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Chapter 4 · Determinants

Exercise 4.4 · Q15: Show A3−6A2+5A+11I=OA^3-6A^2+5A+11I=O⁠; hence find A−1A^{-1}

Exercise 4.4Adjoint and inverse of a matrix · Section 4.5
The question

For the matrix

A=[11112−32−13]A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}

show that A3−6A2+5A+11 I=OA^3 - 6A^2 + 5A + 11\,I = O⁠. Hence, find A−1A^{-1}⁠.

Exam answer

Entries cijc_{ij} of A2=AAA^2 = AA (row ii times column jj⁠):

c11=1+1+2=4c12=1+2−1=2c13=1−3+3=1c21=1+2−6=−3c22=1+4+3=8c23=1−6−9=−14c31=2−1+6=7c32=2−2−3=−3c33=2+3+9=14\begin{aligned} c_{11} &= 1 + 1 + 2 = 4 \\ c_{12} &= 1 + 2 - 1 = 2 \\ c_{13} &= 1 - 3 + 3 = 1 \\ c_{21} &= 1 + 2 - 6 = -3 \\ c_{22} &= 1 + 4 + 3 = 8 \\ c_{23} &= 1 - 6 - 9 = -14 \\ c_{31} &= 2 - 1 + 6 = 7 \\ c_{32} &= 2 - 2 - 3 = -3 \\ c_{33} &= 2 + 3 + 9 = 14 \end{aligned} A2=[421−38−147−314]A^2 = \begin{bmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix}

Entries dijd_{ij} of A3=A2AA^3 = A^2A⁠:

d11=4+2+2=8d12=4+4−1=7d13=4−6+3=1d21=−3+8−28=−23d22=−3+16+14=27d23=−3−24−42=−69d31=7−3+28=32d32=7−6−14=−13d33=7+9+42=58\begin{aligned} d_{11} &= 4 + 2 + 2 = 8 \\ d_{12} &= 4 + 4 - 1 = 7 \\ d_{13} &= 4 - 6 + 3 = 1 \\ d_{21} &= -3 + 8 - 28 = -23 \\ d_{22} &= -3 + 16 + 14 = 27 \\ d_{23} &= -3 - 24 - 42 = -69 \\ d_{31} &= 7 - 3 + 28 = 32 \\ d_{32} &= 7 - 6 - 14 = -13 \\ d_{33} &= 7 + 9 + 42 = 58 \end{aligned} A3=[871−2327−6932−1358]A^3 = \begin{bmatrix} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{bmatrix} A3−6A2+5A+11I=[871−2327−6932−1358]−[24126−1848−8442−1884]+[555510−1510−515]+[110001100011]=[000000000]=O\begin{aligned} &A^3 - 6A^2 + 5A + 11I \\ &= \begin{bmatrix} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{bmatrix} \\ &\quad - \begin{bmatrix} 24 & 12 & 6 \\ -18 & 48 & -84 \\ 42 & -18 & 84 \end{bmatrix} \\ &\quad + \begin{bmatrix} 5 & 5 & 5 \\ 5 & 10 & -15 \\ 10 & -5 & 15 \end{bmatrix} \\ &\quad + \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix} \\ &= \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = O \end{aligned}

Hence A3−6A2+5A+11I=OA^3 - 6A^2 + 5A + 11I = O⁠.

Expanding along R1R_1⁠:

∣A∣=1(6−3)−1(3+6)+1(−1−4)=3−9−5=−11\begin{aligned} |A| &= 1(6 - 3) - 1(3 + 6) \\ &\quad + 1(-1 - 4) \\ &= 3 - 9 - 5 = -11 \end{aligned}

Since ∣A∣≠0|A| \ne 0⁠, A−1A^{-1} exists. Post-multiplying by A−1A^{-1}⁠,

(A3−6A2+5A+11I)A−1=OA−1⇒ A2−6A+5I+11A−1=O⇒ A−1=−111(A2−6A+5I)\begin{aligned} &(A^3 - 6A^2 + 5A + 11I)A^{-1} \\ &\qquad = OA^{-1} \\ &\Rightarrow\ A^2 - 6A + 5I + 11A^{-1} = O \\ &\Rightarrow\ A^{-1} = -\frac{1}{11}(A^2 - 6A + 5I) \end{aligned} A2−6A+5I=[421−38−147−314]−[666612−1812−618]+[500050005]=[3−4−5−914−531]\begin{aligned} &A^2 - 6A + 5I \\ &= \begin{bmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix} \\ &\quad - \begin{bmatrix} 6 & 6 & 6 \\ 6 & 12 & -18 \\ 12 & -6 & 18 \end{bmatrix} \\ &\quad + \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix} \\ &= \begin{bmatrix} 3 & -4 & -5 \\ -9 & 1 & 4 \\ -5 & 3 & 1 \end{bmatrix} \end{aligned}

