Exercise 4.4 · Q7: Inverse of an upper triangular matrix

Basics you need

Solution

1. Compute $|A|$ and decide whether $A^{-1}$ exists

   Column 1 is $(1,\ 0,\ 0)$⁠, with two zeros, so expand along it. The sign pattern is

   $$\begin{matrix} + & - & + \\ - & + & - \\ + & - & + \end{matrix}$$ $$\begin{aligned} |A| &= a_{11}M_{11} - a_{21}M_{21} + a_{31}M_{31} \\ &= 1\cdot M_{11} - 0\cdot M_{21} + 0\cdot M_{31} \\ &= 1\cdot\begin{vmatrix} 2 & 4 \\ 0 & 5 \end{vmatrix} \\ &= 2\cdot 5 - 4\cdot 0 = 10 \end{aligned}$$

   $|A| = 10\neq0$⁠, so $A$ is non-singular and $A^{-1}$ exists.

   The signs down column 1 are $+,\ -,\ +$⁠. The two zero entries kill their terms whatever their minors are, so only $M_{11}$ is needed. Notice $|A| = 1\cdot2\cdot5$⁠, the product of the diagonal entries: expanding an upper triangular matrix down its first column always peels off one diagonal entry at a time (Step 8 shows this for every upper triangular $3\times3$ matrix). Theorem 4 says $A$ is invertible if and only if $|A|\neq0$⁠, so this one number settles the "if it exists" in the question.
   Section 4.2.3, Remarks (i) and (ii): expand along the line with the most zeros, using $+1$ or $-1$ as $i+j$ is even or odd. Lesson: expanding a 3 × 3 determinant

2. Cofactors of the row 1 entries

   Delete row 1, which leaves rows $(0,\ 2,\ 4)$ and $(0,\ 0,\ 5)$⁠, and one column at a time:

   $$\begin{aligned} M_{11} &= \begin{vmatrix} 2 & 4 \\ 0 & 5 \end{vmatrix} = 2\cdot 5 - 4\cdot 0 \\ &= 10 - 0 = 10 \\ A_{11} &= (-1)^{1+1}\,(10) = 10 \\ M_{12} &= \begin{vmatrix} 0 & 4 \\ 0 & 5 \end{vmatrix} = 0\cdot 5 - 4\cdot 0 \\ &= 0 - 0 = 0 \\ A_{12} &= (-1)^{1+2}\,(0) = 0 \\ M_{13} &= \begin{vmatrix} 0 & 2 \\ 0 & 0 \end{vmatrix} = 0\cdot 0 - 2\cdot 0 \\ &= 0 - 0 = 0 \\ A_{13} &= (-1)^{1+3}\,(0) = 0 \end{aligned}$$
   $M_{12}$ and $M_{13}$ are 0 because, once row 1 is gone, column 1 of what is left is $(0,\ 0)$⁠: both of these minors keep that zero column, and a $2\times2$ determinant with a zero column is 0. $M_{11}$ is the minor already used in Step 1.
   Definitions 1 and 2, Section 4.4. Lesson: minors and cofactors

3. Cofactors of the row 2 entries

   Delete row 2, which leaves rows $(1,\ 2,\ 3)$ and $(0,\ 0,\ 5)$⁠:

   $$\begin{aligned} M_{21} &= \begin{vmatrix} 2 & 3 \\ 0 & 5 \end{vmatrix} = 2\cdot 5 - 3\cdot 0 \\ &= 10 - 0 = 10 \\ A_{21} &= (-1)^{2+1}\,(10) = -10 \\ M_{22} &= \begin{vmatrix} 1 & 3 \\ 0 & 5 \end{vmatrix} = 1\cdot 5 - 3\cdot 0 \\ &= 5 - 0 = 5 \\ A_{22} &= (-1)^{2+2}\,(5) = 5 \\ M_{23} &= \begin{vmatrix} 1 & 2 \\ 0 & 0 \end{vmatrix} = 1\cdot 0 - 2\cdot 0 \\ &= 0 - 0 = 0 \\ A_{23} &= (-1)^{2+3}\,(0) = 0 \end{aligned}$$
   The entry $a_{21}$ is 0, but its cofactor $A_{21}=-10$ is not: a cofactor is built from the other entries. $M_{23}$ is 0 because what is left has the zero row $(0,\ 0)$ taken from row 3.

