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Every question from the end of every unit, with hints and answers to open. After you check an answer, mark how you did, and come back to the ones you didn’t get.

76 questions

Unit 6 · Matrix and Determinant

Exercise 6.1

Question 2

a.

R=(321) R = \begin{pmatrix} 3 & 2 & 1 \end{pmatrix}

Answer

RT=(321) R^{\mathrm T} = \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix} : a row matrix becomes a column matrix (1 × 3 to 3 × 1).

b.

W=(−345) W = \begin{pmatrix} -3 \\ 4 \\ 5 \end{pmatrix}

Answer

WT=(−345) W^{\mathrm T} = \begin{pmatrix} -3 & 4 & 5 \end{pmatrix} (3 × 1 to 1 × 3).

c.

X=(−1537) X = \begin{pmatrix} -1 & 5 \\ 3 & 7 \end{pmatrix}

Answer

XT=(−1357) X^{\mathrm T} = \begin{pmatrix} -1 & 3 \\ 5 & 7 \end{pmatrix} : the 5 and the 3 change places.

d.

Y=(3472−110) Y = \begin{pmatrix} 3 & 4 & 7 \\ 2 & -1 & 10 \end{pmatrix}

Answer

YT=(324−1710) Y^{\mathrm T} = \begin{pmatrix} 3 & 2 \\ 4 & -1 \\ 7 & 10 \end{pmatrix} (2 × 3 to 3 × 2).

e.

Z=(015−445) Z = \begin{pmatrix} 0 & 1 \\ 5 & -4 \\ 4 & 5 \end{pmatrix}

Answer

ZT=(0541−45) Z^{\mathrm T} = \begin{pmatrix} 0 & 5 & 4 \\ 1 & -4 & 5 \end{pmatrix} (3 × 2 to 2 × 3).

f.

M=(671052−134) M = \begin{pmatrix} 6 & 7 & 1 \\ 0 & 5 & 2 \\ -1 & 3 & 4 \end{pmatrix}

Answer

MT=(60−1753124) M^{\mathrm T} = \begin{pmatrix} 6 & 0 & -1 \\ 7 & 5 & 3 \\ 1 & 2 & 4 \end{pmatrix} : the diagonal 6, 5, 4 stays; the rest is mirrored across it.

Questions 3 to 7

3.

If P=(9485) P = \begin{pmatrix} 9 & 4 \\ 8 & 5 \end{pmatrix} , prove that (PT)T=P (P^{\mathrm T})^{\mathrm T} = P .

Answer

PT=(9845) P^{\mathrm T} = \begin{pmatrix} 9 & 8 \\ 4 & 5 \end{pmatrix} , so (PT)T=(9485)=P (P^{\mathrm T})^{\mathrm T} = \begin{pmatrix} 9 & 4 \\ 8 & 5 \end{pmatrix} = P . Proved.

4.

If M=(pqrabcxyz) M = \begin{pmatrix} p & q & r \\ a & b & c \\ x & y & z \end{pmatrix} :

  • a.Find MT M^{\mathrm T} and (MT)T (M^{\mathrm T})^{\mathrm T} .
  • b.Write the relation between M and (MT)T (M^{\mathrm T})^{\mathrm T} .
Answer
  • a.MT=(paxqbyrcz) M^{\mathrm T} = \begin{pmatrix} p & a & x \\ q & b & y \\ r & c & z \end{pmatrix} , and (MT)T=(pqrabcxyz) (M^{\mathrm T})^{\mathrm T} = \begin{pmatrix} p & q & r \\ a & b & c \\ x & y & z \end{pmatrix} .
  • b.(MT)T=M (M^{\mathrm T})^{\mathrm T} = M : they are equal.
5.

If A=(13−24) A = \begin{pmatrix} 1 & 3 \\ -2 & 4 \end{pmatrix} and B=(0−325) B = \begin{pmatrix} 0 & -3 \\ 2 & 5 \end{pmatrix} :

  • a.Find AT+BT A^{\mathrm T} + B^{\mathrm T} .
  • b.Find (A+B)T (A + B)^{\mathrm T} .
  • c.Are (A+B)T (A + B)^{\mathrm T} and AT+BT A^{\mathrm T} + B^{\mathrm T} equal? Give a reason.
Answer
  • a.AT+BT=(1−234)+(02−35)= A^{\mathrm T} + B^{\mathrm T} = \begin{pmatrix} 1 & -2 \\ 3 & 4 \end{pmatrix} + \begin{pmatrix} 0 & 2 \\ -3 & 5 \end{pmatrix} = (1009) \begin{pmatrix} 1 & 0 \\ 0 & 9 \end{pmatrix} .
  • b.A+B=(1009) A + B = \begin{pmatrix} 1 & 0 \\ 0 & 9 \end{pmatrix} , so (A+B)T=(1009) (A + B)^{\mathrm T} = \begin{pmatrix} 1 & 0 \\ 0 & 9 \end{pmatrix} .
  • c.Yes. Both are (1009) \begin{pmatrix} 1 & 0 \\ 0 & 9 \end{pmatrix} : the transpose of a sum is always the sum of the transposes.
6.

If A=(1357) A = \begin{pmatrix} 1 & 3 \\ 5 & 7 \end{pmatrix} and B=(2468) B = \begin{pmatrix} 2 & 4 \\ 6 & 8 \end{pmatrix} :

  • a.Prove that (A+B)T=AT+BT (A + B)^{\mathrm T} = A^{\mathrm T} + B^{\mathrm T} .
  • b.Is A + B a symmetric matrix? Give a reason.
Answer

a. A+B=(371115) A + B = \begin{pmatrix} 3 & 7 \\ 11 & 15 \end{pmatrix} , so (A+B)T=(311715) (A + B)^{\mathrm T} = \begin{pmatrix} 3 & 11 \\ 7 & 15 \end{pmatrix} .
AT+BT=(1537)+(2648)=(311715) A^{\mathrm T} + B^{\mathrm T} = \begin{pmatrix} 1 & 5 \\ 3 & 7 \end{pmatrix} + \begin{pmatrix} 2 & 6 \\ 4 & 8 \end{pmatrix} = \begin{pmatrix} 3 & 11 \\ 7 & 15 \end{pmatrix} . Equal: proved.
b. No. (A+B)T≠A+B (A + B)^{\mathrm T} \ne A + B : the 7 and the 11 across the diagonal are different.

7.

