Lecture 16 · Week 4· 17:55

Inverse matrices

Invertibility, Gauss–Jordan inversion.

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In this lecture

Today: undoing a matrix
Invertible matrices
One inverse, and solving A x = b
Two by two inverses
Rules for inverses
One product is enough
Gauss–Jordan inversion
Inverting a three by three matrix
When there is no inverse
Summary

Materials

Check yourself

Try it before you look

Use Theorem 3.5.6 to find the inverse of $[3512]$ , and check one of the two products.
Use Gauss–Jordan inversion to find the inverse of $120011001$ .
Let $A = 101213314$ . Is $A$ invertible? Decide with Theorem 3.5.10, and find a nonzero $x$ with $A x = 0$ if there is one.

Work it out on paper first. Then open the hint if you are stuck, and the answer to check.

1Hint

Everything you need is in Section 3.5 of the course book. Open it at page 29 of the chapter PDF; any theorem the question names is stated there, with an example worked the same way.

2Answer

Here $a d - b c = 3 \cdot 2 - 1 \cdot 5 = 1$ , so the inverse is $[2 - 5 - 1 3]$ . Check: $[3512] [2 - 5 - 1 3] = [6 - 5 10 - 10 - 3 + 3 - 5 + 6] = I_{2}$ .
$R_{2} \to R_{2} - 2 R_{1}$ gives row $2$ $= [010 ∣ - 210]$ ; then $R_{3} \to R_{3} - R_{2}$ gives row $3$ $= [001 ∣ 2 - 1 1]$ . The left block is $I_{3}$ , so the inverse is $1 - 2 2 01 - 1 001$ (Theorem 3.5.10(a)).
No. $R_{3} \to R_{3} - R_{1}$ gives row $3$ $= (0, 1, 1)$ , and $R_{3} \to R_{3} - R_{2}$ makes it a zero row, so the reduced row echelon form is not $I_{3}$ and the matrix is not invertible (Theorem 3.5.10(b)). With $x_{3} = 1$ free, $x_{2} = - 1$ and $x_{1} = - 2 x_{2} - 3 x_{3} = - 1$ : indeed $A (- 1, - 1, 1) = (- 1 - 2 + 3, - 1 + 1, - 1 - 3 + 4) = (0, 0, 0)$ .