Exercise 4.4 · Q8: Inverse of a lower triangular matrix

Basics you need

Solution

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

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

   $$\begin{matrix} + & - & + \\ - & + & - \\ + & - & + \end{matrix}$$ $$\begin{aligned} |A| &= a_{11}M_{11} - a_{12}M_{12} + a_{13}M_{13} \\ &= 1\cdot M_{11} - 0\cdot M_{12} + 0\cdot M_{13} \\ &= 1\cdot\begin{vmatrix} 3 & 0 \\ 2 & -1 \end{vmatrix} \\ &= 3\cdot(-1) - 0\cdot 2 = -3 \end{aligned}$$

   $|A| = -3\neq0$⁠, so $A$ is non-singular and $A^{-1}$ exists.

   The signs along row 1 are $+,\ -,\ +$⁠. The zeros remove two terms, leaving only $M_{11}$⁠. As Step 8 shows for every lower triangular $3\times3$ matrix, $|A| = 1\cdot3\cdot(-1)$ is the product of the diagonal entries. A negative determinant is perfectly fine; only $|A|=0$ would rule out an inverse. 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 $(3,\ 3,\ 0)$ and $(5,\ 2,\ -1)$⁠, and one column at a time:

   $$\begin{aligned} M_{11} &= \begin{vmatrix} 3 & 0 \\ 2 & -1 \end{vmatrix} = 3\cdot (-1) - 0\cdot 2 \\ &= -3 - 0 = -3 \\ A_{11} &= (-1)^{1+1}\,(-3) = -3 \\ M_{12} &= \begin{vmatrix} 3 & 0 \\ 5 & -1 \end{vmatrix} = 3\cdot (-1) - 0\cdot 5 \\ &= -3 - 0 = -3 \\ A_{12} &= (-1)^{1+2}\,(-3) = 3 \\ M_{13} &= \begin{vmatrix} 3 & 3 \\ 5 & 2 \end{vmatrix} = 3\cdot 2 - 3\cdot 5 \\ &= 6 - 15 = -9 \\ A_{13} &= (-1)^{1+3}\,(-9) = -9 \end{aligned}$$
   All three row 1 cofactors are nonzero. Position $(1,2)$ has odd $i+j$⁠, so $M_{12}=-3$ becomes $A_{12}=+3$⁠. In $M_{13}$⁠, $3\cdot2 - 3\cdot5 = 6 - 15 = -9$⁠.
   Definitions 1 and 2, Section 4.4. Lesson: minors and cofactors

3. Cofactors of the row 2 entries

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

   $$\begin{aligned} M_{21} &= \begin{vmatrix} 0 & 0 \\ 2 & -1 \end{vmatrix} = 0\cdot (-1) - 0\cdot 2 \\ &= 0 - 0 = 0 \\ A_{21} &= (-1)^{2+1}\,(0) = 0 \\ M_{22} &= \begin{vmatrix} 1 & 0 \\ 5 & -1 \end{vmatrix} = 1\cdot (-1) - 0\cdot 5 \\ &= -1 - 0 = -1 \\ A_{22} &= (-1)^{2+2}\,(-1) = -1 \\ M_{23} &= \begin{vmatrix} 1 & 0 \\ 5 & 2 \end{vmatrix} = 1\cdot 2 - 0\cdot 5 \\ &= 2 - 0 = 2 \\ A_{23} &= (-1)^{2+3}\,(2) = -2 \end{aligned}$$
   $M_{21}$ is 0 because what is left after deleting row 2 and column 1 has the zero row $(0,\ 0)$ from row 1. The zero entry $a_{23}$ still has the cofactor $A_{23}=-2$⁠.

4. Cofactors of the row 3 entries

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

   $$\begin{aligned} M_{31} &= \begin{vmatrix} 0 & 0 \\ 3 & 0 \end{vmatrix} = 0\cdot 0 - 0\cdot 3 \\ &= 0 - 0 = 0 \\ A_{31} &= (-1)^{3+1}\,(0) = 0 \\ M_{32} &= \begin{vmatrix} 1 & 0 \\ 3 & 0 \end{vmatrix} = 1\cdot 0 - 0\cdot 3 \\ &= 0 - 0 = 0 \\ A_{32} &= (-1)^{3+2}\,(0) = 0 \\ M_{33} &= \begin{vmatrix} 1 & 0 \\ 3 & 3 \end{vmatrix} = 1\cdot 3 - 0\cdot 3 \\ &= 3 - 0 = 3 \\ A_{33} &= (-1)^{3+3}\,(3) = 3 \end{aligned}$$
   $M_{31}$ and $M_{32}$ are 0: deleting row 3 and column 1 leaves the zero row $(0,\ 0)$ from row 1, and deleting row 3 and column 2 leaves the zero column $(0,\ 0)$ from column 3.

