Lecture · 18 min · advanced

Matrices

Row reduction, invertibility, and solving Ax = b

Elimination, $Ax = b$, and the invertible matrix

A matrix is a linear map written down. Multiplying $A$ by $x$ is “apply the map.” Solving $A x = b$ is “find which inputs land on $b$ .” Gaussian elimination is the algorithm; invertibility is the geometric question of whether the map is one-to-one and onto.

On the board today

—Read $A x$ as a linear combination of the columns of $A$ .
—Row-reduce a small system.
—Know equivalent tests for invertibility of a square matrix.

Columns in action

A x = x_{1} a_{1} + x_{2} a_{2} + \dots + x_{n} a_{n}

$A x$ is in the column space. $A x = b$ is solvable iff $b$ is in that span.

For square $A$ , the following are equivalent: $A$ is invertible; $A x = b$ has a unique solution for every $b$ ; the columns are independent; $det A \neq = 0$ ; $0$ is not an eigenvalue. Elimination either produces $n$ pivots (invertible) or a zero row (singular). The inverse, when it exists, is the unique matrix $A^{- 1}$ with $A A^{- 1} = I$ .

Worked example · A $2 \times 2$ inverse

Invert $A = (2111)$ .

1. $det A = 2 - 1 = 1 \neq = 0$ , so invertible.
2.The formula $A^{- 1} = \frac{1}{d e t} (d - c - b a)$ gives $(1 - 1 - 1 2)$ .

Check: $A A^{- 1} = I$ .

Pause the lecture

If $A$ is singular, how many solutions can $A x = b$ have?

Common mistake. Writing $A^{- 1}$ for a non-square matrix, or assuming $A B = I$ implies $B A = I$ without using that $A, B$ are square. For square matrices the two-sided inverse is automatic.

Practice this lecture

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