4. Cofactors of the row 3 entries

   Delete row 3, which leaves rows $(1,\ 2,\ 3)$ and $(0,\ 2,\ 4)$⁠:

   $$\begin{aligned} M_{31} &= \begin{vmatrix} 2 & 3 \\ 2 & 4 \end{vmatrix} = 2\cdot 4 - 3\cdot 2 \\ &= 8 - 6 = 2 \\ A_{31} &= (-1)^{3+1}\,(2) = 2 \\ M_{32} &= \begin{vmatrix} 1 & 3 \\ 0 & 4 \end{vmatrix} = 1\cdot 4 - 3\cdot 0 \\ &= 4 - 0 = 4 \\ A_{32} &= (-1)^{3+2}\,(4) = -4 \\ M_{33} &= \begin{vmatrix} 1 & 2 \\ 0 & 2 \end{vmatrix} = 1\cdot 2 - 2\cdot 0 \\ &= 2 - 0 = 2 \\ A_{33} &= (-1)^{3+3}\,(2) = 2 \end{aligned}$$
   None of the row 3 cofactors is 0. Position $(3,2)$ has odd $i+j=5$⁠, so $M_{32}=4$ becomes $A_{32}=-4$⁠.

5. The matrix of cofactors and $\operatorname{adj}A$

   $$[A_{ij}] = \begin{bmatrix} 10 & 0 & 0 \\ -10 & 5 & 0 \\ 2 & -4 & 2 \end{bmatrix}$$

   Transposing (row $i$ of $[A_{ij}]$ becomes column $i$⁠):

   $$\operatorname{adj}A = \begin{bmatrix} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{bmatrix}$$
   The matrix of cofactors is lower triangular (zeros above the diagonal: $A_{12}=A_{13}=A_{23}=0$⁠). Transposing turns it into an upper triangular matrix, the same shape as $A$⁠.
   Definition 3, Section 4.5.1; transpose, Chapter 3, Section 3.5. Basics: transpose and identity

6. Divide by $|A|$

   $$\begin{aligned} A^{-1} &= \frac{1}{|A|}\operatorname{adj}A \\ &= \frac{1}{10}\begin{bmatrix} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{bmatrix} \end{aligned}$$

   Multiplying every entry by $\frac{1}{10}$⁠:

   $$A^{-1} = \begin{bmatrix} 1 & -1 & \frac{1}{5} \\ 0 & \frac{1}{2} & -\frac{2}{5} \\ 0 & 0 & \frac{1}{5} \end{bmatrix}$$
   This is Theorem 4. Each entry of $\operatorname{adj}A$ is divided by 10: $\frac{2}{10}=\frac{1}{5}$⁠, $\frac{5}{10}=\frac{1}{2}$⁠, $\frac{-4}{10}=-\frac{2}{5}$⁠.
   Scalar multiple: each entry is multiplied by the scalar (Chapter 3, Section 3.4.2). Lesson: inverse of a matrix

7. Verify $AA^{-1} = I$

   Since $A^{-1} = \frac{1}{10}\operatorname{adj}A$⁠, it is enough to show $A(\operatorname{adj}A) = 10\,I$⁠. Write $A(\operatorname{adj}A)=[p_{ij}]$⁠. Row 1 of $A$ is $(1,\ 2,\ 3)$⁠:

   $$\begin{aligned} p_{11} &= 1\cdot 10 + 2\cdot 0 + 3\cdot 0 \\ &= 10 + 0 + 0 = 10 \\ p_{12} &= 1\cdot (-10) + 2\cdot 5 + 3\cdot 0 \\ &= -10 + 10 + 0 = 0 \\ p_{13} &= 1\cdot 2 + 2\cdot (-4) + 3\cdot 2 \\ &= 2 - 8 + 6 = 0 \end{aligned}$$

