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Chapter 4 · Determinants

Exercise 4.4 · Q9: Inverse of a 3×33\times3 matrix

Exercise 4.4Inverse of a matrix · Theorem 4, Section 4.5
The question

Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11.

9.

A=[2134−10−721]A = \begin{bmatrix} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{bmatrix}
Exam answer

Expanding along R2R_2⁠:

∣A∣=−4∣1321∣+(−1)∣23−71∣−0∣21−72∣=−4(1−6)−(2+21)−0=20−23=−3\begin{aligned} |A| &= -4\begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix} \\ &\quad + (-1)\begin{vmatrix} 2 & 3 \\ -7 & 1 \end{vmatrix} \\ &\quad - 0\begin{vmatrix} 2 & 1 \\ -7 & 2 \end{vmatrix} \\ &= -4(1 - 6) - (2 + 21) - 0 \\ &= 20 - 23 = -3 \end{aligned}

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

Cofactors, Aij=(−1)i+jMijA_{ij} = (-1)^{i+j}M_{ij}⁠:

A11=∣−1021∣=−1A12=−∣40−71∣=−4A13=∣4−1−72∣=1\begin{aligned} A_{11} &= \begin{vmatrix} -1 & 0 \\ 2 & 1 \end{vmatrix} = -1 \\ A_{12} &= -\begin{vmatrix} 4 & 0 \\ -7 & 1 \end{vmatrix} = -4 \\ A_{13} &= \begin{vmatrix} 4 & -1 \\ -7 & 2 \end{vmatrix} = 1 \end{aligned} A21=−∣1321∣=5A22=∣23−71∣=23A23=−∣21−72∣=−11\begin{aligned} A_{21} &= -\begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix} = 5 \\ A_{22} &= \begin{vmatrix} 2 & 3 \\ -7 & 1 \end{vmatrix} = 23 \\ A_{23} &= -\begin{vmatrix} 2 & 1 \\ -7 & 2 \end{vmatrix} = -11 \end{aligned} A31=∣13−10∣=3A32=−∣2340∣=12A33=∣214−1∣=−6\begin{aligned} A_{31} &= \begin{vmatrix} 1 & 3 \\ -1 & 0 \end{vmatrix} = 3 \\ A_{32} &= -\begin{vmatrix} 2 & 3 \\ 4 & 0 \end{vmatrix} = 12 \\ A_{33} &= \begin{vmatrix} 2 & 1 \\ 4 & -1 \end{vmatrix} = -6 \end{aligned} adj⁡A=[−153−423121−11−6]\operatorname{adj}A = \begin{bmatrix} -1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6 \end{bmatrix} A−1=1∣A∣adj⁡AA^{-1} = \frac{1}{|A|}\operatorname{adj}A

∴ A−1=−13[−153−423121−11−6]\therefore\ A^{-1} = -\dfrac{1}{3}\begin{bmatrix} -1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6 \end{bmatrix}

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

The one idea this problem is built on

A square matrix has an inverse exactly when ∣A∣≠0|A|\neq0⁠, and then A−1=1∣A∣adj⁡AA^{-1} = \frac{1}{|A|}\operatorname{adj}A (Theorem 4, Section 4.5). The formula is Theorem 1, A(adj⁡A)=∣A∣ IA(\operatorname{adj}A) = |A|\,I⁠, divided by the number ∣A∣|A|⁠. So the work is: one determinant, nine cofactors, one transpose, one division, and a final multiplication to confirm. Here no row or column has two zeros, so expect a little more arithmetic, and a negative determinant to carry through to the end.

Given, to find, and the plan
Given
The 3×33\times3 matrix AA above.
To find
A−1A^{-1}⁠, if it exists.
Plan
(1) Compute ∣A∣|A|⁠, expanding along row 2 (it contains a zero); check ∣A∣≠0|A|\neq0⁠. (2) Find all nine cofactors, row by row. (3) Form [Aij][A_{ij}] and transpose it to get adj⁡A\operatorname{adj}A⁠. (4) Divide by ∣A∣|A|⁠. (5) Verify AA−1=IAA^{-1}=I entry by entry.