If A=(abcd) A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} , B=(4−3−2−5) B = \begin{pmatrix} 4 & -3 \\ -2 & -5 \end{pmatrix} and AT=B A^{\mathrm T} = B , find a, b, c and d.

Answer

AT=(acbd)=(4−3−2−5) A^{\mathrm T} = \begin{pmatrix} a & c \\ b & d \end{pmatrix} = \begin{pmatrix} 4 & -3 \\ -2 & -5 \end{pmatrix} .
Matching elements: a = 4, c = −3, b = −2, d = −5.

Exercise 6.2

Question 2

a.

(54)(13) \begin{pmatrix} 5 & 4 \end{pmatrix} \begin{pmatrix} 1 \\ 3 \end{pmatrix}

Answer

(1 × 2)(2 × 1): possible, 1 × 1. 5 × 1 + 4 × 3 = 17, so the product is (17) \begin{pmatrix} 17 \end{pmatrix} .

b.

(13)(54) \begin{pmatrix} 1 \\ 3 \end{pmatrix} \begin{pmatrix} 5 & 4 \end{pmatrix}

Answer

(2 × 1)(1 × 2): possible, 2 × 2. Each row (one number) times each column (one number):
(1×51×43×53×4)=(541512) \begin{pmatrix} 1 \times 5 & 1 \times 4 \\ 3 \times 5 & 3 \times 4 \end{pmatrix} = \begin{pmatrix} 5 & 4 \\ 15 & 12 \end{pmatrix} . Compare part a: the same two matrices, the other order, a different size.

c.

(321)(−345) \begin{pmatrix} 3 & 2 & 1 \end{pmatrix} \begin{pmatrix} -3 \\ 4 \\ 5 \end{pmatrix}

Answer

(1 × 3)(3 × 1): possible, 1 × 1. 3(−3) + 2(4) + 1(5) = −9 + 8 + 5 = 4: (4) \begin{pmatrix} 4 \end{pmatrix} .

d.

(015−445)(3472−110) \begin{pmatrix} 0 & 1 \\ 5 & -4 \\ 4 & 5 \end{pmatrix} \begin{pmatrix} 3 & 4 & 7 \\ 2 & -1 & 10 \end{pmatrix}

Answer

(3 × 2)(2 × 3): possible, 3 × 3.
(0+20−10+1015−820+435−4012+1016−528+50)=(2−110724−5221178) \begin{pmatrix} 0 + 2 & 0 - 1 & 0 + 10 \\ 15 - 8 & 20 + 4 & 35 - 40 \\ 12 + 10 & 16 - 5 & 28 + 50 \end{pmatrix} = \begin{pmatrix} 2 & -1 & 10 \\ 7 & 24 & -5 \\ 22 & 11 & 78 \end{pmatrix}

e.

(−1537)(6182) \begin{pmatrix} -1 & 5 \\ 3 & 7 \end{pmatrix} \begin{pmatrix} 6 & 1 \\ 8 & 2 \end{pmatrix}

Answer

(2 × 2)(2 × 2): possible.
(−6+40−1+1018+563+14)=(3497417) \begin{pmatrix} -6 + 40 & -1 + 10 \\ 18 + 56 & 3 + 14 \end{pmatrix} = \begin{pmatrix} 34 & 9 \\ 74 & 17 \end{pmatrix}

f.

(1423)(−340612) \begin{pmatrix} 1 & 4 \\ 2 & 3 \end{pmatrix} \begin{pmatrix} -3 & 4 \\ 0 & 6 \\ 1 & 2 \end{pmatrix}

Answer

(2 × 2)(3 × 2): the first has 2 columns, the second 3 rows. Not possible. (The other order, (3 × 2)(2 × 2), could be found.)

Questions 3 to 8

3.

If Q=(36−2−4) Q = \begin{pmatrix} 3 & 6 \\ -2 & -4 \end{pmatrix} , find Q2 Q^2 .

Answer

Q2=(36−2−4)(36−2−4)=(9−1218−24−6+8−12+16)=(−3−624) Q^2 = \begin{pmatrix} 3 & 6 \\ -2 & -4 \end{pmatrix} \begin{pmatrix} 3 & 6 \\ -2 & -4 \end{pmatrix} = \begin{pmatrix} 9 - 12 & 18 - 24 \\ -6 + 8 & -12 + 16 \end{pmatrix} = \begin{pmatrix} -3 & -6 \\ 2 & 4 \end{pmatrix}
Q2=−Q Q^2 = -Q , a curious result: check it with the Cayley–Hamilton fact above (a + d = −1, ad − bc = 0).

4.
  • a.If M=(−23−12) M = \begin{pmatrix} -2 & 3 \\ -1 & 2 \end{pmatrix} and N=(4−350) N = \begin{pmatrix} 4 & -3 \\ 5 & 0 \end{pmatrix} , find MN and NM.
  • b.If C=(3501) C = \begin{pmatrix} 3 & 5 \\ 0 & 1 \end{pmatrix} , D=(−24−13) D = \begin{pmatrix} -2 & 4 \\ -1 & 3 \end{pmatrix} and E=(2−364) E = \begin{pmatrix} 2 & -3 \\ 6 & 4 \end{pmatrix} , find C(D + E).
Answer
  • a.MN=(−8+156+0−4+103+0)= MN = \begin{pmatrix} -8 + 15 & 6 + 0 \\ -4 + 10 & 3 + 0 \end{pmatrix} = (7663) \begin{pmatrix} 7 & 6 \\ 6 & 3 \end{pmatrix} ; NM=(−8+312−6−10+015+0)= NM = \begin{pmatrix} -8 + 3 & 12 - 6 \\ -10 + 0 & 15 + 0 \end{pmatrix} = (−56−1015) \begin{pmatrix} -5 & 6 \\ -10 & 15 \end{pmatrix} .
  • b.D+E=(0157) D + E = \begin{pmatrix} 0 & 1 \\ 5 & 7 \end{pmatrix} , so C(D+E)=(0+253+350+50+7)= C(D + E) = \begin{pmatrix} 0 + 25 & 3 + 35 \\ 0 + 5 & 0 + 7 \end{pmatrix} = (253857) \begin{pmatrix} 25 & 38 \\ 5 & 7 \end{pmatrix} .
5.
  • a.If A=(1−1−11) A = \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix} , prove that A2=2A A^2 = 2A .
  • b.If X=(1−13−3) X = \begin{pmatrix} 1 & -1 \\ 3 & -3 \end{pmatrix} and Y=(2−52−5) Y = \begin{pmatrix} 2 & -5 \\ 2 & -5 \end{pmatrix} , prove that XY is a zero matrix.
  • c.If R=(2−31−2) R = \begin{pmatrix} 2 & -3 \\ 1 & -2 \end{pmatrix} , prove that R2=I R^2 = I , the 2 × 2 identity matrix.
Answer
  • a.A2=(1+1−1−1−1−11+1)=(2−2−22)=2(1−1−11)=2A A^2 = \begin{pmatrix} 1 + 1 & -1 - 1 \\ -1 - 1 & 1 + 1 \end{pmatrix} = \begin{pmatrix} 2 & -2 \\ -2 & 2 \end{pmatrix} = 2\begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix} = 2A . Proved.
  • b.XY=(2−2−5+56−6−15+15)=(0000) XY = \begin{pmatrix} 2 - 2 & -5 + 5 \\ 6 - 6 & -15 + 15 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} . Proved: two matrices that aren’t zero with a zero product.
  • c.R2=(4−3−6+62−2−3+4)=(1001)=I R^2 = \begin{pmatrix} 4 - 3 & -6 + 6 \\ 2 - 2 & -3 + 4 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I . Proved: R undoes itself.
6.

