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Chapter 3 · Matrices

Exercise 3.2 · Q15: Find A2−5A+6IA^2-5A+6I for a 3×33\times3 matrix

Exercise 3.2Matrix multiplication, scalar multiples, identity matrix · Sections 3.3, 3.4
The question

Find A2−5A+6IA^2 - 5A + 6I⁠, if

A=[2012131−10].A = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix}.
Exam answer A2=AA=[2012131−10][2012131−10]\begin{aligned} &A^2 = AA \\ &= \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix}\begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix} \end{aligned}

Multiplying row by column, the entries cijc_{ij} of A2A^2 are

c11=2⋅2+0⋅2+1⋅1=5c12=2⋅0+0⋅1+1(−1)=−1c13=2⋅1+0⋅3+1⋅0=2c21=2⋅2+1⋅2+3⋅1=9c22=2⋅0+1⋅1+3(−1)=−2c23=2⋅1+1⋅3+3⋅0=5c31=1⋅2+(−1)⋅2+0⋅1=0c32=1⋅0+(−1)⋅1+0(−1)=−1c33=1⋅1+(−1)⋅3+0⋅0=−2\begin{aligned} c_{11} &= 2\cdot2 + 0\cdot2 + 1\cdot1 = 5 \\ c_{12} &= 2\cdot0 + 0\cdot1 + 1(-1) = -1 \\ c_{13} &= 2\cdot1 + 0\cdot3 + 1\cdot0 = 2 \\ c_{21} &= 2\cdot2 + 1\cdot2 + 3\cdot1 = 9 \\ c_{22} &= 2\cdot0 + 1\cdot1 + 3(-1) = -2 \\ c_{23} &= 2\cdot1 + 1\cdot3 + 3\cdot0 = 5 \\ c_{31} &= 1\cdot2 + (-1)\cdot2 + 0\cdot1 = 0 \\ c_{32} &= 1\cdot0 + (-1)\cdot1 + 0(-1) \\ &= -1 \\ c_{33} &= 1\cdot1 + (-1)\cdot3 + 0\cdot0 = -2 \end{aligned} ∴ A2=[5−129−250−1−2]\therefore\ A^2 = \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix} A2−5A+6I=[5−129−250−1−2]−[1005105155−50]+[600060006]\begin{aligned} &A^2 - 5A + 6I \\ &= \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix} \\ &\quad - \begin{bmatrix} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{bmatrix} \\ &\quad + \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix} \end{aligned}

∴ A2−5A+6I=[1−1−3−1−1−10−544]\begin{aligned} &\therefore\ A^2 - 5A + 6I \\ &= \begin{bmatrix} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{bmatrix} \end{aligned}

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

The one idea this problem is built on

A "polynomial in a matrix" is evaluated just like a polynomial in a number, with three translations: A2A^2 means the matrix product AAAA⁠; 5A5A means multiply every entry by 5; and the constant term 6 becomes 6I6I⁠, where II is the identity matrix of the same order as AA⁠, so that all three terms can be added. Then combine the three matrices position by position.

Given, to find, and the plan
Given
The 3×33\times3 matrix AA above; II is the identity matrix.
To find
The single matrix A2−5A+6IA^2 - 5A + 6I⁠.
Plan
(1) Check orders and fix the order of II⁠. (2) Compute A2=AAA^2 = AA⁠, every entry. (3) Compute 5A5A and 6I6I⁠. (4) Combine entry by entry. (5) Check.

Basics you need

1Matrix multiplication

For 3×33\times3 matrices, entry (i,k)(i,k) of ABAB is ai1b1k+ai2b2k+ai3b3ka_{i1}b_{1k} + a_{i2}b_{2k} + a_{i3}b_{3k}⁠: row ii of the first times column kk of the second. Basics: pair up, multiply, add

2Powers of a square matrix

A2=AAA^2 = AA⁠. It is defined only for a square matrix, because the columns of the first factor must match the rows of the second. It is not the matrix of squared entries.

3The identity matrix

InI_n is the n×nn\times n matrix with 1 on the main diagonal and 0 everywhere else. 6I6I therefore has 6 on the diagonal and 0 elsewhere. Basics: identity and undo

4Scalar multiples, sums and differences

kAkA multiplies every entry by kk⁠; A−B=A+(−1)BA - B = A + (-1)B subtracts entries in the same position, for matrices of the same order. Basics: adding only like with like

Solution

  1. Check the orders and fix the order of II

    AA is 3×33\times3⁠, a square matrix. So AAAA is defined (3 columns of the first factor, 3 rows of the second) and A2A^2 is 3×33\times3⁠. 5A5A is also 3×33\times3⁠. For A2−5A+6IA^2 - 5A + 6I to be defined, 6I6I must be 3×33\times3 too, so II here means I3I_3⁠.