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

   $$[A_{ij}] = \begin{bmatrix} -3 & 3 & -9 \\ 0 & -1 & -2 \\ 0 & 0 & 3 \end{bmatrix}$$

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

   $$\operatorname{adj}A = \begin{bmatrix} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{bmatrix}$$
   The matrix of cofactors is upper triangular ($A_{21}=A_{31}=A_{32}=0$⁠); its transpose $\operatorname{adj}A$ is lower triangular, the same shape as $A$⁠.
   Definition 3, Section 4.5.1; transpose, Chapter 3, Section 3.5. Basics: transpose and identity

6. Divide by $|A|$

   $$\begin{aligned} A^{-1} &= \frac{1}{|A|}\operatorname{adj}A \\ &= -\frac{1}{3}\begin{bmatrix} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{bmatrix} \end{aligned}$$

   Multiplying every entry by $-\frac{1}{3}$⁠:

   $$A^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ -1 & \frac{1}{3} & 0 \\ 3 & \frac{2}{3} & -1 \end{bmatrix}$$
   This is Theorem 4. Dividing by $|A| = -3$ multiplies every entry by $-\frac{1}{3}$⁠, which flips every sign: for example $-3\cdot(-\frac{1}{3}) = 1$ and $-2\cdot(-\frac{1}{3}) = \frac{2}{3}$⁠.
   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 $A^{-1} = -\frac{1}{3}\operatorname{adj}A$⁠, it is enough to show $A(\operatorname{adj}A) = -3\,I$⁠. Write $A(\operatorname{adj}A)=[p_{ij}]$⁠. Row 1 of $A$ is $(1,\ 0,\ 0)$⁠:

   $$\begin{aligned} p_{11} &= 1\cdot (-3) + 0\cdot 3 + 0\cdot (-9) \\ &= -3 + 0 + 0 = -3 \\ p_{12} &= 1\cdot 0 + 0\cdot (-1) + 0\cdot (-2) \\ &= 0 + 0 + 0 = 0 \\ p_{13} &= 1\cdot 0 + 0\cdot 0 + 0\cdot 3 \\ &= 0 + 0 + 0 = 0 \end{aligned}$$

   Row 2 of $A$ is $(3,\ 3,\ 0)$⁠:

   $$\begin{aligned} p_{21} &= 3\cdot (-3) + 3\cdot 3 + 0\cdot (-9) \\ &= -9 + 9 + 0 = 0 \\ p_{22} &= 3\cdot 0 + 3\cdot (-1) + 0\cdot (-2) \\ &= 0 - 3 + 0 = -3 \\ p_{23} &= 3\cdot 0 + 3\cdot 0 + 0\cdot 3 \\ &= 0 + 0 + 0 = 0 \end{aligned}$$

   Row 3 of $A$ is $(5,\ 2,\ -1)$⁠:

   $$\begin{aligned} p_{31} &= 5\cdot (-3) + 2\cdot 3 + (-1)\cdot (-9) \\ &= -15 + 6 + 9 = 0 \\ p_{32} &= 5\cdot 0 + 2\cdot (-1) + (-1)\cdot (-2) \\ &= 0 - 2 + 2 = 0 \\ p_{33} &= 5\cdot 0 + 2\cdot 0 + (-1)\cdot 3 \\ &= 0 + 0 - 3 = -3 \end{aligned}$$ $$A(\operatorname{adj}A) = \begin{bmatrix} -3 & 0 & 0 \\ 0 & -3 & 0 \\ 0 & 0 & -3 \end{bmatrix} = -3\,I$$

   Hence $AA^{-1} = -\frac{1}{3}\,(-3\,I) = I$⁠. 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 $-3\,I$ 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 lower triangular

   For any lower triangular $3\times3$ matrix ($a_{12}=a_{13}=a_{23}=0$⁠), the three cofactors below the diagonal vanish:

   • $M_{21}$ keeps rows 1 and 3 in columns 2 and 3; its first row is $(a_{12},\ a_{13}) = (0,\ 0)$⁠, so it is 0.
   • $M_{31}$ keeps rows 1 and 2 in columns 2 and 3; its first row is again $(0,\ 0)$⁠, so it is 0.
   • $M_{32}$ keeps rows 1 and 2 in columns 1 and 3; its second column is $(a_{13},\ a_{23}) = (0,\ 0)$⁠, so it is 0.

   So $A_{21}=A_{31}=A_{32}=0$⁠: $[A_{ij}]$ is upper triangular, and its transpose $\operatorname{adj}A$ is lower triangular.

   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 above the diagonal, which is 0. Expanding along row 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 above 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 $1, \frac{1}{3}, -1$⁠, matching Step 6.

   This is the mirror image of Exercise 4.4 Q7, where an upper triangular matrix had an upper triangular inverse. Knowing the shape in advance is a free check on the answer.
   A determinant with a zero row or column is 0. Lesson: expanding a 3 × 3 determinant

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