Basics you need

1Expanding a 3×33\times3 determinant

Along any row or column: each entry times its sign (−1)i+j(-1)^{i+j} times its minor, added up. All six choices give the same value, so choose a line with zeros (Section 4.2.3, Remark (i)). Lesson: expanding a 3 × 3 determinant

2Minor, cofactor and sign pattern

MijM_{ij} is the 2×22\times2 determinant left after deleting row ii and column jj⁠, and Aij=(−1)i+jMijA_{ij}=(-1)^{i+j}M_{ij} (Definitions 1 and 2, Section 4.4). Basics: the + − + sign pattern

3Adjoint

adj⁡A\operatorname{adj}A is the transpose of the matrix of cofactors (Definition 3, Section 4.5.1). Lesson: the adjoint

4Inverse via the adjoint

If ∣A∣≠0|A|\neq0⁠, A−1=1∣A∣adj⁡AA^{-1}=\frac{1}{|A|}\operatorname{adj}A⁠; if ∣A∣=0|A|=0⁠, no inverse exists (Theorem 4, Section 4.5). Basics: why the inverse divides by |A|

5Matrix product and scalar multiple

The (i,j)(i,j) entry of PQPQ is row ii of PP times column jj of QQ⁠; kBkB multiplies every entry of BB by kk⁠. Basics: matrix facts the chapter uses

Solution

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

    Row 2 is (4, −1, 0)(4,\ -1,\ 0)⁠; its zero removes one term, so expand along it. The sign pattern is

    +−+−+−+−+\begin{matrix} + & - & + \\ - & + & - \\ + & - & + \end{matrix} ∣A∣=−a21M21+a22M22−a23M23=−4∣1321∣+(−1)∣23−71∣−0=−4(1−6)−(2+21)=20−23=−3\begin{aligned} |A| &= -a_{21}M_{21} + a_{22}M_{22} - a_{23}M_{23} \\ &= -4\begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix} + (-1)\begin{vmatrix} 2 & 3 \\ -7 & 1 \end{vmatrix} - 0 \\ &= -4(1 - 6) - (2 + 21) \\ &= 20 - 23 = -3 \end{aligned}

    ∣A∣=−3≠0|A| = -3\neq0⁠, so AA is non-singular and A−1A^{-1} exists.

    The signs along row 2 are −, +, −-,\ +,\ -⁠. Column 3, (3, 0, 1)(3,\ 0,\ 1)⁠, also has one zero and would do equally well; any row or column gives the same value (Section 4.2.3). In the second minor, 2⋅1−3⋅(−7)=2+21=232\cdot1 - 3\cdot(-7) = 2 + 21 = 23⁠. Theorem 4 says AA is invertible if and only if ∣A∣≠0|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+1 or −1-1 as i+ji+j is even or odd. Lesson: expanding a 3 × 3 determinant

  2. Cofactors of the row 1 entries

    Delete row 1, which leaves rows (4, −1, 0)(4,\ -1,\ 0) and (−7, 2, 1)(-7,\ 2,\ 1)⁠, and one column at a time:

