A.E: Linear Algebra (Exercises)
( \newcommand{\kernel}{\mathrm{null}\,}\)
A.1: Vectors, Mappings, and Matrices
On a piece of graph paper draw the vectors:
- [25]
- [−2−4]
- (3,−4)
On a piece of graph paper draw the vector (1,2) starting at (based at) the given point:
- based at (0,0)
- based at (1,2)
- based at (0,−1)
On a piece of graph paper draw the following operations. Draw and label the vectors involved in the operations as well as the result:
- [1−4]+[23]
- [−32]−[13]
- 3[21]
Compute the magnitude of
- [72]
- [−231]
- (1,3,−4)
Compute
- [23]+[7−8]
- [−23]−[6−4]
- −[−32]
- 4[−15]
- 5[10]+9[01]
- 3[1−8]−2[3−1]
Find the unit vector in the direction of the given vector
- [1−3]
- [21−1]
- (3,1,−2)
If →x=(1,2) and →y are added together, we find →x+→y=(0,2). What is →y?
Write (1,2,3) as a linear combination of the standard basis vectors →e1, →e2, and →e3.
If the magnitude of →x is 4, what is the magnitude of
- 0→x
- 3→x
- −→x
- −4→x
- →x+→x
- →x−→x
Suppose a linear mapping F:R2→R2 takes (1,0) to (2,−1) and it takes (0,1) to (3,3). Where does it take
- (1,1)
- (2,0)
- (2,−1)
Suppose a linear mapping F:R3→R2 takes (1,0,0) to (2,1), it takes (0,1,0) to (3,4), and it takes (0,0,1) to (5,6). Write down the matrix representing the mapping F.
Suppose that a mapping F:R2→R2 takes (1,0) to (1,2), (0,1) to (3,4), and (1,1) to (0,−1). Explain why F is not linear.
Let R3 represent the space of quadratic polynomials in t: a point (a0,a1,a2) in R3 represents the polynomial a0+a1t+a2t2. Consider the derivative ddt as a mapping of R3 to R3, and note that ddt is linear. Write down ddt as a 3×3 matrix.
Compute the magnitude of
- [13]
- [23−1]
- (−2,1,−2)
- Answer
-
- √10
- √14
- 3
Find the unit vector in the direction of the given vector
- [−11]
- [1−12]
- (2,−5,2)
- Answer
-
- [−1√21√2]
- [1√6−1√62√6]
- (2√33,−5√33,2√33)
Compute
- [31]+[6−3]
- [−12]−[2−1]
- −[−53]
- 2[−24]
- 3[10]+7[01]
- 2[2−3]−6[2−1]
- Answer
-
- [9−2]
- [−33]
- [5−3]
- [−48]
- [37]
- [−83]
If the magnitude of →x is 5, what is the magnitude of
- 4→x
- −2→x
- −4→x
- Answer
-
- 20
- 10
- 20
Suppose a linear mapping F:R2→R2 takes (1,0) to (1,−1) and it takes (0,1) to (2,0). Where does it take
- (1,1)
- (0,2)
- (1,−1)
- Answer
-
- (3,−1)
- (4,0)
- (−1,−1)
A.2: Matrix Algebra
Add the following matrices
- [−12258−1]+[323835]
- [124231051]+[2−8−33106−41]
Compute
- 3[03−22]+6[15−15]
- 2[−3122]−3[2−132]
Multiply the following matrices
- [−123158][3−131832−3]
- [123311103][2317123−11−130]
- [416356504660][25123556]
- [114051][221064]
Compute the inverse of the given matrices
- [−3]
- [0−110]
- [1413]
- [2214]
Compute the inverse of the given matrices
- [−2001]
- [3000−20001]
- [10000−100000.010000−5]
Add the following matrices
- [21011−1]+[534125]
- [6−237338−12]+[−1−1−3673−94−1]
- Answer
-
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Compute
- 2[1234]+3[−1312]
- 3[2−113]−2[21−12]
- Answer
-
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Multiply the following matrices
- [214344][246335]
- [0332−2135−2][662460204]
- [3412−104−15][025020523616]
- [−2−25321][0313]
- Answer
-
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Compute the inverse of the given matrices
- [2]
- [0110]
- [1235]
- [4244]
- Answer
-
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Compute the inverse of the given matrices
- [2003]
- [40005000−1]
- [−1000020000300000.1]
- Answer
-
Add texts here. Do not delete this text first.