∴ A−1=111[−3459−1−45−3−1]\therefore\ A^{-1} = \dfrac{1}{11}\begin{bmatrix} -3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1 \end{bmatrix}

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Once the equation is known to hold, multiply it by A−1A^{-1}⁠: each power of AA drops by one and 11I11I becomes 11A−111A^{-1}⁠. That leaves

A2−6A+5I+11A−1=O,A^2 - 6A + 5I + 11A^{-1} = O,

so A−1=−111(A2−6A+5I)A^{-1} = -\frac{1}{11}(A^2 - 6A + 5I)⁠, built from A2A^2⁠, which we will already have computed. No cofactors are needed.

Given, to find, and the plan
Given
The 3×33\times3 matrix AA above; II and OO are the 3×33\times3 identity and zero matrices.
To show / find
A3−6A2+5A+11I=OA^3 - 6A^2 + 5A + 11I = O⁠; then A−1A^{-1}⁠.
Plan
(1) A2A^2⁠. (2) A3=A2AA^3 = A^2A⁠. (3) Combine the four terms entry by entry. (4) ∣A∣≠0|A| \neq 0⁠, so A−1A^{-1} exists. (5) Multiply the equation by A−1A^{-1}⁠. (6) Compute A2−6A+5IA^2 - 6A + 5I and A−1A^{-1}⁠. (7) Check AA−1=IAA^{-1} = I⁠.

Basics you need

1Powers of a matrix

A2=AAA^2 = AA and A3=A2AA^3 = A^2A⁠, matrix products computed row times column. Basics: matrix facts the chapter uses

2Scalar multiples and sums

6A26A^2 multiplies every entry by 6; 11I11I has 1111 on the diagonal only; same-order matrices add entry by entry. Basics: scaling and adding matrices

3Expanding a 3×33\times3 determinant

Along any row: each element times its cofactor, summed (Section 4.2.3). Lesson: expanding a 3 × 3 determinant

4Inverse

AA−1=A−1A=IAA^{-1} = A^{-1}A = I⁠; A−1A^{-1} exists exactly when ∣A∣≠0|A| \neq 0 (Theorem 4). The method is the one in Example 15. Basics: why the inverse divides by |A|

Solution

  1. Compute A2=A⋅AA^2 = A\cdot A

    AA is 3×33\times3⁠, so A⋅AA\cdot A is defined and is 3×33\times3⁠. The rows of AA are (1,1,1)(1,1,1)⁠, (1,2,−3)(1,2,-3)⁠, (2,−1,3)(2,-1,3)⁠; its columns are (1,1,2)(1,1,2)⁠, (1,2,−1)(1,2,-1)⁠, (1,−3,3)(1,-3,3)⁠.

    c11=1⋅1+1⋅1+1⋅2=4c12=1⋅1+1⋅2+1(−1)=2c13=1⋅1+1(−3)+1⋅3=1c21=1⋅1+2⋅1+(−3)2=−3c22=1⋅1+2⋅2+(−3)(−1)=8c23=1⋅1+2(−3)+(−3)3=−14c31=2⋅1+(−1)1+3⋅2=7c32=2⋅1+(−1)2+3(−1)=−3c33=2⋅1+(−1)(−3)+3⋅3=14\begin{aligned} c_{11} &= 1\cdot1 + 1\cdot1 + 1\cdot2 = 4\\ c_{12} &= 1\cdot1 + 1\cdot2 + 1(-1) = 2\\ c_{13} &= 1\cdot1 + 1(-3) + 1\cdot3 = 1\\[1mm] c_{21} &= 1\cdot1 + 2\cdot1 + (-3)2 = -3\\ c_{22} &= 1\cdot1 + 2\cdot2 + (-3)(-1) = 8\\ c_{23} &= 1\cdot1 + 2(-3) + (-3)3 = -14\\[1mm] c_{31} &= 2\cdot1 + (-1)1 + 3\cdot2 = 7\\ c_{32} &= 2\cdot1 + (-1)2 + 3(-1) = -3\\ c_{33} &= 2\cdot1 + (-1)(-3) + 3\cdot3 = 14 \end{aligned} A2=[421−38−147−314]A^2 = \begin{bmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix}
    Listing the rows and columns of AA first makes each entry a mechanical "row ii times column jj". For example c23c_{23} pairs row 2, (1,2,−3)(1, 2, -3)⁠, with column 3, (1,−3,3)(1, -3, 3)⁠: 1−6−9=−141 - 6 - 9 = -14⁠.
    Pair up, multiply, add. Basics: matrix facts the chapter uses