    M11=∣−1021∣=(−1)⋅1−0⋅2=−1−0=−1A11=(−1)1+1 (−1)=−1M12=∣40−71∣=4⋅1−0⋅(−7)=4−0=4A12=(−1)1+2 (4)=−4M13=∣4−1−72∣=4⋅2−(−1)⋅(−7)=8−7=1A13=(−1)1+3 (1)=1\begin{aligned} M_{11} &= \begin{vmatrix} -1 & 0 \\ 2 & 1 \end{vmatrix} = (-1)\cdot 1 - 0\cdot 2 \\ &= -1 - 0 = -1 \\ A_{11} &= (-1)^{1+1}\,(-1) = -1 \\ M_{12} &= \begin{vmatrix} 4 & 0 \\ -7 & 1 \end{vmatrix} = 4\cdot 1 - 0\cdot (-7) \\ &= 4 - 0 = 4 \\ A_{12} &= (-1)^{1+2}\,(4) = -4 \\ M_{13} &= \begin{vmatrix} 4 & -1 \\ -7 & 2 \end{vmatrix} = 4\cdot 2 - (-1)\cdot (-7) \\ &= 8 - 7 = 1 \\ A_{13} &= (-1)^{1+3}\,(1) = 1 \end{aligned}
    All three are nonzero. The middle one has sign −-⁠: M12=4M_{12} = 4 gives A12=−4A_{12} = -4⁠. In M13M_{13}⁠, 4⋅2−(−1)⋅(−7)=8−7=14\cdot2 - (-1)\cdot(-7) = 8 - 7 = 1⁠: the product of two negatives is positive and is then subtracted.
    Definitions 1 and 2, Section 4.4. Lesson: minors and cofactors

  3. Cofactors of the row 2 entries

    Delete row 2, which leaves rows (2, 1, 3)(2,\ 1,\ 3) and (−7, 2, 1)(-7,\ 2,\ 1)⁠:

    M21=∣1321∣=1⋅1−3⋅2=1−6=−5A21=(−1)2+1 (−5)=5M22=∣23−71∣=2⋅1−3⋅(−7)=2+21=23A22=(−1)2+2 (23)=23M23=∣21−72∣=2⋅2−1⋅(−7)=4+7=11A23=(−1)2+3 (11)=−11\begin{aligned} M_{21} &= \begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix} = 1\cdot 1 - 3\cdot 2 \\ &= 1 - 6 = -5 \\ A_{21} &= (-1)^{2+1}\,(-5) = 5 \\ M_{22} &= \begin{vmatrix} 2 & 3 \\ -7 & 1 \end{vmatrix} = 2\cdot 1 - 3\cdot (-7) \\ &= 2 + 21 = 23 \\ A_{22} &= (-1)^{2+2}\,(23) = 23 \\ M_{23} &= \begin{vmatrix} 2 & 1 \\ -7 & 2 \end{vmatrix} = 2\cdot 2 - 1\cdot (-7) \\ &= 4 + 7 = 11 \\ A_{23} &= (-1)^{2+3}\,(11) = -11 \end{aligned}
    M21M_{21} and M22M_{22} are the two minors already met in Step 1, which is a useful cross-check. M23=2⋅2−1⋅(−7)=4+7=11M_{23} = 2\cdot2 - 1\cdot(-7) = 4 + 7 = 11⁠, and its minus sign gives A23=−11A_{23} = -11⁠.

  4. Cofactors of the row 3 entries

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

    M31=∣13−10∣=1⋅0−3⋅(−1)=0+3=3A31=(−1)3+1 (3)=3M32=∣2340∣=2⋅0−3⋅4=0−12=−12A32=(−1)3+2 (−12)=12M33=∣214−1∣=2⋅(−1)−1⋅4=−2−4=−6A33=(−1)3+3 (−6)=−6\begin{aligned} M_{31} &= \begin{vmatrix} 1 & 3 \\ -1 & 0 \end{vmatrix} = 1\cdot 0 - 3\cdot (-1) \\ &= 0 + 3 = 3 \\ A_{31} &= (-1)^{3+1}\,(3) = 3 \\ M_{32} &= \begin{vmatrix} 2 & 3 \\ 4 & 0 \end{vmatrix} = 2\cdot 0 - 3\cdot 4 \\ &= 0 - 12 = -12 \\ A_{32} &= (-1)^{3+2}\,(-12) = 12 \\ M_{33} &= \begin{vmatrix} 2 & 1 \\ 4 & -1 \end{vmatrix} = 2\cdot (-1) - 1\cdot 4 \\ &= -2 - 4 = -6 \\ A_{33} &= (-1)^{3+3}\,(-6) = -6 \end{aligned}
    Position (3,2)(3,2) has odd i+ji+j⁠, so the negative minor M32=−12M_{32} = -12 gives the positive cofactor A32=12A_{32} = 12⁠.

