The inverse of a matrix

We want the point that $A$ sends to $(3, 2)$, so we call it $(p, q)$ and write out $A(p, q)$ with Day 4's column picture:

That gives $2p + q = 3$ and $p + q = 2$. Subtracting the second equation from the first removes $q$ and leaves $p = 1$. Then $q = 2 - 1 = 1$.

So $\yellow{\mathbf x} = (1, 1)$. Solving worked, but we would have to solve again for every new target. It would be better to have one matrix that runs $A$ backwards on any point.

If $C$ undoes $A$, then doing $A$ and then $C$ should change nothing. By Day 6, "do $A$, then $C$" is the product $CA$. We compute it with the row–column rule:

That is the identity $I$, the matrix that leaves every point where it is. The other order checks that doing $C$ first and then $A$ also changes nothing:

Both products are $I$, so the claim holds. As a check, $C(3, 2) = (3 - 2, -3 + 4) = (1, 1)$, which is the point from Problem 1.

Call the point $(p, q)$. We need $2p + q = 1$ and $p + q = 0$. Subtracting the second equation from the first gives $p = 1$, and then $q = -1$.

The point is $(1, -1)$, which is exactly the first column of $C$. That is no accident. The first column of $A^{-1}$ is $A^{-1}\hat\imath$, and $A^{-1}$ carries $\hat\imath$ back to the point that $A$ sent onto $\hat\imath$.

We multiply with the row–column rule, one entry at a time:

The result is $2I$, which is almost the identity. Dividing the new matrix by $2$ divides the product by $2$ and gives exactly $I$:

The $2$ we divided by is $4 \cdot 2 - 3 \cdot 2$. The same swap and sign flip works for any $2 \times 2$ matrix, and the leftover number is always $ad - bc$.

The system is $A\mathbf x = \mathbf b$ with $A = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}$ and $\teal{\mathbf b} = (4, 3)$. We want $\mathbf x$ alone, so we multiply both sides on the left by $A^{-1}$, which cancels the $A$:

Now compute the right side:

Check: $2(1) + 2 = 4$ and $1 + 2 = 3$.

We want to remove the $A$ that sits on the right of $X$. Multiplying on the left would put $A^{-1}$ in front of $X$, where it cancels nothing. So we multiply both sides on the right by $A^{-1}$:

The inverse has to go on the same side as the $A$ it cancels, and it has to go on that side of both equations.

The determinant of $RS$ is $0 \cdot 1 - (-1)(1) = 1$. Swapping the diagonal and flipping the other signs gives

Now the two candidates. The formula gives $S^{-1} = \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}$ and $R^{-1} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$.

So $(RS)^{-1} = S^{-1}R^{-1}$. Read right to left, it undoes the turn first and the shear second. The turn was the last move made, so it has to be the first move undone, just as you take off your shoes before your socks.

Use the entries as weights on the columns $(1, 2)$ and $(2, 4)$:

All three points land on $(1, 2)$. A matrix that undoes $M$ would have to send $(1, 2)$ back to all three of them at once, and a matrix sends each input to only one output. So no inverse exists. Its determinant confirms this: $1 \cdot 4 - 2 \cdot 2 = 0$.

The three inputs differ by multiples of $(-2, 1)$, and $M(-2, 1) = -2(1, 2) + 1(2, 4) = \mathbf 0$. Adding a vector that lands on $\mathbf 0$ never changes where a point lands.

The matrix is singular exactly when its determinant is zero, so we compute the determinant with $h$ left as a letter: $2h - 1 \cdot 4 = 2h - 4$. Setting $2h - 4 = 0$ gives $h = 2$.

At $h = 2$ the columns are $(2, 4)$ and $(1, 2)$. The first is twice the second, so they lie on one line. For every other $h$, the columns point in different directions, and the matrix is invertible.