A.3: Elimination
Compute the reduced row echelon form for the following matrices:
- [131011]
- [336−3]
- [36−2−3]
- [66771101]
- [930286367979]
- [213−3600−1−2443]
- [6650−22656]
- [020−166−3362−35]
Compute the inverse of the given matrices
- [100001010]
- [111021001]
- [123201021]
Solve (find all solutions), or show no solution exists
- 4x1+3x2=−2−x1+3x2=4
- x1+5x2+3x3=78x1+7x2+8x3=84x1+8x2+6x3=4
- 4x1+8x2+2x3=3−x1−2x2+3x3=14x1+8x2+3x3=2
- x+2y+3z=42x−2y+3z=13x+2y+6z=6
By computing the inverse, solve the following systems for →x.
- [41−13]→x=[1326]
- [3334]→x=[2−1]
Compute the rank of the given matrices
- [635141776]
- [5−2−1306245]
- [123−1−2−3246]
For the matrices in Exercise A.E.A.3.5, find a linearly independent set of row vectors that span the row space (they don’t need to be rows of the matrix).
For the matrices in Exercise A.E.A.3.5, find a linearly independent set of columns that span the column space. That is, find the pivot columns of the matrices.
Find a linearly independent subset of the following vectors that has the same span. [−112],[2−2−4],[−241],[−13−2]
Compute the reduced row echelon form for the following matrices:
- [101010]
- [1234]
- [11−2−2]
- [1−3146−2−26−2]
- [22521−24−1031−2]
- [−264360−3042−11]
- [00000000]
- [12331235]
- Answer
-
- [101010]
- [1001]
- [1100]
- [10001−13000]
- [1007715010−215001−85]
- [10−1200112120000]
- [00000000]
- [12300001]
Compute the inverse of the given matrices
- [010−100001]
- [111110100]
- [240223241]
- Answer
-
- [0−10100001]
- [00101−11−10]
- [521−3−1−1232−101]
Solve (find all solutions), or show no solution exists
- 4x1+3x2=−15x1+6x2=4
- 5x+6y+5z=76x+8y+6z=−15x+2y+5z=2
- a+5b+6c=−1a+5b+6c=−1−2a+5b+6c=8
- −2x1+2x2+8x3=6x2+8x3=2x1+4x2+8x3=7
- Answer
-
- x1=−2,x2=73
- no solution
- a=−3,b=10,c=−8
- x3 is free, x1=−1+3x3, x2=2−x3
By computing the inverse, solve the following systems for →x.
- [−1133]→x=[46]
- [2716]→x=[13]
- Answer
-
- [−13]
- [−31]
Compute the rank of the given matrices
- [7−16777762]
- [111111222]
- [03−163147−1]
- Answer
-
- 3
- 1
- 2
For the matrices in Exercise A.E.A.3.13, find a linearly independent set of row vectors that span the row space (they don’t need to be rows of the matrix).
- Answer
-
- [100], [010], [001]
- [111]
- [1013], [01−13]
For the matrices in Exercise A.E.A.3.13, find a linearly independent set of columns that span the column space. That is, find the pivot columns of the matrices.
- Answer
-
- [777], [−176], [762]
- [112]
- [064], [337]
Find a linearly independent subset of the following vectors that has the same span. [000],[31−5],[03−1],[−324]
- Answer
-
[31−5], [03−1]
A.4: Subspaces, Dimension, and The Kernel
For the following sets of vectors, find a basis for the subspace spanned by the vectors, and find the dimension of the subspace.
- [111],[−1−1−1]
- [105],[010],[0−10]
- [−4−35],[233],[202]
- [130],[022],[−1−12]
- [13],[02],[−1−1]
- [313],[24−4],[−5−5−2]
For the following matrices, find a basis for the kernel (nullspace).
- [11111511−4]
- [2−1−340−4−112]
- [−444−111−555]
- [−2111−42221043]
Suppose a 5×5 matrix A has rank 3. What is the nullity?