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

    [Aij]=[−1−41523−11312−6][A_{ij}] = \begin{bmatrix} -1 & -4 & 1 \\ 5 & 23 & -11 \\ 3 & 12 & -6 \end{bmatrix}

    Transposing (row ii of [Aij][A_{ij}] becomes column ii⁠):

    adj⁡A=[−153−423121−11−6]\operatorname{adj}A = \begin{bmatrix} -1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6 \end{bmatrix}
    Row 1 of [Aij][A_{ij}]⁠, (−1, −4, 1)(-1,\ -4,\ 1)⁠, becomes column 1 of adj⁡A\operatorname{adj}A⁠; row 2, (5, 23, −11)(5,\ 23,\ -11)⁠, becomes column 2; row 3, (3, 12, −6)(3,\ 12,\ -6)⁠, becomes column 3.
    Definition 3, Section 4.5.1; transpose, Chapter 3, Section 3.5. Basics: transpose and identity

  6. Divide by ∣A∣|A|

    A−1=1∣A∣adj⁡A=−13[−153−423121−11−6]\begin{aligned} A^{-1} &= \frac{1}{|A|}\operatorname{adj}A \\ &= -\frac{1}{3}\begin{bmatrix} -1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6 \end{bmatrix} \end{aligned}

    Multiplying every entry by −13-\frac{1}{3}⁠:

    A−1=[13−53−143−233−4−131132]A^{-1} = \begin{bmatrix} \frac{1}{3} & -\frac{5}{3} & -1 \\ \frac{4}{3} & -\frac{23}{3} & -4 \\ -\frac{1}{3} & \frac{11}{3} & 2 \end{bmatrix}
    This is Theorem 4. Since ∣A∣=−3|A| = -3⁠, every entry of adj⁡A\operatorname{adj}A is multiplied by −13-\frac{1}{3}⁠, so every sign flips: for example 23⋅(−13)=−23323\cdot(-\frac{1}{3}) = -\frac{23}{3} and (−6)⋅(−13)=2(-6)\cdot(-\frac{1}{3}) = 2⁠.
    Scalar multiple: each entry is multiplied by the scalar (Chapter 3, Section 3.4.2). Lesson: inverse of a matrix

  7. Verify AA−1=IAA^{-1} = I

    Since A−1=−13adj⁡AA^{-1} = -\frac{1}{3}\operatorname{adj}A⁠, it is enough to show A(adj⁡A)=−3 IA(\operatorname{adj}A) = -3\,I⁠. Write A(adj⁡A)=[pij]A(\operatorname{adj}A)=[p_{ij}]⁠. Row 1 of AA is (2, 1, 3)(2,\ 1,\ 3)⁠:

    p11=2⋅(−1)+1⋅(−4)+3⋅1=−2−4+3=−3p12=2⋅5+1⋅23+3⋅(−11)=10+23−33=0p13=2⋅3+1⋅12+3⋅(−6)=6+12−18=0\begin{aligned} p_{11} &= 2\cdot (-1) + 1\cdot (-4) + 3\cdot 1 \\ &= -2 - 4 + 3 = -3 \\ p_{12} &= 2\cdot 5 + 1\cdot 23 + 3\cdot (-11) \\ &= 10 + 23 - 33 = 0 \\ p_{13} &= 2\cdot 3 + 1\cdot 12 + 3\cdot (-6) \\ &= 6 + 12 - 18 = 0 \end{aligned}

    Row 2 of AA is (4, −1, 0)(4,\ -1,\ 0)⁠:

    p21=4⋅(−1)+(−1)⋅(−4)+0⋅1=−4+4+0=0p22=4⋅5+(−1)⋅23+0⋅(−11)=20−23+0=−3p23=4⋅3+(−1)⋅12+0⋅(−6)=12−12+0=0\begin{aligned} p_{21} &= 4\cdot (-1) + (-1)\cdot (-4) + 0\cdot 1 \\ &= -4 + 4 + 0 = 0 \\ p_{22} &= 4\cdot 5 + (-1)\cdot 23 + 0\cdot (-11) \\ &= 20 - 23 + 0 = -3 \\ p_{23} &= 4\cdot 3 + (-1)\cdot 12 + 0\cdot (-6) \\ &= 12 - 12 + 0 = 0 \end{aligned}