If A=(214−1) A = \begin{pmatrix} 2 & 1 \\ 4 & -1 \end{pmatrix} , prove that A2−A−6I=O A^2 - A - 6I = O , where I and O are the identity and zero matrices of order 2 × 2.

Answer

A2=(4+42−18−44+1)=(8145) A^2 = \begin{pmatrix} 4 + 4 & 2 - 1 \\ 8 - 4 & 4 + 1 \end{pmatrix} = \begin{pmatrix} 8 & 1 \\ 4 & 5 \end{pmatrix} .
A2−A−6I=(8−2−61−1−04−4−05+1−6)=(0000)=O A^2 - A - 6I = \begin{pmatrix} 8 - 2 - 6 & 1 - 1 - 0 \\ 4 - 4 - 0 & 5 + 1 - 6 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = O
Proved. (By Cayley–Hamilton: a + d = 1 and ad − bc = −2 − 4 = −6, so the equation is A2−1A+(−6)I=O A^2 - 1A + (-6)I = O .)

7.

If U=(3004) U = \begin{pmatrix} 3 & 0 \\ 0 & 4 \end{pmatrix} , V=(pq0r) V = \begin{pmatrix} p & q \\ 0 & r \end{pmatrix} and UV = U + V, find p, q and r.

Answer

UV=(3p3q04r) UV = \begin{pmatrix} 3p & 3q \\ 0 & 4r \end{pmatrix} and U+V=(3+pq04+r) U + V = \begin{pmatrix} 3 + p & q \\ 0 & 4 + r \end{pmatrix} .
3p = 3 + p, 3q = q, 4r = 4 + r: p=32 p = \dfrac{3}{2} , q = 0, r=43 r = \dfrac{4}{3} .

8.

If X=(63−20) X = \begin{pmatrix} 6 & 3 \\ -2 & 0 \end{pmatrix} and Y=(251−4) Y = \begin{pmatrix} 2 & 5 \\ 1 & -4 \end{pmatrix} , prove that (XY)T=YTXT (XY)^{\mathrm T} = Y^{\mathrm T} X^{\mathrm T} .

Answer

XY=(12+330−12−4+0−10+0)=(1518−4−10) XY = \begin{pmatrix} 12 + 3 & 30 - 12 \\ -4 + 0 & -10 + 0 \end{pmatrix} = \begin{pmatrix} 15 & 18 \\ -4 & -10 \end{pmatrix} , so (XY)T=(15−418−10) (XY)^{\mathrm T} = \begin{pmatrix} 15 & -4 \\ 18 & -10 \end{pmatrix} .
YTXT=(215−4)(6−230)=(12+3−4+030−12−10+0)=(15−418−10) Y^{\mathrm T} X^{\mathrm T} = \begin{pmatrix} 2 & 1 \\ 5 & -4 \end{pmatrix} \begin{pmatrix} 6 & -2 \\ 3 & 0 \end{pmatrix} = \begin{pmatrix} 12 + 3 & -4 + 0 \\ 30 - 12 & -10 + 0 \end{pmatrix} = \begin{pmatrix} 15 & -4 \\ 18 & -10 \end{pmatrix} . Equal: proved.

Question 9

a.

Prove that AB ≠ BA.

Answer

AB=(0+80+123+125+18)=(8121523) AB = \begin{pmatrix} 0 + 8 & 0 + 12 \\ 3 + 12 & 5 + 18 \end{pmatrix} = \begin{pmatrix} 8 & 12 \\ 15 & 23 \end{pmatrix} ; BA=(0+56+150+68+18)=(521626) BA = \begin{pmatrix} 0 + 5 & 6 + 15 \\ 0 + 6 & 8 + 18 \end{pmatrix} = \begin{pmatrix} 5 & 21 \\ 6 & 26 \end{pmatrix} . Different, so AB ≠ BA.

b.

Prove that A(BC) = (AB)C.

Answer

BC=(18+3524+4524+4232+54)=(53696686) BC = \begin{pmatrix} 18 + 35 & 24 + 45 \\ 24 + 42 & 32 + 54 \end{pmatrix} = \begin{pmatrix} 53 & 69 \\ 66 & 86 \end{pmatrix} , so A(BC)=(0+1320+17253+19869+258)=(132172251327) A(BC) = \begin{pmatrix} 0 + 132 & 0 + 172 \\ 53 + 198 & 69 + 258 \end{pmatrix} = \begin{pmatrix} 132 & 172 \\ 251 & 327 \end{pmatrix} .
(AB)C=(48+8464+10890+161120+207)=(132172251327) (AB)C = \begin{pmatrix} 48 + 84 & 64 + 108 \\ 90 + 161 & 120 + 207 \end{pmatrix} = \begin{pmatrix} 132 & 172 \\ 251 & 327 \end{pmatrix} . Equal: proved.

c.

Prove that A(B + C) = AB + AC.

