9514 1404 393
Answer:
they are not
Step-by-step explanation:
The inverse matrix is the transpose of the cofactor matrix, divided by the determinant.
The cofactor matrix for a 2×2 matrix is ...
![\left[\begin{array}{cc}a_{22}&-a_{21}\\-a_{12}&a_{11}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da_%7B22%7D%26-a_%7B21%7D%5C%5C-a_%7B12%7D%26a_%7B11%7D%5Cend%7Barray%7D%5Cright%5D)
The transpose of this will have the off-diagonal terms swapped, so the inverse matrix is ...
![\displaystyle\frac{\left[\begin{array}{cc}a_{22}&-a_{12}\\-a_{21}&a_{11}\end{array}\right]}{a_{11}a_{22}-a_{21}a_{12}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cfrac%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da_%7B22%7D%26-a_%7B12%7D%5C%5C-a_%7B21%7D%26a_%7B11%7D%5Cend%7Barray%7D%5Cright%5D%7D%7Ba_%7B11%7Da_%7B22%7D-a_%7B21%7Da_%7B12%7D%7D)
We see that the second matrix is the transpose of the cofactor matrix, but the determinant is (5)(2)-(3)(4) = -2, so there has clearly been no division by the determinant. The actual inverse matrix of the first one shown is ...
![\left[\begin{array}{cc}-1&2\\1.5&-2.5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-1%262%5C%5C1.5%26-2.5%5Cend%7Barray%7D%5Cright%5D)
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You can compute the matrix product to see if you get an identity matrix. Here, the upper left term in the product is ...
(5)(2) +(4)(-3) = -2 . . . . . not 1, so the product matrix is not an identity matrix
The matrices are not inverses.