3.1E: The Cofactor Expansion Exercises
( \newcommand{\kernel}{\mathrm{null}\,}\)
Exercises for 1
solutions
2
Compute the determinants of the following matrices.
[2−132] [69812] [a2ababb2] [a+1aaa−1] [cosθ−sinθsinθcosθ] [20−3125030] [123456789] [0a0bcd0e0] [1bcbc1c1b] [0aba0cbc0] [01−10300201215007] [10312260−10−3141120] [31−5213011052112−1] [4−13−13102012212−11] [1−1553124−1−380112−1] [000a00bp0cqkdstu]
- 0
- −1
- −39
- 0
- 2abc
- 0
- −56
- 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.
[1−123112−13] [−1312531−21] [−1−1102113011213−12] [231102−1305111125]
- −17
- 106
Evaluate by cursory inspection:
- det[abca+1b+1c+1a−1b−1c−1]
- det[abca+b2bc+b222]
- 0
If det[abcpqrxyz]=−1 compute:
- det[−x−y−z3p+a3q+b3r+c2p2q2r]
- det[−2a−2b−2c2p+x2q+y2r+z3x3y3z]
- 12
Show that:
- det[p+xq+yr+za+xb+yc+za+pb+qc+r]=2det[abcpqrxyz]
- det[2a+p2b+q2c+r2p+x2q+y2r+z2x+a2y+b2z+c]=9det[abcpqrxyz]
-
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+zp−aq−br−cx−py−qz−r]
=3det[3x3y3zp−aq−br−cx−py−qz−r]⋯
In each case either prove the statement or give an example showing that it is false:
- det(A+B)=detA+detB.
- If detA=0, then A has two equal rows.
- If A is 2×2, then det(AT)=detA.
- If R is the reduced row-echelon form of A, then detA=detR.
- If A is 2×2, then det(7A)=49detA.
- det(AT)=−detA.
- det(−A)=−detA.
- If detA=detB where A and B are the same size, then A = B.
- False. A=[1122]
- False. A=[2001]→R=[1001]
- False. A=[1101]
- False. A=[1101] and B=[1011]
Compute the determinant of each matrix, using Theorem [thm:007890].
- [1−120−201041115000003−100011]
- [12030−131400021100−10200301]
- 35
If detA=2,detB=−1, and detC=3, find:
det[AXY0BZ00C] det[A00XB0YZC] det[AXY0B00ZC] det[AX00B0YZC]
- −6
- −6
If A has three columns with only the top two entries nonzero, show that detA=0.
- Find detA if A is 3×3 and det(2A)=6.
- Under what conditions is det(−A)=detA?
Evaluate by first adding all other rows to the first row.
- det[x−1232−3x−2−2x−2]
- det[x−1−312−1x−1−3x+2−2]
- −(x−2)(x2+2x−12)
- Find b if det[5−1x26y−54z]=ax+by+cz.
- Find c if det[2x−11y3−3z4]=ax+by+cz.
- −7
Find the real numbers x and y such that detA=0 if:
A=[0xyy0xxy0] A=[1xx−x−2x−x−x−3] A=[1xx2x3xx2x31x2x31xx31xx2] A=[xy000xy000xyy00x]
- ±√62
- x=±y
Show that
det[011110xx1x0x1xx0]=−3x2
Show that
det[1xx2x3a1xx2pb1xqrc1]=(1−ax)(1−bx)(1−cx).
[ex:3.1.19] Given the polynomial p(x)=a+bx+cx2+dx3+x4, the matrix C=[010000100001−a−b−c−d] is called the companion matrix of p(x). Show that det(xI−C)=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=[x1x2⋮xn], y=[y1y2⋮yn] and A=[c1⋯x+y⋯cn] where x+y is in column j. Expanding detA along column j (the one containing x+y):
T(x+y)=detA=n∑i=1(xi+yi)cij(A)=n∑i=1xicij(A)+n∑i=1yicij(A)=T(x)+T(y)
Similarly for T(ax)=aT(x).
Show that
det[00⋯0a100⋯a2∗⋮⋮⋮⋮0an−1⋯∗∗an∗⋯∗∗]=(−1)ka1a2⋯an
where either n=2k or n=2k+1, and ∗-entries are arbitrary.
By expanding along the first column, show that:
det[1100⋯000110⋯000011⋯00⋮⋮⋮⋮⋮⋮0000⋯111000⋯01]=1+(−1)n+1
if the matrix is n×n,n≥2.
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)=2−3x+5x2, then
p(A)=2I−3A+5A2. The characteristic polynomial of A is defined to be cA(x)=det[xI−A], and the Cayley-Hamilton theorem asserts that cA(A)=0 for any matrix A.
- 2
- A=[321−1]
- A=[1−11010822]
- Prove the theorem for A=[abcd]