Answer

B+C=(9131115) B + C = \begin{pmatrix} 9 & 13 \\ 11 & 15 \end{pmatrix} , so A(B+C)=(0+220+309+3313+45)=(22304258) A(B + C) = \begin{pmatrix} 0 + 22 & 0 + 30 \\ 9 + 33 & 13 + 45 \end{pmatrix} = \begin{pmatrix} 22 & 30 \\ 42 & 58 \end{pmatrix} .
AC=(0+140+186+218+27)=(14182735) AC = \begin{pmatrix} 0 + 14 & 0 + 18 \\ 6 + 21 & 8 + 27 \end{pmatrix} = \begin{pmatrix} 14 & 18 \\ 27 & 35 \end{pmatrix} , and AB+AC=(8+1412+1815+2723+35)=(22304258) AB + AC = \begin{pmatrix} 8 + 14 & 12 + 18 \\ 15 + 27 & 23 + 35 \end{pmatrix} = \begin{pmatrix} 22 & 30 \\ 42 & 58 \end{pmatrix} . Equal: proved.

d.

Prove that IA = AI = A.

Answer

IA=(0+02+00+10+3)=(0213)=A IA = \begin{pmatrix} 0 + 0 & 2 + 0 \\ 0 + 1 & 0 + 3 \end{pmatrix} = \begin{pmatrix} 0 & 2 \\ 1 & 3 \end{pmatrix} = A and AI=(0+00+21+00+3)=(0213)=A AI = \begin{pmatrix} 0 + 0 & 0 + 2 \\ 1 + 0 & 0 + 3 \end{pmatrix} = \begin{pmatrix} 0 & 2 \\ 1 & 3 \end{pmatrix} = A . Proved.

e.

Prove that (AB)T=BTAT (AB)^{\mathrm T} = B^{\mathrm T} A^{\mathrm T} .

Answer

From a, (AB)T=(8151223) (AB)^{\mathrm T} = \begin{pmatrix} 8 & 15 \\ 12 & 23 \end{pmatrix} .
BTAT=(3456)(0123)=(0+83+120+125+18)=(8151223) B^{\mathrm T} A^{\mathrm T} = \begin{pmatrix} 3 & 4 \\ 5 & 6 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 2 & 3 \end{pmatrix} = \begin{pmatrix} 0 + 8 & 3 + 12 \\ 0 + 12 & 5 + 18 \end{pmatrix} = \begin{pmatrix} 8 & 15 \\ 12 & 23 \end{pmatrix} . Equal: proved.

Exercise 6.3

Question 2

a.

det⁡(2) \det \begin{pmatrix} 2 \end{pmatrix}

Answer

2.

b.

det⁡(−7) \det \begin{pmatrix} -7 \end{pmatrix}

Answer

−7: the determinant of a 1 × 1 matrix is its element, sign and all.

c.

∣2314∣ \begin{vmatrix} 2 & 3 \\ 1 & 4 \end{vmatrix}

Answer

8 − 3 = 5.

d.

∣sin⁡A−cos⁡Acos⁡Asin⁡A∣ \begin{vmatrix} \sin A & -\cos A \\ \cos A & \sin A \end{vmatrix}

Answer

sin⁡Asin⁡A−(−cos⁡A)cos⁡A=sin⁡2A+cos⁡2A= \sin A \sin A - (-\cos A)\cos A = \sin^2 A + \cos^2 A = 1.

e.

∣x+yx−yx−yx+y∣ \begin{vmatrix} x + y & x - y \\ x - y & x + y \end{vmatrix}

Answer

(x+y)2−(x−y)2=(x2+2xy+y2)−(x2−2xy+y2)= (x + y)^2 - (x - y)^2 = (x^2 + 2xy + y^2) - (x^2 - 2xy + y^2) = 4xy.

f.

∣x+yx−yx+yx−y∣ \begin{vmatrix} x + y & x - y \\ x + y & x - y \end{vmatrix}

Answer

(x + y)(x − y) − (x − y)(x + y) = 0: the two rows are the same, and a determinant with two equal rows is always 0.

g.

∣3−4−35∣ \begin{vmatrix} 3 & -4 \\ -3 & 5 \end{vmatrix}

Answer

15 − 12 = 3.

h.

∣x+yx−yx2+xy+y2x2−xy+y2∣ \begin{vmatrix} x + y & x - y \\ x^2 + xy + y^2 & x^2 - xy + y^2 \end{vmatrix}

Hint

(x+y)(x2−xy+y2)=x3+y3 (x + y)(x^2 - xy + y^2) = x^3 + y^3 and (x−y)(x2+xy+y2)=x3−y3 (x - y)(x^2 + xy + y^2) = x^3 - y^3 .

Answer

(x+y)(x2−xy+y2)−(x−y)(x2+xy+y2)=(x3+y3)−(x3−y3) (x + y)(x^2 - xy + y^2) - (x - y)(x^2 + xy + y^2) = (x^3 + y^3) - (x^3 - y^3)
=2y3 = 2y^3 .

Question 3

a.

(10) \begin{pmatrix} 10 \end{pmatrix}

Answer

10.

b.

(−8) \begin{pmatrix} -8 \end{pmatrix}

Answer

−8.

c.

(xyyx) \begin{pmatrix} x & y \\ y & x \end{pmatrix}

Answer

x2−y2 x^2 - y^2 .

d.

(1−324) \begin{pmatrix} 1 & -3 \\ 2 & 4 \end{pmatrix}

Answer

4 − (−6) = 10.

e.

(10542) \begin{pmatrix} 10 & 5 \\ 4 & 2 \end{pmatrix}

Answer

20 − 20 = 0: a singular matrix (the first row is 2.5 times the second).

f.

(−7112−4) \begin{pmatrix} -7 & 11 \\ 2 & -4 \end{pmatrix}

Answer

28 − 22 = 6.

Question 4

a.

∣3x28∣=20 \begin{vmatrix} 3 & x \\ 2 & 8 \end{vmatrix} = 20

Answer

24 − 2x = 20, so 2x = 4: x = 2.

b.

∣5−22x−3∣=1 \begin{vmatrix} 5 & -2 \\ 2x & -3 \end{vmatrix} = 1

Answer

−15 − (−2)(2x) = 1, so −15 + 4x = 1, 4x = 16: x = 4.

c.

∣3x2−43∣=8 \begin{vmatrix} 3x & 2 \\ -4 & 3 \end{vmatrix} = 8

Answer

9x + 8 = 8, so 9x = 0: x = 0.

d.

∣x−2x3x4x∣=100 \begin{vmatrix} x & -2x \\ 3x & 4x \end{vmatrix} = 100

Answer

4x2+6x2=100 4x^2 + 6x^2 = 100 , so 10x2=100 10x^2 = 100 , x2=10 x^2 = 10 : x=±10 x = \pm\sqrt{10} .

