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Mathematics LibreTexts

3.1E: The Cofactor Expansion Exercises

( \newcommand{\kernel}{\mathrm{null}\,}\)

Exercises for 1

solutions

2

Compute the determinants of the following matrices.

[2132] [69812] [a2ababb2] [a+1aaa1] [cosθsinθsinθcosθ] [203125030] [123456789] [0a0bcd0e0] [1bcbc1c1b] [0aba0cbc0] [0110300201215007] [10312260103141120] [3152130110521121] [4131310201221211] [1155312413801121] [000a00bp0cqkdstu]

  1. 0
  2. 1
  3. 39
  4. 0
  5. 2abc
  6. 0
  7. 56
  8. abcd

Show that detA=0 if A has a row or column consisting of zeros.

Show that the sign of the position in the last row and the last column of A is always +1.

Show that detI=1 for any identity matrix I.

Evaluate the determinant of each matrix by reducing it to upper triangular form.

[112311213] [131253121] [1110211301121312] [2311021305111125]

  1. 17
  2. 106

Evaluate by cursory inspection:

  1. det[abca+1b+1c+1a1b1c1]
  2. det[abca+b2bc+b222]
  1. 0

If det[abcpqrxyz]=1 compute:

  1. det[xyz3p+a3q+b3r+c2p2q2r]
  2. det[2a2b2c2p+x2q+y2r+z3x3y3z]
  1. 12

Show that:

  1. det[p+xq+yr+za+xb+yc+za+pb+qc+r]=2det[abcpqrxyz]
  2. det[2a+p2b+q2c+r2p+x2q+y2r+z2x+a2y+b2z+c]=9det[abcpqrxyz]
  1. det[2a+p2b+q2c+r2p+x2q+y2r+z2x+a2y+b2z+c]
    =3det[a+p+xb+q+yc+r+z2p+x2q+y2r+z2x+a2y+b2z+c]
    =3det[a+p+xb+q+yc+r+zpaqbrcxpyqzr]
    =3det[3x3y3zpaqbrcxpyqzr]

In each case either prove the statement or give an example showing that it is false:

  1. det(A+B)=detA+detB.
  2. If detA=0, then A has two equal rows.
  3. If A is 2×2, then det(AT)=detA.
  4. If R is the reduced row-echelon form of A, then detA=detR.
  5. If A is 2×2, then det(7A)=49detA.
  6. det(AT)=detA.
  7. det(A)=detA.
  8. If detA=detB where A and B are the same size, then A = B.
  1. False. A=[1122]
  2. False. A=[2001]R=[1001]
  3. False. A=[1101]
  4. False. A=[1101] and B=[1011]

Compute the determinant of each matrix, using Theorem [thm:007890].

  1. [1120201041115000003100011]
  2. [1203013140002110010200301]
  1. 35

If detA=2,detB=1, and detC=3, find:

det[AXY0BZ00C] det[A00XB0YZC] det[AXY0B00ZC] det[AX00B0YZC]

  1. 6
  2. 6

If A has three columns with only the top two entries nonzero, show that detA=0.

  1. Find detA if A is 3×3 and det(2A)=6.
  2. Under what conditions is det(A)=detA?

Evaluate by first adding all other rows to the first row.

  1. det[x12323x22x2]
  2. det[x13121x13x+22]
  1. (x2)(x2+2x12)
  1. Find b if det[51x26y54z]=ax+by+cz.
  2. Find c if det[2x11y33z4]=ax+by+cz.
  1. 7

Find the real numbers x and y such that detA=0 if:

A=[0xyy0xxy0] A=[1xxx2xxx3] A=[1xx2x3xx2x31x2x31xx31xx2] A=[xy000xy000xyy00x]

  1. ±62
  2. x=±y

Show that
det[011110xx1x0x1xx0]=3x2

Show that
det[1xx2x3a1xx2pb1xqrc1]=(1ax)(1bx)(1cx).

[ex:3.1.19] Given the polynomial p(x)=a+bx+cx2+dx3+x4, the matrix C=[010000100001abcd] is called the companion matrix of p(x). Show that det(xIC)=p(x).

Show that
det[a+xb+xc+xb+xc+xa+xc+xa+xb+x]=(a+b+c+3x)[(ab+ac+bc)(a2+b2+c2)]

[ex:3_1_21]. Prove Theorem [thm:007914]. [Hint: Expand the determinant along column j.]

Let x=[x1x2xn], y=[y1y2yn] and A=[c1x+ycn] where x+y is in column j. Expanding detA along column j (the one containing x+y):

T(x+y)=detA=ni=1(xi+yi)cij(A)=ni=1xicij(A)+ni=1yicij(A)=T(x)+T(y)

Similarly for T(ax)=aT(x).

Show that

det[000a100a20an1an]=(1)ka1a2an

where either n=2k or n=2k+1, and -entries are arbitrary.

By expanding along the first column, show that:

det[110000011000001100000011100001]=1+(1)n+1

if the matrix is n×n,n2.

Form matrix B from a matrix A by writing the columns of A in reverse order. Express detB in terms of detA.

If A is n×n, then detB=(1)kdetA where n=2k or n=2k+1.

Prove property 3 of Theorem [thm:007779] by expanding along the row (or column) in question.

Show that the line through two distinct points (x1,y1) and (x2,y2) in the plane has equation

det[xy1x1y11x2y21]=0

Let A be an n×n matrix. Given a polynomial p(x)=a0+a1x++amxm, we write
p(A)=a0I+a1A++amAm.

For example, if p(x)=23x+5x2, then
p(A)=2I3A+5A2. The characteristic polynomial of A is defined to be cA(x)=det[xIA], and the Cayley-Hamilton theorem asserts that cA(A)=0 for any matrix A.

  1. 2
    1. A=[3211]
    2. A=[111010822]
  2. Prove the theorem for A=[abcd]

3.1E: The Cofactor Expansion Exercises is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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