  2. Compute A3=A2⋅AA^3 = A^2\cdot A

    A2A^2 is 3×33\times3 and AA is 3×33\times3⁠, so the product is 3×33\times3⁠. Rows of A2A^2 times the same columns of AA⁠:

    d11=4⋅1+2⋅1+1⋅2=8d12=4⋅1+2⋅2+1(−1)=7d13=4⋅1+2(−3)+1⋅3=1d21=(−3)1+8⋅1+(−14)2=−23d22=(−3)1+8⋅2+(−14)(−1)=27d23=(−3)1+8(−3)+(−14)3=−69d31=7⋅1+(−3)1+14⋅2=32d32=7⋅1+(−3)2+14(−1)=−13d33=7⋅1+(−3)(−3)+14⋅3=58\begin{aligned} d_{11} &= 4\cdot1 + 2\cdot1 + 1\cdot2 = 8\\ d_{12} &= 4\cdot1 + 2\cdot2 + 1(-1) = 7\\ d_{13} &= 4\cdot1 + 2(-3) + 1\cdot3 = 1\\[1mm] d_{21} &= (-3)1 + 8\cdot1 + (-14)2 = -23\\ d_{22} &= (-3)1 + 8\cdot2 + (-14)(-1) = 27\\ d_{23} &= (-3)1 + 8(-3) + (-14)3 = -69\\[1mm] d_{31} &= 7\cdot1 + (-3)1 + 14\cdot2 = 32\\ d_{32} &= 7\cdot1 + (-3)2 + 14(-1) = -13\\ d_{33} &= 7\cdot1 + (-3)(-3) + 14\cdot3 = 58 \end{aligned} A3=[871−2327−6932−1358]A^3 = \begin{bmatrix} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{bmatrix}
    A3A^3 means AAAAAA⁠. Matrix multiplication is associative, so (AA)A(AA)A is the same as A(AA)A(AA)⁠; computing A2⋅AA^2\cdot A reuses the work of Step 1.

  3. Combine A3−6A2+5A+11IA^3 - 6A^2 + 5A + 11I

    6A2=[24126−1848−8442−1884]6A^2 = \begin{bmatrix} 24 & 12 & 6 \\ -18 & 48 & -84 \\ 42 & -18 & 84 \end{bmatrix} 5A=[555510−1510−515]5A = \begin{bmatrix} 5 & 5 & 5 \\ 5 & 10 & -15 \\ 10 & -5 & 15 \end{bmatrix}

    11I11I adds 1111 to the diagonal entries only. All four matrices are 3×33\times3⁠. Entry by entry (A3A^3⁠, minus 6A26A^2⁠, plus 5A5A⁠, plus 11I11I⁠):

    (1,1): 8−24+5+11=0(1,2): 7−12+5+0=0(1,3): 1−6+5+0=0(2,1): −23+18+5+0=0(2,2): 27−48+10+11=0(2,3): −69+84−15+0=0(3,1): 32−42+10+0=0(3,2): −13+18−5+0=0(3,3): 58−84+15+11=0\begin{aligned} (1,1)&:\ 8 - 24 + 5 + 11 = 0\\ (1,2)&:\ 7 - 12 + 5 + 0 = 0\\ (1,3)&:\ 1 - 6 + 5 + 0 = 0\\[1mm] (2,1)&:\ -23 + 18 + 5 + 0 = 0\\ (2,2)&:\ 27 - 48 + 10 + 11 = 0\\ (2,3)&:\ -69 + 84 - 15 + 0 = 0\\[1mm] (3,1)&:\ 32 - 42 + 10 + 0 = 0\\ (3,2)&:\ -13 + 18 - 5 + 0 = 0\\ (3,3)&:\ 58 - 84 + 15 + 11 = 0 \end{aligned}

    Every entry is 00⁠, so A3−6A2+5A+11I=OA^3 - 6A^2 + 5A + 11I = O.