Question 5

a.

A − 3B

Answer

A−3B=(2−34+153−182−6)=(−119−15−4) A - 3B = \begin{pmatrix} 2 - 3 & 4 + 15 \\ 3 - 18 & 2 - 6 \end{pmatrix} = \begin{pmatrix} -1 & 19 \\ -15 & -4 \end{pmatrix} ; determinant 4 + 285 = 289.

b.

2A + B

Answer

2A+B=(53126) 2A + B = \begin{pmatrix} 5 & 3 \\ 12 & 6 \end{pmatrix} ; determinant 30 − 36 = −6.

c.

3A − 2B

Answer

3A−2B=(6−212+109−126−4)=(422−32) 3A - 2B = \begin{pmatrix} 6 - 2 & 12 + 10 \\ 9 - 12 & 6 - 4 \end{pmatrix} = \begin{pmatrix} 4 & 22 \\ -3 & 2 \end{pmatrix} ; determinant 8 + 66 = 74.

d.

A + 2B

Answer

A+2B=(4−6156) A + 2B = \begin{pmatrix} 4 & -6 \\ 15 & 6 \end{pmatrix} ; determinant 24 + 90 = 114.

Question 6

a.

∣x2y2yx∣=(x−y)(x2+xy+y2) \begin{vmatrix} x^2 & y^2 \\ y & x \end{vmatrix} = (x - y)(x^2 + xy + y^2)

Answer

Left side =x2⋅x−y2⋅y=x3−y3=(x−y)(x2+xy+y2) = x^2 \cdot x - y^2 \cdot y = x^3 - y^3 = (x - y)(x^2 + xy + y^2) , the difference of two cubes. Proved.

b.

∣x+yy+zz−yx−y∣=x2−z2 \begin{vmatrix} x + y & y + z \\ z - y & x - y \end{vmatrix} = x^2 - z^2

Answer

Left side =(x+y)(x−y)−(y+z)(z−y)=(x2−y2)−(z2−y2)=x2−z2 = (x + y)(x - y) - (y + z)(z - y) = (x^2 - y^2) - (z^2 - y^2) = x^2 - z^2 . Proved.

c.

∣a−1a2−a+1a+1a2+a+1∣=−2 \begin{vmatrix} a - 1 & a^2 - a + 1 \\ a + 1 & a^2 + a + 1 \end{vmatrix} = -2

Answer

Left side =(a−1)(a2+a+1)−(a+1)(a2−a+1)=(a3−1)−(a3+1)=−2 = (a - 1)(a^2 + a + 1) - (a + 1)(a^2 - a + 1) = (a^3 - 1) - (a^3 + 1) = -2 . Proved, whatever a is.

d.

∣1+x111+y∣=xy(1+1x+1y) \begin{vmatrix} 1 + x & 1 \\ 1 & 1 + y \end{vmatrix} = xy\left(1 + \dfrac{1}{x} + \dfrac{1}{y}\right)

Answer

Left side = (1 + x)(1 + y) − 1 = 1 + x + y + xy − 1 = xy + x + y.
Right side =xy+xy⋅1x+xy⋅1y=xy+y+x = xy + xy \cdot \dfrac{1}{x} + xy \cdot \dfrac{1}{y} = xy + y + x . Equal: proved (for x ≠ 0, y ≠ 0, where the right side is defined).

Question 7

7.

If M=(5243) M = \begin{pmatrix} 5 & 2 \\ 4 & 3 \end{pmatrix} and N=(1−362) N = \begin{pmatrix} 1 & -3 \\ 6 & 2 \end{pmatrix} , verify that |MN| = |M||N|.

Answer

|M| = 15 − 8 = 7 and |N| = 2 + 18 = 20, so |M||N| = 140.
MN=(5+12−15+44+18−12+6)=(17−1122−6) MN = \begin{pmatrix} 5 + 12 & -15 + 4 \\ 4 + 18 & -12 + 6 \end{pmatrix} = \begin{pmatrix} 17 & -11 \\ 22 & -6 \end{pmatrix} , and |MN| = −102 + 242 = 140. Verified: |MN| = |M||N| = 140.

Exercise 6.4

Question 2

a.

A=(3121) A = \begin{pmatrix} 3 & 1 \\ 2 & 1 \end{pmatrix} and B=(1−1−23) B = \begin{pmatrix} 1 & -1 \\ -2 & 3 \end{pmatrix}

Answer

AB=(3−2−3+32−2−2+3)=I AB = \begin{pmatrix} 3 - 2 & -3 + 3 \\ 2 - 2 & -2 + 3 \end{pmatrix} = I and BA=(3−21−1−6+6−2+3)=I BA = \begin{pmatrix} 3 - 2 & 1 - 1 \\ -6 + 6 & -2 + 3 \end{pmatrix} = I . So they are inverses.

b.

A=(5−3−32) A = \begin{pmatrix} 5 & -3 \\ -3 & 2 \end{pmatrix} and B=(2335) B = \begin{pmatrix} 2 & 3 \\ 3 & 5 \end{pmatrix}

Answer

AB=(10−915−15−6+6−9+10)=I AB = \begin{pmatrix} 10 - 9 & 15 - 15 \\ -6 + 6 & -9 + 10 \end{pmatrix} = I and BA=(10−9−6+615−15−9+10)=I BA = \begin{pmatrix} 10 - 9 & -6 + 6 \\ 15 - 15 & -9 + 10 \end{pmatrix} = I . Inverses.

c.

A=(2153) A = \begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix} and B=(3−1−52) B = \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}

Answer

AB=(6−5−2+215−15−5+6)=I AB = \begin{pmatrix} 6 - 5 & -2 + 2 \\ 15 - 15 & -5 + 6 \end{pmatrix} = I and BA=(6−53−3−10+10−5+6)=I BA = \begin{pmatrix} 6 - 5 & 3 - 3 \\ -10 + 10 & -5 + 6 \end{pmatrix} = I . Inverses.

d.

A=(4759) A = \begin{pmatrix} 4 & 7 \\ 5 & 9 \end{pmatrix} and B=(9−7−54) B = \begin{pmatrix} 9 & -7 \\ -5 & 4 \end{pmatrix}

Answer

AB=(36−35−28+2845−45−35+36)=I AB = \begin{pmatrix} 36 - 35 & -28 + 28 \\ 45 - 45 & -35 + 36 \end{pmatrix} = I and BA=(36−3563−63−20+20−35+36)=I BA = \begin{pmatrix} 36 - 35 & 63 - 63 \\ -20 + 20 & -35 + 36 \end{pmatrix} = I . Inverses.