    Row 3 of AA is (−7, 2, 1)(-7,\ 2,\ 1)⁠:

    p31=(−7)⋅(−1)+2⋅(−4)+1⋅1=7−8+1=0p32=(−7)⋅5+2⋅23+1⋅(−11)=−35+46−11=0p33=(−7)⋅3+2⋅12+1⋅(−6)=−21+24−6=−3\begin{aligned} p_{31} &= (-7)\cdot (-1) + 2\cdot (-4) + 1\cdot 1 \\ &= 7 - 8 + 1 = 0 \\ p_{32} &= (-7)\cdot 5 + 2\cdot 23 + 1\cdot (-11) \\ &= -35 + 46 - 11 = 0 \\ p_{33} &= (-7)\cdot 3 + 2\cdot 12 + 1\cdot (-6) \\ &= -21 + 24 - 6 = -3 \end{aligned} A(adj⁡A)=[−3000−3000−3]=−3 IA(\operatorname{adj}A) = \begin{bmatrix} -3 & 0 & 0 \\ 0 & -3 & 0 \\ 0 & 0 & -3 \end{bmatrix} = -3\,I

    Hence AA−1=−13 (−3 I)=IAA^{-1} = -\frac{1}{3}\,(-3\,I) = I⁠. By Theorem 1, (adj⁡A)A=∣A∣ I(\operatorname{adj}A)A = |A|\,I as well, so A−1A=IA^{-1}A = I too.

    The definition of the inverse (Chapter 3, Definition 6) asks for AB=BA=IAB = BA = I⁠. The diagonal entries p11,p22,p33p_{11}, p_{22}, p_{33} are expansions of ∣A∣|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-3\,I confirms all nine cofactors at once.
    A(kB)=k(AB)A(kB) = k(AB) for a scalar kk⁠. Basics: matrix facts the chapter uses
Result

∣A∣=−3≠0|A| = -3\neq0⁠, so A−1A^{-1} exists and

A−1=−13[−153−423121−11−6]=[13−53−143−233−4−131132]\begin{aligned} A^{-1} &= -\frac{1}{3}\begin{bmatrix} -1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6 \end{bmatrix} \\ &= \begin{bmatrix} \frac{1}{3} & -\frac{5}{3} & -1 \\ \frac{4}{3} & -\frac{23}{3} & -4 \\ -\frac{1}{3} & \frac{11}{3} & 2 \end{bmatrix} \end{aligned}

(Textbook answer: −13-\frac{1}{3} times the same matrix.)

Check it with numbers

The determinant a second way. Along row 1, with the cofactors of Step 2:

∣A∣=2⋅(−1)+1⋅(−4)+3⋅1=−2−4+3=−3\begin{aligned} |A| &= 2\cdot(-1) + 1\cdot(-4) + 3\cdot 1 \\ &= -2 - 4 + 3 = -3 \end{aligned}

as in Step 1.

Undo a multiplication. Take x=(1, 1, 1)\mathbf{x} = (1,\ 1,\ 1) as a column and compute b=Ax\mathbf{b} = A\mathbf{x} (each entry is a row sum of AA⁠):

b=Ax=[2134−10−721][111]=[2+1+34−1+0−7+2+1]=[63−4]\begin{aligned} \mathbf{b} = A\mathbf{x} &= \begin{bmatrix} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{bmatrix}\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \\ &= \begin{bmatrix} 2 + 1 + 3 \\ 4 - 1 + 0 \\ -7 + 2 + 1 \end{bmatrix} \\ &= \begin{bmatrix} 6 \\ 3 \\ -4 \end{bmatrix} \end{aligned}