    Subtracting a negative entry adds: in (2,1)(2,1)⁠, −6⋅(−3)=+18-6\cdot(-3) = +18⁠. That sign is where most slips happen, so each row writes the four contributions in the same order.
    Basics: scaling and adding matrices

  4. ∣A∣≠0|A| \neq 0⁠, so A−1A^{-1} exists

    No row or column contains a zero; expand along row 1, whose entries are all 1. Signs +,−,++, -, +⁠.

    M11=∣2−3−13∣=2⋅3−(−3)(−1)=3M12=∣1−323∣=1⋅3−(−3)2=9M13=∣122−1∣=1(−1)−2⋅2=−5\begin{aligned} M_{11} &= \begin{vmatrix} 2 & -3 \\ -1 & 3 \end{vmatrix}\\ &= 2\cdot3-(-3)(-1) = 3\\[1mm] M_{12} &= \begin{vmatrix} 1 & -3 \\ 2 & 3 \end{vmatrix}\\ &= 1\cdot3-(-3)2 = 9\\[1mm] M_{13} &= \begin{vmatrix} 1 & 2 \\ 2 & -1 \end{vmatrix}\\ &= 1(-1)-2\cdot2 = -5 \end{aligned} ∣A∣=1⋅3−1⋅9+1⋅(−5)=3−9−5=−11\begin{aligned} |A| &= 1\cdot3 - 1\cdot9 + 1\cdot(-5)\\ &= 3 - 9 - 5 = -11 \end{aligned}

    ∣A∣=−11≠0|A| = -11 \neq 0⁠, so AA is non-singular and A−1A^{-1} exists.

    We are about to multiply by A−1A^{-1}⁠, which is only legitimate if it exists (Theorem 4). The number 11 in the term 11I11I matching ∣A∣=−11|A| = -11 is not chance: the cubic obtained from ∣xI−A∣|xI - A| by putting AA in place of xx always ends in −∣A∣ I-|A|\,I⁠. That result (the Cayley–Hamilton theorem) is beyond this book and is not needed here.
    Lesson: expanding a 3 × 3 determinant

  5. Multiply the equation by A−1A^{-1}

    Post-multiply every term of AAA−6AA+5A+11I=OAAA - 6AA + 5A + 11I = O by A−1A^{-1}⁠, and use AA−1=IAA^{-1} = I⁠:

    AA(AA−1)−6A(AA−1)+5AA−1+11IA−1=OA−1A2−6A+5I+11A−1=O\begin{aligned} AA(AA^{-1}) - 6A(AA^{-1}) &\\ + 5AA^{-1} + 11IA^{-1} &= OA^{-1}\\ A^2 - 6A + 5I + 11A^{-1} &= O \end{aligned}

    So

    A−1=−111(A2−6A+5I)A^{-1} = -\frac{1}{11}\left(A^2 - 6A + 5I\right)
    The distributive law lets us multiply each term separately, all on the right, and associativity lets us regroup, for example (AAA)A−1(AAA)A^{-1} as AA(AA−1)AA(AA^{-1})⁠. Then AA−1=IAA^{-1} = I⁠, XI=XXI = X⁠, IA−1=A−1IA^{-1} = A^{-1} and OA−1=OOA^{-1} = O⁠. Finally 11A−1=−(A2−6A+5I)11A^{-1} = -(A^2 - 6A + 5I)⁠, and dividing by the number 11 gives the formula.
    Basics: why the inverse divides by |A|

  6. Compute A2−6A+5IA^2 - 6A + 5I and A−1A^{-1}

    Entry by entry (A2A^2⁠, minus 6A6A⁠, plus 5I5I on the diagonal):

    (1,1): 4−6+5=3(1,2): 2−6+0=−4(1,3): 1−6+0=−5(2,1): −3−6+0=−9(2,2): 8−12+5=1(2,3): −14+18+0=4(3,1): 7−12+0=−5(3,2): −3+6+0=3(3,3): 14−18+5=1\begin{aligned} (1,1)&:\ 4 - 6 + 5 = 3\\ (1,2)&:\ 2 - 6 + 0 = -4\\ (1,3)&:\ 1 - 6 + 0 = -5\\[1mm] (2,1)&:\ -3 - 6 + 0 = -9\\ (2,2)&:\ 8 - 12 + 5 = 1\\ (2,3)&:\ -14 + 18 + 0 = 4\\[1mm] (3,1)&:\ 7 - 12 + 0 = -5\\ (3,2)&:\ -3 + 6 + 0 = 3\\ (3,3)&:\ 14 - 18 + 5 = 1 \end{aligned} A2−6A+5I=[3−4−5−914−531]A^2 - 6A + 5I = \begin{bmatrix} 3 & -4 & -5 \\ -9 & 1 & 4 \\ -5 & 3 & 1 \end{bmatrix}

