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Chapter 3 · Matrices

Miscellaneous · Q5: Show A2−5A+7I=OA^2-5A+7I=O

Miscellaneous ExerciseMatrix multiplication · Section 3.4.5
The question

If

A=[31−12],A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix},

show that A2−5A+7I=0A^2 - 5A + 7I = 0⁠.

Exam answer A2=[31−12][31−12]=[9−13+2−3−2−1+4]=[85−53]\begin{aligned} A^2 &= \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \\ &= \begin{bmatrix} 9 - 1 & 3 + 2 \\ -3 - 2 & -1 + 4 \end{bmatrix} \\ &= \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} \end{aligned} LHS=A2−5A+7I=[85−53]−[155−510]+[7007]=[8−15+75−5+0−5+5+03−10+7]=[0000]=O\begin{aligned} \text{LHS} &= A^2 - 5A + 7I \\ &= \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} \\ &\quad + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} \\ &= \begin{bmatrix} 8 - 15 + 7 & 5 - 5 + 0 \\ -5 + 5 + 0 & 3 - 10 + 7 \end{bmatrix} \\ &= \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O \end{aligned}

∴ A2−5A+7I=O\therefore\ A^2 - 5A + 7I = O = RHS. Hence proved.

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

The one idea this problem is built on

A "matrix polynomial" is evaluated just like an ordinary one, term by term: A2A^2 means AAAA (a matrix product, worked out with row-times-column sums), 5A5A multiplies every entry by 5, and 7I7I is the matrix with 7 on the diagonal and 0 elsewhere. All three are 2×22\times2⁠, so they can be combined entry by entry, and every entry comes out 0. (The "0" on the right is the 2×22\times2 zero matrix O.)

Given, to show, and the plan
Given
The 2×22\times2 matrix A above; I is the 2×22\times2 identity.
To show
A2−5A+7I=OA^2 - 5A + 7I = O⁠, the 2×22\times2 zero matrix.
Plan
(1) Compute A2=AAA^2 = AA⁠, entry by entry. (2) Compute 5A5A and 7I7I⁠. (3) Combine the three matrices entry by entry. (4) Check.

Basics you need

1Row times column

The (i,j)(i,j) entry of ABAB is row ii of A times column jj of B. A2A^2 means AAAA⁠, not "square every entry". Basics: pair up, multiply, add

2Scalar multiple

kAkA multiplies every entry of A by kk⁠. Basics: adding only like with like

3Identity matrix

I2I_2 has 1 in both diagonal positions and 0 in the other two, so 7I7I has 7 on the diagonal and 0 elsewhere. Basics: identity and undo

4Adding matrices

Matrices of the same order are added (or subtracted) entry by entry. Lesson: addition and scalar multiplication

Solution

  1. Compute A2=AAA^2 = AA

    A is 2×22\times2⁠, so AAAA is (2×2)(2×2)(2\times2)(2\times2)⁠: defined, of order 2×22\times2⁠. The rows of A are (3,1)(3, 1) and (−1,2)(-1, 2)⁠; its columns are (3,−1)(3, -1) and (1,2)(1, 2)⁠.

    c11=3⋅3+1⋅(−1)=8c12=3⋅1+1⋅2=5c21=(−1)⋅3+2⋅(−1)=−5c22=(−1)⋅1+2⋅2=3\begin{aligned} c_{11} &= 3\cdot3 + 1\cdot(-1) = 8 \\ c_{12} &= 3\cdot1 + 1\cdot2 = 5 \\ c_{21} &= (-1)\cdot3 + 2\cdot(-1) = -5 \\ c_{22} &= (-1)\cdot1 + 2\cdot2 = 3 \end{aligned} A2=[85−53].A^2 = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}.
    Each entry pairs one row of the first factor with one column of the second. The sign in c21c_{21} needs care: it has two negative terms, (−1)⋅3=−3(-1)\cdot3 = -3 and 2⋅(−1)=−22\cdot(-1) = -2⁠. Squaring each entry instead would give rows (9,1)(9, 1) and (1,4)(1, 4)⁠, which is not A2A^2⁠.
    Entry formula of Section 3.4.5. Basics: pair up, multiply, add

