** = (1, -2, 1, 1)- **u₂** = (-2, 1, 3, 1)- **u₃** = (0, -3, -3, -3)The equation to solve is:\[ \mathbf{v} = \text{(\_\_\_)}\mathbf{u}_1 + \text{(\_\_\_)}\mathbf{u}_2 + \text{(\_\_\_)}\mathbf{u}_3 \]The goal is to find the scalar multipliers for **u₁**, **u₂**, and **u₃** that can combine to form **v**.

Here, A=102400-1-221122113 adjA=CijT Cij=-1i+jMij A=10-1-2112113-00-1-2212213+200-2212213-400-1211211 where, 0-1-2112113=0+13-2-21-1=1 00-2212213=0-0+-22-2=0 00-1211211=0-0-12-2=0 Thus , A=11-0+20-40=1 C11=0-1-2112113=0+13-2-21-1=1 C12=-0-1-2212213=-0+16-4-22-2=-2 C13=00-2212213=0-0-22-2=0 C14=-00-1211211=-0-0-12-2=0 C21=-024112113=-0-23-2+41-1=2 C22=124212213=13-2-26-4+42-2=-3 C23=-104212213=-13-2-0+42-2=-1C24=102211211=11-1-0+22-2=0 C31=0240-1-2113=0-20+2+40+1=0 C32=-1240-1-2213=-1-3+2-20+4+40+2=1 C33=10400-2213=10+2-0+40=2 C34=-10200-1211=-10+1-0+20=-1C41=-0240-1-2112=-0-20+2+40+1=0 C42=1240-1-2212=1-2+2-20+4+40+2=0 C43=-10400-2212=-10+2-0+40-0=-2 C44=10200-1211=10+1-0+20=1Cij=1-2002-3-10012-100-21 AdjA=1200-2-3100-12-200-11 A-1=1AAdjA =111200-2-3100-12-200-11 Thus ,A-1=1200-2-3100-12-200-11