Question 3

a.

A=(2134) A = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}

Answer

|A| = 8 − 3 = 5: A−1=15(4−1−32) A^{-1} = \dfrac{1}{5} \begin{pmatrix} 4 & -1 \\ -3 & 2 \end{pmatrix} .

b.

B=(7432) B = \begin{pmatrix} 7 & 4 \\ 3 & 2 \end{pmatrix}

Answer

|B| = 14 − 12 = 2: B−1=12(2−4−37)= B^{-1} = \dfrac{1}{2} \begin{pmatrix} 2 & -4 \\ -3 & 7 \end{pmatrix} = (1−2−3/27/2) \begin{pmatrix} 1 & -2 \\ -3/2 & 7/2 \end{pmatrix} .

c.

P=(2031) P = \begin{pmatrix} 2 & 0 \\ 3 & 1 \end{pmatrix}

Answer

|P| = 2 − 0 = 2: P−1=12(10−32)= P^{-1} = \dfrac{1}{2} \begin{pmatrix} 1 & 0 \\ -3 & 2 \end{pmatrix} = (1/20−3/21) \begin{pmatrix} 1/2 & 0 \\ -3/2 & 1 \end{pmatrix} .

d.

R=(cos⁡A−sin⁡Asin⁡Acos⁡A) R = \begin{pmatrix} \cos A & -\sin A \\ \sin A & \cos A \end{pmatrix}

Answer

∣R∣=cos⁡2A+sin⁡2A=1 |R| = \cos^2 A + \sin^2 A = 1 : R−1=(cos⁡Asin⁡A−sin⁡Acos⁡A) R^{-1} = \begin{pmatrix} \cos A & \sin A \\ -\sin A & \cos A \end{pmatrix} . It is the turn through −A: turning back undoes turning.

Question 4

a.

(a152) \begin{pmatrix} a & 1 \\ 5 & 2 \end{pmatrix} and (2−1−5b) \begin{pmatrix} 2 & -1 \\ -5 & b \end{pmatrix}

Answer

Their product is I: (2a−5−a+b10−10−5+2b)=(1001) \begin{pmatrix} 2a - 5 & -a + b \\ 10 - 10 & -5 + 2b \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} .
2a − 5 = 1 gives a = 3; 2b − 5 = 1 gives b = 3; and −a + b = 0 ✓. a = 3, b = 3.

b.

(p2p−9−q3) \begin{pmatrix} p & 2p - 9 \\ -q & 3 \end{pmatrix} and (35qp) \begin{pmatrix} 3 & 5 \\ q & p \end{pmatrix}

Hint

Multiply them and set the result equal to I. One of the four equations has only p in it.

Answer

(3p+(2p−9)q5p+(2p−9)p−3q+3q−5q+3p)=(1001) \begin{pmatrix} 3p + (2p - 9)q & 5p + (2p - 9)p \\ -3q + 3q & -5q + 3p \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}
Top right: 5p+2p2−9p=0 5p + 2p^2 - 9p = 0 , so 2p2−4p=0 2p^2 - 4p = 0 , 2p(p − 2) = 0: p = 0 or p = 2.
Bottom right: 3p − 5q = 1. With p = 2: 6 − 5q = 1, q = 1. With p = 0: q = −1/5, but then the top left is 0 + (−9)(−1/5) = 9/5, not 1, so it is rejected.
Check p = 2, q = 1 in the top left: 6 + (4 − 9)(1) = 1 ✓. p = 2, q = 1.

Questions 5 and 6

5.

If A−1=(3254) A^{-1} = \begin{pmatrix} 3 & 2 \\ 5 & 4 \end{pmatrix} , find A.

Answer

A=(A−1)−1 A = (A^{-1})^{-1} . ∣A−1∣=12−10=2 |A^{-1}| = 12 - 10 = 2 , so
A=12(4−2−53)=(2−1−5232) A = \frac{1}{2} \begin{pmatrix} 4 & -2 \\ -5 & 3 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ -\frac{5}{2} & \frac{3}{2} \end{pmatrix}
Check: (3254)(2−1−5/23/2)=(6−5−3+310−10−5+6)=I \begin{pmatrix} 3 & 2 \\ 5 & 4 \end{pmatrix} \begin{pmatrix} 2 & -1 \\ -5/2 & 3/2 \end{pmatrix} = \begin{pmatrix} 6 - 5 & -3 + 3 \\ 10 - 10 & -5 + 6 \end{pmatrix} = I ✓.

6.

If A=(2053) A = \begin{pmatrix} 2 & 0 \\ 5 & 3 \end{pmatrix} and B=(4172) B = \begin{pmatrix} 4 & 1 \\ 7 & 2 \end{pmatrix} :

  • a.Find |A| and |B|.
  • b.Find A−1 A^{-1} and B−1 B^{-1} .
  • c.Find (AB)−1 (AB)^{-1} .
  • d.Verify that (AB)−1=B−1A−1 (AB)^{-1} = B^{-1}A^{-1} .
Answer
  • a.|A| = 6 − 0 = 6, |B| = 8 − 7 = 1.
  • b.A−1=16(30−52)= A^{-1} = \dfrac{1}{6} \begin{pmatrix} 3 & 0 \\ -5 & 2 \end{pmatrix} = (1/20−5/61/3) \begin{pmatrix} 1/2 & 0 \\ -5/6 & 1/3 \end{pmatrix} ; B−1=(2−1−74) B^{-1} = \begin{pmatrix} 2 & -1 \\ -7 & 4 \end{pmatrix} .
  • c.AB=(8220+215+6)=(824111) AB = \begin{pmatrix} 8 & 2 \\ 20 + 21 & 5 + 6 \end{pmatrix} = \begin{pmatrix} 8 & 2 \\ 41 & 11 \end{pmatrix} , |AB| = 88 − 82 = 6, so (AB)−1=16(11−2−418)= (AB)^{-1} = \dfrac{1}{6} \begin{pmatrix} 11 & -2 \\ -41 & 8 \end{pmatrix} = (11/6−1/3−41/64/3) \begin{pmatrix} 11/6 & -1/3 \\ -41/6 & 4/3 \end{pmatrix} .
  • d.B−1A−1=(2−1−74)⋅16(30−52)=16(6+5−2−21−208)=16(11−2−418) B^{-1}A^{-1} = \begin{pmatrix} 2 & -1 \\ -7 & 4 \end{pmatrix} \cdot \dfrac{1}{6} \begin{pmatrix} 3 & 0 \\ -5 & 2 \end{pmatrix} = \dfrac{1}{6} \begin{pmatrix} 6 + 5 & -2 \\ -21 - 20 & 8 \end{pmatrix} = \dfrac{1}{6} \begin{pmatrix} 11 & -2 \\ -41 & 8 \end{pmatrix} , the same as c. Verified. (And |AB| = 6 = |A||B|, as 6.4 promised.)