    Multiplying by −111-\frac{1}{11} changes every sign and divides by 11:

    A−1=111[−3459−1−45−3−1]A^{-1} = \frac{1}{11}\begin{bmatrix} -3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1 \end{bmatrix}
    6A6A has rows 6,6,66, 6, 6 / 6,12,−186, 12, -18 / 12,−6,1812, -6, 18⁠, so for example entry (2,3)(2,3) is −14−(−18)=4-14 - (-18) = 4⁠. There is a cross-check with the adjoint. By Theorem 4, A−1=1−11adj⁡AA^{-1} = \frac{1}{-11}\operatorname{adj}A⁠, and we found A−1=−111(A2−6A+5I)A^{-1} = -\frac{1}{11}(A^2 - 6A + 5I)⁠; so the matrix A2−6A+5IA^2 - 6A + 5I must be exactly adj⁡A\operatorname{adj}A⁠. Its first row should hold the cofactors of column 1 of AA⁠: A11=M11=3A_{11} = M_{11} = 3 (Step 4); A21=−M21A_{21} = -M_{21}⁠, where M21=1⋅3−1⋅(−1)=4M_{21} = 1\cdot3 - 1\cdot(-1) = 4⁠, so A21=−4A_{21} = -4⁠; A31=M31A_{31} = M_{31}⁠, where M31=1(−3)−1⋅2=−5M_{31} = 1(-3) - 1\cdot2 = -5⁠. That is 3,−4,−53, -4, -5 ✓.

  7. Check: AA−1=IAA^{-1} = I

    Multiply AA by the integer matrix, then divide by 11:

    (1,1): 1(−3)+1⋅9+1⋅5=11(1,2): 1⋅4+1(−1)+1(−3)=0(1,3): 1⋅5+1(−4)+1(−1)=0(2,1): 1(−3)+2⋅9+(−3)5=0(2,2): 1⋅4+2(−1)+(−3)(−3)=11(2,3): 1⋅5+2(−4)+(−3)(−1)=0(3,1): 2(−3)+(−1)9+3⋅5=0(3,2): 2⋅4+(−1)(−1)+3(−3)=0(3,3): 2⋅5+(−1)(−4)+3(−1)=11\begin{aligned} (1,1)&:\ 1(-3) + 1\cdot9 + 1\cdot5 = 11\\ (1,2)&:\ 1\cdot4 + 1(-1) + 1(-3) = 0\\ (1,3)&:\ 1\cdot5 + 1(-4) + 1(-1) = 0\\[1mm] (2,1)&:\ 1(-3) + 2\cdot9 + (-3)5 = 0\\ (2,2)&:\ 1\cdot4 + 2(-1) + (-3)(-3) = 11\\ (2,3)&:\ 1\cdot5 + 2(-4) + (-3)(-1) = 0\\[1mm] (3,1)&:\ 2(-3) + (-1)9 + 3\cdot5 = 0\\ (3,2)&:\ 2\cdot4 + (-1)(-1) + 3(-3) = 0\\ (3,3)&:\ 2\cdot5 + (-1)(-4) + 3(-1) = 11 \end{aligned} AA−1=111[110001100011]=IAA^{-1} = \frac{1}{11}\begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix} = I
    The product before dividing is 11I11I⁠, which is −∣A∣ I-|A|\,I⁠. That is what Theorem 1 predicts: the integer matrix here is −adj⁡A-\operatorname{adj}A (Step 6), and A(adj⁡A)=∣A∣ IA(\operatorname{adj}A) = |A|\,I⁠, so A(−adj⁡A)=−∣A∣ I=11IA(-\operatorname{adj}A) = -|A|\,I = 11I⁠.
Result

A3−6A2+5A+11I=OA^3 - 6A^2 + 5A + 11I = O holds, ∣A∣=−11|A| = -11⁠, and

A−1=111[−3459−1−45−3−1]A^{-1} = \frac{1}{11}\begin{bmatrix} -3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1 \end{bmatrix}

(Textbook answer: the same matrix.)