Applying adj⁡A\operatorname{adj}A and then dividing by ∣A∣|A| should return x\mathbf{x}⁠. Entry ii of (adj⁡A) b(\operatorname{adj}A)\,\mathbf{b} is row ii of adj⁡A\operatorname{adj}A times b\mathbf{b}⁠, written with the products already worked out:

(adj⁡A) b=[−6+15−12−24+69−486−33+24]=[−3−3−3]\begin{aligned} (\operatorname{adj}A)\,\mathbf{b} &= \begin{bmatrix} -6 + 15 - 12 \\ -24 + 69 - 48 \\ 6 - 33 + 24 \end{bmatrix} \\ &= \begin{bmatrix} -3 \\ -3 \\ -3 \end{bmatrix} \end{aligned} A−1b=−13[−3−3−3]=[111]A^{-1}\mathbf{b} = -\frac{1}{3}\begin{bmatrix} -3 \\ -3 \\ -3 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}

We are back at x\mathbf{x}⁠, as the inverse must give.

Common mistakes
  1. Losing the sign of ∣A∣|A|⁠. ∣A∣=−3|A| = -3⁠, so A−1=−13adj⁡AA^{-1} = -\frac{1}{3}\operatorname{adj}A⁠. Using 13\frac{1}{3} gives −A−1-A^{-1}⁠.
  2. Sign slips inside minors. M22=2⋅1−3⋅(−7)=2+21M_{22} = 2\cdot1 - 3\cdot(-7) = 2 + 21⁠, and M23=2⋅2−1⋅(−7)=4+7M_{23} = 2\cdot2 - 1\cdot(-7) = 4 + 7⁠.
  3. Forgetting the transpose. For example A12=−4A_{12} = -4 belongs in position (2,1)(2,1) of adj⁡A\operatorname{adj}A⁠, not (1,2)(1,2)⁠.
  4. Skipping the final check. The product A(adj⁡A)A(\operatorname{adj}A) must come out as −3I-3I⁠; any wrong cofactor shows up there.
What to take away
  1. When no row or column has two zeros, expand along a line with one zero (here row 2) and work carefully; Step 1's minors M21M_{21} and M22M_{22} are reused in Step 3.
  2. A negative determinant flips every sign when you divide.
  3. A(adj⁡A)=∣A∣ IA(\operatorname{adj}A) = |A|\,I checks all nine cofactors in one product.

Every fact used, and where it comes from

FactUsed inSource
Expansion along any row or column; signs ±1\pm1⁠; choose zerosStep 1Section 4.2.3 and its Remarks (i), (ii) (Lesson: expanding)
2×22\times2 determinant =ad−bc= ad-bcSteps 1–4Section 4.2.2
Non-singular: ∣A∣≠0|A|\neq0Step 1Definition 5, Section 4.5
Minor and cofactorSteps 2–4Definitions 1 and 2, Section 4.4 (Lesson: minors and cofactors)
Adjoint = transpose of [Aij][A_{ij}]Step 5Definition 3, Section 4.5.1 (Lesson: the adjoint)
Invertible   ⟺  ∣A∣≠0\iff |A|\neq0⁠; A−1=1∣A∣adj⁡AA^{-1}=\frac{1}{|A|}\operatorname{adj}ASteps 1, 6Theorem 4, Section 4.5 (Lesson: inverse of a matrix)
Scalar multiple; product of matricesSteps 6, 7Chapter 3, Sections 3.4.2 and 3.4.5 (Basics: matrix facts)
Own cofactors give ∣A∣|A|⁠; another row's give 0Step 7, CheckRemark after Example 10 and Note, Section 4.4
A(adj⁡A)=(adj⁡A)A=∣A∣ IA(\operatorname{adj}A)=(\operatorname{adj}A)A=|A|\,IStep 7Theorem 1, Section 4.5.1
Definition of the inverseStep 7Chapter 3, Definition 6, Section 3.7
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