Miscellaneous exercise

1.

In the mapping diagram, the function g(x) is from set A to set B, and h(x) from B to C.

A123g(x)B246h(x)C357
Figure 6.5 The functions g from A to B and h from B to C.
  • a.Write the composite function from A to C in symbols.
  • b.Write it as ordered pairs.
  • c.Is the inverse of this composite function defined? If it is, write it as ordered pairs.
Answer
  • a.h ∘ g, that is h(g(x)): g first, then h.
  • b.1 → 2 → 3, 2 → 4 → 5, 3 → 6 → 7: h ∘ g = {(1, 3), (2, 5), (3, 7)}.
  • c.Yes: h ∘ g is one-to-one (different inputs, different outputs) and onto C (every element of C is used). Its inverse, from C to A: (h∘g)−1={(3,1),(5,2),(7,3)} (h \circ g)^{-1} = \{(3, 1), (5, 2), (7, 3)\} .
2.

The functions are f(x) = 2x + 3 and g(x) = x − 2.

  • a.Find the composite functions f(g(x)) and g−1(f(x)) g^{-1}(f(x)) .
  • b.When f(x) is divided by g(x), find the quotient and the remainder.
  • c.Show the linear inequality 2x + 3y > 6 on a graph.
  • d.What is a surd equation? Solve 2x+3=1 \sqrt{2x + 3} = 1 and check the answer.
Answer
  • a.f(g(x)) = 2(x − 2) + 3 = 2x − 1. g−1(x)=x+2 g^{-1}(x) = x + 2 , so g−1(f(x))=(2x+3)+2= g^{-1}(f(x)) = (2x + 3) + 2 = 2x + 5.
  • b.2x + 3 = 2(x − 2) + 7: quotient 2, remainder 7. (By the remainder theorem, the remainder is f(2) = 7.)
  • c.The boundary 2x + 3y = 6 passes through (3, 0) and (0, 2); it is drawn dashed, since points on it are not included (>, not ≥). Test O(0, 0): 0 > 6 is false, so the region away from the origin is shaded.
12341XYO2x + 3y = 6
Figure 6.6 The graph of 2x + 3y > 6: the dashed line 2x + 3y = 6, and the side not containing the origin shaded.

d. An equation in which the unknown is under a root sign is a surd equation. Square both sides: 2x + 3 = 1, so 2x = −2 and x = −1. Check: 2(−1)+3=1=1 \sqrt{2(-1) + 3} = \sqrt{1} = 1 ✓.

3.

The polynomials are f(x)=x3−2x2−4x+8 f(x) = x^3 - 2x^2 - 4x + 8 and d(x) = x + 2.

  • a.Write the remainder when f(x) is divided by d(x).
  • b.Is d(x) a factor of f(x)? Give a reason.
  • c.Find the factors of f(x) using the factor theorem.
  • d.If d(x) is a function, find d−1(x) d^{-1}(x) .
Answer
  • a.By the remainder theorem, the remainder is f(−2) = −8 − 8 + 8 + 8 = 0.
  • b.Yes, because the remainder is 0.
  • c.f(x)=(x+2)(x2−4x+4)= f(x) = (x + 2)(x^2 - 4x + 4) = (x + 2)(x − 2)(x − 2), that is (x+2)(x−2)2 (x + 2)(x - 2)^2 . (Or by grouping: x2(x−2)−4(x−2)=(x−2)(x2−4) x^2(x - 2) - 4(x - 2) = (x - 2)(x^2 - 4) .)
  • d.y = x + 2; swap and solve: x = y + 2, y = x − 2. d−1(x)=x−2 d^{-1}(x) = x - 2 .
4.

Look at the graph of the system and answer the questions.

14124XYOABC
Figure 6.7 The feasible region OABC.
  • a.Write the coordinates of the points O, A, B and C.
  • b.From the feasible values, find the maximum and minimum of the objective function f(x, y) = 20x + 15y.
  • c.Write the four inequalities of the shaded region.
  • d.Taking the equation of line AB as f(x) and that of line BC as g(x), find f ∘ g(x).
Answer
  • a.O(0, 0), A(2, 0), B(1, 2), C(0, 3).
  • b.At the corners: O: 0; A: 40; B: 20 + 30 = 50; C: 45. Maximum 50 at B(1, 2); minimum 0 at O(0, 0).
  • c.AB through (2, 0) and (1, 2) is 2x + y = 4; BC through (1, 2) and (0, 3) is x + y = 3. The origin side of both is shaded:

x ≥ 0, y ≥ 0, x + y ≤ 3, 2x + y ≤ 4.
d. f(x) = 4 − 2x and g(x) = 3 − x. f ∘ g(x) = f(3 − x) = 4 − 2(3 − x) = 2x − 2.

5.

Look at the graph and answer the questions.

−3−11123XYOVPQ
Figure 6.8 A curve with its lowest point V, through P and Q.
  • a.What is the curve called?
  • b.Write the coordinates of its vertex.
  • c.Find the equation of its axis of symmetry.
  • d.Find the equation of the curve.
Answer
  • a.A parabola.
  • b.V(−1, −1).
  • c.The upright line through the vertex: x = −1.
  • d.Vertex form: y=a(x+1)2−1 y = a(x + 1)^2 - 1 . Through P(1, 3): 3=a(2)2−1 3 = a(2)^2 - 1 , so 4a = 4 and a = 1. y=(x+1)2−1 y = (x + 1)^2 - 1 , that is y=x2+2x y = x^2 + 2x .
6.