  2. Compute 5A5A and 7I7I

    5A=[5⋅35⋅15⋅(−1)5⋅2]=[155−510]7I=[7⋅17⋅07⋅07⋅1]=[7007]\begin{aligned} 5A &= \begin{bmatrix} 5\cdot3 & 5\cdot1 \\ 5\cdot(-1) & 5\cdot2 \end{bmatrix} = \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} \\ 7I &= \begin{bmatrix} 7\cdot1 & 7\cdot0 \\ 7\cdot0 & 7\cdot1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} \end{aligned}
    I is taken of order 2 so that 7I7I has the same order as A2A^2 and 5A5A⁠; otherwise the sum would not be defined. The identity plays the role of the number 1 in an ordinary polynomial: the constant term 7 becomes 7I7I⁠.
    Section 3.4.2: kA=[k aij]kA = [k\,a_{ij}]⁠. Lesson: addition and scalar multiplication

  3. Combine entry by entry

    All three matrices are 2×22\times2⁠, so A2−5A+7IA^2 - 5A + 7I is defined. Take each position in turn:

    (1,1):8−15+7=0(1,2):5−5+0=0(2,1):−5−(−5)+0=0(2,2):3−10+7=0\begin{aligned} (1,1) &: 8 - 15 + 7 = 0 \\ (1,2) &: 5 - 5 + 0 = 0 \\ (2,1) &: -5 - (-5) + 0 = 0 \\ (2,2) &: 3 - 10 + 7 = 0 \end{aligned} A2−5A+7I=[0000]=O.A^2 - 5A + 7I = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O.

    This is what we had to show. (The book prints the right side as 0; it means the 2×22\times2 zero matrix O.)

    Subtracting 5A5A means adding (−1)(5A)(-1)(5A)⁠, so in position (2,1)(2,1) we subtract −5-5⁠, which is the same as adding 5. A sign slip here is the most common way to get a non-zero entry.
    Addition and subtraction of matrices of the same order, Section 3.4.1. Basics: adding only like with like
Result A2=[85−53],A^2 = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix},

and A2−5A+7IA^2 - 5A + 7I has every entry equal to 0, so A2−5A+7I=OA^2 - 5A + 7I = O. (Proofs have no entry in the answer key.)

Check it with numbers

A neat independent check: for any 2×22\times2 matrix with rows (a,b)(a, b) and (c,d)(c, d)⁠, expanding the products shows

A2−(a+d)A+(ad−bc)I=O.A^2 - (a+d)A + (ad - bc)I = O.

Here a+d=3+2=5a + d = 3 + 2 = 5 and ad−bc=3⋅2−1⋅(−1)=7ad - bc = 3\cdot2 - 1\cdot(-1) = 7⁠, exactly the numbers 5 and 7 in the question ✓. (This general fact is beyond the syllabus of this chapter; the verification script checks it symbolically, and the direct computation above does not depend on it.)

Common mistakes
  1. Squaring each entry to get A2A^2⁠. Matrix squaring is a product AAAA⁠.
  2. Writing 7 instead of 7I7I. A number cannot be added to a matrix; the constant term must be a matrix of the same order.
  3. Sign errors in c21c_{21} or in −5−(−5)-5 - (-5)⁠.
What to take away
  1. To evaluate a polynomial in A: compute the powers by multiplication, scale, replace the constant kk by kIkI⁠, and combine entry by entry.
  2. The same technique appears in Example 18 and Exercise 3.2 Q15.
  3. Write out 5A5A and 7I7I in full before combining; most errors happen when several operations are done in one go.

Every fact used, and where it comes from

FactUsed inSource
Product of matrices, entry formulaStep 1Section 3.4.5 (Lesson: matrix multiplication)
Scalar multiple kAkAStep 2Section 3.4.2 (Lesson: addition and scalar multiplication)
Identity matrix I2I_2Step 2Section 3.3 (vi) (Lesson: types of matrices)
Sum and difference of matricesStep 3Section 3.4.1
A2−(a+d)A+(ad−bc)I=OA^2 - (a+d)A + (ad-bc)I = O for 2×22\times2 ACheckVerified symbolically in the site's verification script (not from this chapter)
All matricesSteps 1–3Recomputed with sympy in the site's verification script
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