Suppose that X is the set of all the vectors of R3 whose third component is zero. Is X a subspace? And if so, find a basis and the dimension.
Consider a square matrix A, and suppose that →x is a nonzero vector such that A→x=→0. What does the Fredholm alternative say about invertibility of A.
Consider M=[1232??−1??]. If the nullity of this matrix is 2, fill in the question marks. Hint: What is the rank?
For the following sets of vectors, find a basis for the subspace spanned by the vectors, and find the dimension of the subspace.
- [12],[11]
- [111],[222],[112]
- [531],[5−15],[−13−4]
- [224],[223],[44−3]
- [10],[20],[30]
- [100],[200],[012]
- Answer
-
- [12], [11] dimension 2,
- [111], [112] dimension 2,
- [531], [5−15], [−13−4] dimension 3,
- [224], [223] dimension 2,
- [11] dimension 1,
- [100], [012] dimension 2
For the following matrices, find a basis for the kernel (nullspace).
- [261913293909]
- [2−2−5−115−55−3]
- [1−5−4235−352]
- [044011055]
- Answer
-
- [3−100], [303−1]
- [−1−10]
- [11−1]
- [−100], [01−1]
Suppose the column space of a 9×5 matrix A of dimension 3. Find
- Rank of A.
- Nullity of A.
- Dimension of the row space of A.
- Dimension of the nullspace of A.
- Size of the maximum subset of linearly independent rows of A.
- Answer
-
- 3
- 2
- 3
- 2
- 3
A.5: Inner Product and Projections
Find the s that makes the following vectors orthogonal: (1,2,3), (1,1,s).
Find the angle θ between (1,3,1), (2,1,−1).
Given that ⟨→v,→w⟩=3 and ⟨→v,→u⟩=−1 compute
- ⟨→u,2→v⟩
- ⟨→v,2→w+3→u⟩
- ⟨→w+3→u,→v⟩
Suppose →v=(1,1,−1). Find
- proj→v((1,0,0))
- proj→v((1,2,3))
- proj→v((1,−1,0))
Consider the vectors (1,2,3), (−3,0,1), (1,−5,3).
- Check that the vectors are linearly independent and so form a basis.
- Check that the vectors are mutually orthogonal, and are therefore an orthogonal basis.
- Represent (1,1,1) as a linear combination of this basis.
- Make the basis orthonormal.
Let S be the subspace spanned by (1,3,−1), (1,1,1). Find an orthogonal basis of S by the Gram-Schmidt process.
Starting with (1,2,3), (1,1,1), (2,2,0), follow the Gram-Schmidt process to find an orthogonal basis of R3.
Find an orthogonal basis of R3 such that (3,1,−2) is one of the vectors. Hint: First find two extra vectors to make a linearly independent set.
Using cosines and sines of θ, find a unit vector →u in R2 that makes angle θ with →ı=(1,0). What is ⟨→ı,→u⟩?
Find the s that makes the following vectors orthogonal: (1,1,1), (1,s,1).
- Answer
-
s=−2
Find the angle θ between (1,2,3), (1,1,1).
- Answer
-
θ≈0.3876
Given that ⟨→v,→w⟩=1 and ⟨→v,→u⟩=−1 and ‖ and
- \langle 3 \vec{u} , 5 \vec{v} \rangle
- \langle \vec{v} , 2 \vec{w} + 3 \vec{u} \rangle
- \langle \vec{w} + 3 \vec{v}, \vec{v} \rangle
- Answer
-
- -15
- -1
- 28
Suppose \vec{v} = (1,0,-1). Find
- \operatorname{proj}_{\vec{v}}\bigl( (0,2,1) \bigr)
- \operatorname{proj}_{\vec{v}}\bigl( (1,0,1) \bigr)
- \operatorname{proj}_{\vec{v}}\bigl( (4,-1,0) \bigr)
- Answer
-
- \left(-\frac{1}{2},0,\frac{1}{2}\right)
- (0,0,0)
- (2,0,-2)
The vectors (1,1,-1), (2,-1,1), (1,-5,3) form an orthogonal basis. Represent the following vectors in terms of this basis:
- (1,-8,4)
- (5,-7,5)
- (0,-6,2)
- Answer
-
- (1,1,-1)-(2,-1,1)+2(1,-5,3)
- 2(2,-1,1)+(1,-5,3)
- 2(1,1,-1)-2(2,-1,1)+2(1,-5,3)
Let S be the subspace spanned by (2,-1,1), (2,2,2). Find an orthogonal basis of S by the Gram-Schmidt process.