If A=(2−3−47) A = \begin{pmatrix} 2 & -3 \\ -4 & 7 \end{pmatrix} :

  • a.Say whether A−1 A^{-1} is defined, with a reason.
  • b.If A−1 A^{-1} is defined, find it.
  • c.Which of (491649) \begin{pmatrix} 4 & 9 \\ 16 & 49 \end{pmatrix} and (16−27−3661) \begin{pmatrix} 16 & -27 \\ -36 & 61 \end{pmatrix} is A2 A^2 ? Find it.
Answer
  • a.|A| = 14 − 12 = 2 ≠ 0: A is non-singular, so A−1 A^{-1} is defined.
  • b.A−1=12(7342) A^{-1} = \dfrac{1}{2} \begin{pmatrix} 7 & 3 \\ 4 & 2 \end{pmatrix} .
  • c.A2=(4+12−6−21−8−2812+49)= A^2 = \begin{pmatrix} 4 + 12 & -6 - 21 \\ -8 - 28 & 12 + 49 \end{pmatrix} = (16−27−3661) \begin{pmatrix} 16 & -27 \\ -36 & 61 \end{pmatrix} . The other matrix is each element squared, which is not A2 A^2 .
7.

If A=(1235) A = \begin{pmatrix} 1 & 2 \\ 3 & 5 \end{pmatrix} and B=(4172) B = \begin{pmatrix} 4 & 1 \\ 7 & 2 \end{pmatrix} :

  • a.What is a transpose matrix? Write AT A^{\mathrm T} .
  • b.Verify that (A+B)T=BT+AT (A + B)^{\mathrm T} = B^{\mathrm T} + A^{\mathrm T} .
  • c.Can AB be found? Give a reason.
  • d.If AB can be found, verify that (AB)−1=B−1A−1 (AB)^{-1} = B^{-1}A^{-1} .
Answer
  • a.The matrix made by writing the rows as columns. AT=(1325) A^{\mathrm T} = \begin{pmatrix} 1 & 3 \\ 2 & 5 \end{pmatrix} .
  • b.A+B=(53107) A + B = \begin{pmatrix} 5 & 3 \\ 10 & 7 \end{pmatrix} , so (A+B)T=(51037) (A + B)^{\mathrm T} = \begin{pmatrix} 5 & 10 \\ 3 & 7 \end{pmatrix} ; BT+AT=(4712)+(1325)=(51037) B^{\mathrm T} + A^{\mathrm T} = \begin{pmatrix} 4 & 7 \\ 1 & 2 \end{pmatrix} + \begin{pmatrix} 1 & 3 \\ 2 & 5 \end{pmatrix} = \begin{pmatrix} 5 & 10 \\ 3 & 7 \end{pmatrix} . Equal.
  • c.Yes: A has 2 columns and B has 2 rows.
  • d.AB=(4+141+412+353+10)=(1854713) AB = \begin{pmatrix} 4 + 14 & 1 + 4 \\ 12 + 35 & 3 + 10 \end{pmatrix} = \begin{pmatrix} 18 & 5 \\ 47 & 13 \end{pmatrix} , |AB| = 234 − 235 = −1, so (AB)−1=−1(13−5−4718)=(−13547−18) (AB)^{-1} = -1 \begin{pmatrix} 13 & -5 \\ -47 & 18 \end{pmatrix} = \begin{pmatrix} -13 & 5 \\ 47 & -18 \end{pmatrix} .

|A| = 5 − 6 = −1, A−1=(−523−1) A^{-1} = \begin{pmatrix} -5 & 2 \\ 3 & -1 \end{pmatrix} ; |B| = 8 − 7 = 1, B−1=(2−1−74) B^{-1} = \begin{pmatrix} 2 & -1 \\ -7 & 4 \end{pmatrix} .
B−1A−1=(−10−34+135+12−14−4)=(−13547−18) B^{-1}A^{-1} = \begin{pmatrix} -10 - 3 & 4 + 1 \\ 35 + 12 & -14 - 4 \end{pmatrix} = \begin{pmatrix} -13 & 5 \\ 47 & -18 \end{pmatrix} . Verified.

8.

P=(2m759) P = \begin{pmatrix} 2m & 7 \\ 5 & 9 \end{pmatrix} , and its inverse is Q=(9n−54) Q = \begin{pmatrix} 9 & n \\ -5 & 4 \end{pmatrix} .

  • a.Write the determinant of an identity matrix.
  • b.Find m and n.
  • c.Find the determinant of P.
  • d.Find QT Q^{\mathrm T} .
Answer
  • a.|I| = 1 × 1 − 0 × 0 = 1.
  • b.PQ = I, as in Example 2 of 6.5: m = 2, n = −7.
  • c.P=(4759) P = \begin{pmatrix} 4 & 7 \\ 5 & 9 \end{pmatrix} : |P| = 36 − 35 = 1.
  • d.Q=(9−7−54) Q = \begin{pmatrix} 9 & -7 \\ -5 & 4 \end{pmatrix} , so QT=(9−5−74) Q^{\mathrm T} = \begin{pmatrix} 9 & -5 \\ -7 & 4 \end{pmatrix} .
9.

If A=(4−2−32) A = \begin{pmatrix} 4 & -2 \\ -3 & 2 \end{pmatrix} and B=(−11−54) B = \begin{pmatrix} -1 & 1 \\ -5 & 4 \end{pmatrix} :

  • a.Define an inverse matrix.
  • b.Is ((A+B)T)T ((A + B)^{\mathrm T})^{\mathrm T} equal to A + B? Show it by calculation.
  • c.Find the determinant of BA.
Answer
  • a.If AB = BA = I for square matrices A and B, each is the inverse of the other.
  • b.A+B=(3−1−86) A + B = \begin{pmatrix} 3 & -1 \\ -8 & 6 \end{pmatrix} , (A+B)T=(3−8−16) (A + B)^{\mathrm T} = \begin{pmatrix} 3 & -8 \\ -1 & 6 \end{pmatrix} , and ((A+B)T)T=(3−1−86) ((A + B)^{\mathrm T})^{\mathrm T} = \begin{pmatrix} 3 & -1 \\ -8 & 6 \end{pmatrix} . Yes, equal to A + B.
  • c.BA=(−4−32+2−20−1210+8)=(−74−3218) BA = \begin{pmatrix} -4 - 3 & 2 + 2 \\ -20 - 12 & 10 + 8 \end{pmatrix} = \begin{pmatrix} -7 & 4 \\ -32 & 18 \end{pmatrix} , |BA| = −126 + 128 = 2. (Check: |B||A| = 1 × 2 = 2.)