Question

PLEASE HELP! find the inverse matrix​

Answer 1

Answer:

see explanation

Step-by-step explanation:

The inverse of a matrix A = $\left[\begin{array}{ccc}a&b\\c&d\\\end{array}\right]$ is

$A^{-1}$ = $\frac{1}{ad-bc}$ $\left[\begin{array}{ccc}d&-b\\-c&a\\\end{array}\right]$

If ad - bc = 0 then the matrix has no inverse

A = $\left[\begin{array}{ccc}11&-5\\3&-1\\\end{array}\right]$

ad - bc = (11 × - 1) - (- 5 × 3) = - 11 - (- 15) = - 11 + 15 = 4 , then

$A^{-1}$ = $\frac{1}{4}$ $\left[\begin{array}{ccc}-1&3\\-5&11\\\end{array}\right]$ = $\left[\begin{array}{ccc}-0.25&0.75\\-1.25&2.75\\\end{array}\right]$

Answer 2

Answer:

Step-by-step explanation:

$A=\begin{bmatrix}11&-5\\3&-1\end{bmatrix}\\det(A)=11*(-1)-3*(-5)=-11+15=4\\minor=\begin{bmatrix}-1&3\\-5&11\end{bmatrix}\\cofactor=\begin{bmatrix}-1&-3\\5&11\end{bmatrix}\\Transposed\ cofactor=\begin{bmatrix}-1&5\\-3&11\end{bmatrix}\\A^{-1}=\dfrac{Transposed\ cofactorx}{det(A)} =\begin{bmatrix}\dfrac{-1}{4} &\dfrac{5}{4} \\\\ \dfrac{-3}{4} &\dfrac{11}{4}\end{bmatrix}$

$Proof:\\A*A^{-1}=\begin{bmatrix}11&-5\\\\3&-1\end{bmatrix}*\begin{bmatrix}\dfrac{-1}{4} &\dfrac{5}{4}\\\\ \dfrac{-3}{4}&\dfrac{11}{4}\end{bmatrix}\\\\\\=\begin{bmatrix}\dfrac{-11+15}{4} &\dfrac{55-55}{4}\\\\ \dfrac{-3+3}{4}&\dfrac{15-11}{4}\end{bmatrix}\\\\\\=\begin{bmatrix}1&0\\\\ 0&1\end{bmatrix}$

Bonus: method of Gauss 's pivot jointed