   Row 2 of $A$ is $(0,\ 2,\ 4)$⁠:

   $$\begin{aligned} p_{21} &= 0\cdot 10 + 2\cdot 0 + 4\cdot 0 \\ &= 0 + 0 + 0 = 0 \\ p_{22} &= 0\cdot (-10) + 2\cdot 5 + 4\cdot 0 \\ &= 0 + 10 + 0 = 10 \\ p_{23} &= 0\cdot 2 + 2\cdot (-4) + 4\cdot 2 \\ &= 0 - 8 + 8 = 0 \end{aligned}$$

   Row 3 of $A$ is $(0,\ 0,\ 5)$⁠:

   $$\begin{aligned} p_{31} &= 0\cdot 10 + 0\cdot 0 + 5\cdot 0 \\ &= 0 + 0 + 0 = 0 \\ p_{32} &= 0\cdot (-10) + 0\cdot 5 + 5\cdot 0 \\ &= 0 + 0 + 0 = 0 \\ p_{33} &= 0\cdot 2 + 0\cdot (-4) + 5\cdot 2 \\ &= 0 + 0 + 10 = 10 \end{aligned}$$ $$A(\operatorname{adj}A) = \begin{bmatrix} 10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \end{bmatrix} = 10\,I$$

   Hence $AA^{-1} = \frac{1}{10}\,(10\,I) = I$⁠. By Theorem 1, $(\operatorname{adj}A)A = |A|\,I$ as well, so $A^{-1}A = I$ too.

   The definition of the inverse (Chapter 3, Definition 6) asks for $AB = BA = I$⁠. The diagonal entries $p_{11}, p_{22}, p_{33}$ are expansions of $|A|$ along rows 1, 2, 3; the off-diagonal entries pair a row with the cofactors of another row, and the Note in Section 4.4 says such sums are 0. Getting exactly $10\,I$ confirms all nine cofactors at once.
   $A(kB) = k(AB)$ for a scalar $k$⁠. Basics: matrix facts the chapter uses

8. Why the inverse is upper triangular

   The zeros we found are not luck. For any upper triangular $3\times3$ matrix (so $a_{21}=a_{31}=a_{32}=0$⁠):

   • $M_{12}$ and $M_{13}$ keep column 1 of rows 2 and 3, which is $(a_{21},\ a_{31}) = (0,\ 0)$⁠: a zero column, so both are 0.
   • $M_{23}$ keeps rows 1 and 3 in columns 1 and 2; its second row is $(a_{31},\ a_{32}) = (0,\ 0)$⁠, so it is 0.

   So $A_{12}=A_{13}=A_{23}=0$⁠: the matrix of cofactors has zeros above the diagonal, and its transpose $\operatorname{adj}A$ has zeros below it.

   On the diagonal, $A_{11} = a_{22}a_{33}$⁠, $A_{22} = a_{11}a_{33}$⁠, $A_{33} = a_{11}a_{22}$⁠: in each of these minors the other product ($a_{23}a_{32}$⁠, $a_{13}a_{31}$⁠, $a_{12}a_{21}$⁠) contains an entry below the diagonal, which is 0. Expanding along column 1, as in Step 1, gives $|A| = a_{11}A_{11} = a_{11}a_{22}a_{33}$⁠.

   So $A^{-1}$ exists exactly when $a_{11}a_{22}a_{33}\neq0$⁠, that is, when no diagonal entry is 0. Then $A^{-1} = \frac{1}{|A|}\operatorname{adj}A$ has the same zeros below the diagonal as $\operatorname{adj}A$⁠, and its diagonal entries are

   $$\frac{a_{22}a_{33}}{a_{11}a_{22}a_{33}} = \frac{1}{a_{11}},\quad \frac{1}{a_{22}},\quad \frac{1}{a_{33}}.$$

   Here that is $1, \frac{1}{2}, \frac{1}{5}$⁠, exactly what Step 6 gave.

   Seeing the pattern gives a fast sanity check on any answer: an upper triangular matrix whose inverse has a nonzero entry below the diagonal, or whose inverse's diagonal is not the reciprocals, has been inverted wrongly.
   A determinant with a zero row or a zero column is 0: expand along that line and every term has a factor 0. Lesson: expanding a 3 × 3 determinant

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