    Sums and differences need matrices of the same order. The question writes just II⁠; the only order that makes the expression meaningful is 3×33\times3⁠.
    Lesson: types of matrices (identity matrix)

  2. Compute A2=AAA^2 = AA⁠, row 1

    Rows of AA⁠: (2,0,1)(2, 0, 1)⁠, (2,1,3)(2, 1, 3)⁠, (1,−1,0)(1, -1, 0)⁠. Columns of AA⁠, top to bottom: (2,2,1)(2, 2, 1)⁠, (0,1,−1)(0, 1, -1)⁠, (1,3,0)(1, 3, 0)⁠. Write A2=[pik]A^2 = [p_{ik}]⁠. Row 1 of AA with each column:

    p11=2⋅2+0⋅2+1⋅1=5p12=2⋅0+0⋅1+1⋅(−1)=−1p13=2⋅1+0⋅3+1⋅0=2\begin{aligned} p_{11} &= 2\cdot 2 + 0\cdot 2 + 1\cdot 1 = 5\\ p_{12} &= 2\cdot 0 + 0\cdot 1 + 1\cdot(-1) = -1\\ p_{13} &= 2\cdot 1 + 0\cdot 3 + 1\cdot 0 = 2 \end{aligned}
    In AAAA the same matrix supplies both the rows (first factor) and the columns (second factor). For example p11=4+0+1=5p_{11} = 4 + 0 + 1 = 5⁠.
    pik=ai1a1k+ai2a2k+ai3a3kp_{ik} = a_{i1}a_{1k} + a_{i2}a_{2k} + a_{i3}a_{3k}⁠. Basics: pair up, multiply, add

  3. A2A^2⁠, rows 2 and 3

    Row 2 of AA is (2,1,3)(2, 1, 3)⁠:

    p21=2⋅2+1⋅2+3⋅1=9p22=2⋅0+1⋅1+3⋅(−1)=−2p23=2⋅1+1⋅3+3⋅0=5\begin{aligned} p_{21} &= 2\cdot 2 + 1\cdot 2 + 3\cdot 1 = 9\\ p_{22} &= 2\cdot 0 + 1\cdot 1 + 3\cdot(-1) = -2\\ p_{23} &= 2\cdot 1 + 1\cdot 3 + 3\cdot 0 = 5 \end{aligned}

    Row 3 of AA is (1,−1,0)(1, -1, 0)⁠:

    p31=1⋅2+(−1)⋅2+0⋅1=0p32=1⋅0+(−1)⋅1+0⋅(−1)=−1p33=1⋅1+(−1)⋅3+0⋅0=−2\begin{aligned} p_{31} &= 1\cdot 2 + (-1)\cdot 2 + 0\cdot 1 = 0\\ p_{32} &= 1\cdot 0 + (-1)\cdot 1 + 0\cdot(-1) = -1\\ p_{33} &= 1\cdot 1 + (-1)\cdot 3 + 0\cdot 0 = -2 \end{aligned}

    Assembling the nine entries:

    A2=[5−129−250−1−2]A^2 = \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix}
    For instance p22=0+1−3=−2p_{22} = 0 + 1 - 3 = -2 and p33=1−3+0=−2p_{33} = 1 - 3 + 0 = -2⁠. Every one of the nine row-times-column sums is needed; there is no shortcut.

  4. Compute 5A5A and 6I6I

    Multiply every entry of AA by 5:

    5A=[5⋅25⋅05⋅15⋅25⋅15⋅35⋅15⋅(−1)5⋅0]5A = \begin{bmatrix} 5\cdot2 & 5\cdot0 & 5\cdot1 \\ 5\cdot2 & 5\cdot1 & 5\cdot3 \\ 5\cdot1 & 5\cdot(-1) & 5\cdot0 \end{bmatrix} 5A=[1005105155−50]\phantom{5A} = \begin{bmatrix} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{bmatrix}

    Multiply every entry of I3I_3 by 6:

    6I=6[100010001]=[600060006]6I = 6\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix}
    6I6I is how the number 6 enters a matrix expression: it is a scalar matrix that adds 6 to each diagonal entry and leaves the rest unchanged.
    kA=[k aij]kA = [k\,a_{ij}]⁠. Lesson: scalar multiplication

  5. Combine position by position

    Entry (i,k)(i,k) of A2−5A+6IA^2 - 5A + 6I is pik−5aik+6δikp_{ik} - 5a_{ik} + 6\delta_{ik}⁠, where δik\delta_{ik} is 1 on the diagonal and 0 off it:

    (1,1):5−10+6=1(1,2):−1−0+0=−1(1,3):2−5+0=−3(2,1):9−10+0=−1(2,2):−2−5+6=−1(2,3):5−15+0=−10(3,1):0−5+0=−5(3,2):−1−(−5)+0=4(3,3):−2−0+6=4\begin{aligned} (1,1)&: 5 - 10 + 6 = 1\\ (1,2)&: -1 - 0 + 0 = -1\\ (1,3)&: 2 - 5 + 0 = -3\\ (2,1)&: 9 - 10 + 0 = -1\\ (2,2)&: -2 - 5 + 6 = -1\\ (2,3)&: 5 - 15 + 0 = -10\\ (3,1)&: 0 - 5 + 0 = -5\\ (3,2)&: -1 - (-5) + 0 = 4\\ (3,3)&: -2 - 0 + 6 = 4 \end{aligned} A2−5A+6I=[1−1−3−1−1−10−544]A^2 - 5A + 6I = \begin{bmatrix} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{bmatrix}
    All three matrices are 3×33\times3⁠, so subtraction and addition are done entry by entry. Take care with (3,2)(3,2)⁠: subtracting −5-5 is adding 5, so −1+5=4-1 + 5 = 4⁠.
    A−B+C=[aij−bij+cij]A - B + C = [a_{ij} - b_{ij} + c_{ij}] for matrices of the same order. Basics: adding like with like
Result A2−5A+6I=[1−1−3−1−1−10−544]A^2 - 5A + 6I = \begin{bmatrix} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{bmatrix}

(Textbook answer: the same matrix.)

Check it with numbers

Use the factorisation x2−5x+6=(x−2)(x−3)x^2 - 5x + 6 = (x - 2)(x - 3)⁠. It carries over to matrices because AA commutes with II⁠: (A−2I)(A−3I)=A2−3A−2A+6I(A - 2I)(A - 3I) = A^2 - 3A - 2A + 6I⁠. Here

A−2I=[0012−131−1−2],A - 2I = \begin{bmatrix} 0 & 0 & 1 \\ 2 & -1 & 3 \\ 1 & -1 & -2 \end{bmatrix}, A−3I=[−1012−231−1−3].A - 3I = \begin{bmatrix} -1 & 0 & 1 \\ 2 & -2 & 3 \\ 1 & -1 & -3 \end{bmatrix}.

Row 1 of A−2IA - 2I is (0,0,1)(0, 0, 1)⁠, so row 1 of the product is row 3 of A−3IA - 3I⁠: (1,−1,−3)(1, -1, -3) ✓. Row 2 gives 2(−1)+(−1)(2)+3(1)=−12(-1) + (-1)(2) + 3(1) = -1⁠, 2(0)+(−1)(−2)+3(−1)=−12(0) + (-1)(-2) + 3(-1) = -1⁠, 2(1)+(−1)(3)+3(−3)=−102(1) + (-1)(3) + 3(-3) = -10 ✓. Row 3 gives 1(−1)+(−1)(2)+(−2)(1)=−51(-1) + (-1)(2) + (-2)(1) = -5⁠, 0+2+2=40 + 2 + 2 = 4⁠, 1−3+6=41 - 3 + 6 = 4 ✓. The same matrix, by a different route.

Common mistakes
  1. Squaring each entry. A2A^2 is AAAA⁠; squaring entries would give 44 in position (1,1)(1,1) instead of the correct 55⁠.
  2. Writing +6+6 instead of +6I+6I⁠. A number cannot be added to a matrix; 6 must become 6I6I⁠, and it only changes the diagonal.
  3. Adding 6 to every entry. 6I6I has zeros off the diagonal, so the off-diagonal entries get +0+0⁠.
  4. Sign slip in (3,2)(3,2)⁠. −1−5(−1)=−1+5=4-1 - 5(-1) = -1 + 5 = 4⁠.
What to take away
  1. To evaluate a polynomial in a square matrix: powers are products, coefficients are scalar multiples, the constant becomes a multiple of II of the same order.
  2. Show each entry of a product as a row-times-column sum; most errors are sign errors in these sums.
  3. A polynomial in AA factorises as for numbers, because powers of AA and II commute with each other; that gives an independent check.

Every fact used, and where it comes from

FactUsed inSource
Identity matrix InI_n⁠: 1 on the diagonal, 0 elsewhereSteps 1, 4Section 3.3 (vi) (Lesson: types)
ABAB defined when columns of AA = rows of BB⁠; entry = row times columnSteps 1 to 3Section 3.4.5 (Lesson: multiplication)
A2=AAA^2 = AA for a square matrixSteps 2, 3Example 18 (A2=A.AA^2 = A.A⁠)
kA=[k aij]kA = [k\,a_{ij}]Step 4Section 3.4.2
Sum and difference of matrices of the same order, entry by entryStep 5Section 3.4.1
IA=AI=AIA = AI = A⁠; distributive lawCheckSection 3.4.6
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