Check it with numbers

Test the equation on the column (1,0,0)(1, 0, 0)⁠: AA sends it to column 1 of AA⁠, (1,1,2)(1, 1, 2)⁠; A2A^2 to column 1 of A2A^2⁠, (4,−3,7)(4, -3, 7)⁠; A3A^3 to (8,−23,32)(8, -23, 32)⁠. Then the first component of A3−6A2+5A+11IA^3 - 6A^2 + 5A + 11I applied to it is 8−24+5+11=08 - 24 + 5 + 11 = 0⁠, the second −23+18+5=0-23 + 18 + 5 = 0⁠, the third 32−42+10=032 - 42 + 10 = 0 ✓. Also ∣A−1∣|A^{-1}| should be 1∣A∣=−111\frac{1}{|A|} = -\frac{1}{11}⁠. Call the integer matrix in the answer PP and expand along its row 1:

∣P∣=−3 (1−12)−4 (−9+20)+5 (−27+5)=33−44−110=−121\begin{aligned} |P| &= -3\,(1 - 12) - 4\,(-9 + 20)\\ &\quad + 5\,(-27 + 5)\\ &= 33 - 44 - 110 = -121 \end{aligned}

Now A−1=111PA^{-1} = \frac{1}{11}P⁠, which is the product (111I)P\bigl(\frac{1}{11}I\bigr)P⁠. By Theorem 3, ∣A−1∣=∣111I∣ ∣P∣|A^{-1}| = \bigl|\frac{1}{11}I\bigr|\,|P|⁠, and ∣111I∣=1113\bigl|\frac{1}{11}I\bigr| = \frac{1}{11^3} (expand along row 1, as in Exercise 4.4 Q17, Step 3). So ∣A−1∣=−1211331=−111|A^{-1}| = -\frac{121}{1331} = -\frac{1}{11} ✓.

Common mistakes
  1. Forgetting the minus sign. From 11A−1=−(A2−6A+5I)11A^{-1} = -(A^2 - 6A + 5I) the inverse is −111(… )-\frac{1}{11}(\dots)⁠, not +111(… )+\frac{1}{11}(\dots)⁠.
  2. Adding 5I5I or 11I11I to every entry. The identity contributes only on the diagonal.
  3. Reducing A3A^3 to A2A^2 incorrectly. Multiplying A3A^3 by A−1A^{-1} gives A2A^2⁠, and 5A5A gives 5I5I⁠, not 55⁠.
  4. Sign slips with −6-6 times a negative entry. −6⋅(−3)=+18-6\cdot(-3) = +18 and −6⋅(−14)=+84-6\cdot(-14) = +84⁠.
What to take away
  1. If A3+pA2+qA+rI=OA^3 + pA^2 + qA + rI = O with r≠0r \neq 0⁠, then A−1=−1r(A2+pA+qI)A^{-1} = -\frac{1}{r}(A^2 + pA + qI)⁠.
  2. Check that A−1A^{-1} exists before multiplying by it; here ∣A∣=−11|A| = -11⁠.
  3. For long matrix computations, write each entry as its own short line; it keeps the signs under control.

Every fact used, and where it comes from

FactUsed inSource
Row-times-column product; associativitySteps 1, 2, 7Chapter 3, Sections 3.4.5 and 3.4.6
Scalar multiples, sums, distributive lawSteps 3, 5, 6Chapter 3, Sections 3.4.2–3.4.6
Expansion of a 3×33\times3 determinant along a rowStep 4Section 4.2.3 (Lesson: expanding)
AA invertible   ⟺  ∣A∣≠0\iff |A|\neq0Step 4Theorem 4, Section 4.5 (Lesson: inverse)
Post-multiplying by A−1A^{-1} to find the inverseStep 5Example 15
A−1=1∣A∣adj⁡AA^{-1} = \frac{1}{|A|}\operatorname{adj}A⁠, A(adj⁡A)=∣A∣ IA(\operatorname{adj}A) = |A|\,ISteps 6, 7Theorems 1 and 4, Section 4.5
∣XY∣=∣X∣∣Y∣|XY| = |X||Y|CheckTheorem 3, Section 4.5 (stated without proof)
Every entry of A2A^2⁠, A3A^3⁠, the combinations and the checkSteps 1–7, CheckRecomputed in the site's verification script
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