- Answer
-
(2,-1,1), \left(\frac{2}{3},\frac{8}{3},\frac{4}{3}\right)
Starting with (1,1,-1), (2,3,-1), (1,-1,1), follow the Gram-Schmidt process to find an orthogonal basis of {\mathbb{R}}^3.
- Answer
-
(1,1,-1), (0,1,1), \left(\frac{4}{3},-\frac{2}{3},\frac{2}{3}\right)
A.6: Determinant
Compute the determinant of the following matrices:
- \begin{bmatrix} 3 \end{bmatrix}
- \begin{bmatrix} 1 & 3 \\ 2 & 1 \end{bmatrix}
- \begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix}
- \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix}
- \begin{bmatrix} 2 & 1 & 0 \\ -2 & 7 & -3 \\ 0 & 2 & 0 \end{bmatrix}
- \begin{bmatrix} 2 & 1 & 3 \\ 8 & 6 & 3 \\ 7 & 9 & 7 \end{bmatrix}
- \begin{bmatrix} 0 & 2 & 5 & 7 \\ 0 & 0 & 2 & -3 \\ 3 & 4 & 5 & 7 \\ 0 & 0 & 2 & 4 \end{bmatrix}
- \begin{bmatrix} 0 & 1 & 2 & 0 \\ 1 & 1 & -1 & 2 \\ 1 & 1 & 2 & 1 \\ 2 & -1 & -2 & 3 \end{bmatrix}
For which x are the following matrices singular (not invertible).
- \begin{bmatrix} 2 & 3 \\ 2 & x \end{bmatrix}
- \begin{bmatrix} 2 & x \\ 1 & 2 \end{bmatrix}
- \begin{bmatrix} x & 1 \\ 4 & x \end{bmatrix}
- \begin{bmatrix} x & 0 & 1 \\ 1 & 4 & 2 \\ 1 & 6 & 2 \end{bmatrix}
Compute \det \left( \begin{bmatrix} 2 & 1 & 2 & 3 \\ 0 & 8 & 6 & 5 \\ 0 & 0 & 3 & 9 \\ 0 & 0 & 0 & 1 \end{bmatrix}^{-1} \right) \nonumber without computing the inverse.
Suppose L = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 2 & 1 & 0 & 0 \\ 7 & \pi & 1 & 0 \\ 2^8 & 5 & -99 & 1 \end{bmatrix} \qquad \text{and} \qquad U = \begin{bmatrix} 5 & 9 & 1 & -\sin(1) \\ 0 & 1 & 88 & -1 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 1 \end{bmatrix} . \nonumber Let A = LU. Compute \det(A) in a simple way, without computing what is A. Hint: First read off \det(L) and \det(U).
Consider the linear mapping from {\mathbb R}^2 to {\mathbb R}^2 given by the matrix A = \left[ \begin{smallmatrix} 1 & x \\ 2 & 1 \end{smallmatrix} \right] for some number x. You wish to make A such that it doubles the area of every geometric figure. What are the possibilities for x (there are two answers).
Suppose A and S are n \times n matrices, and S is invertible. Suppose that \det(A) = 3. Compute \det(S^{-1}AS) and \det(SAS^{-1}). Justify your answer using the theorems in this section.
Let A be an n \times n matrix such that \det(A)=1. Compute \det(x A) given a number x. Hint: First try computing \det(xI), then note that xA = (xI)A.
Compute the determinant of the following matrices:
- \begin{bmatrix} -2 \end{bmatrix}
- \begin{bmatrix} 2 & -2 \\ 1 & 3 \end{bmatrix}
- \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix}
- \begin{bmatrix} 2 & 9 & -11 \\ 0 & -1 & 5 \\ 0 & 0 & 3 \end{bmatrix}
- \begin{bmatrix} 2 & 1 & 0 \\ -2 & 7 & 3 \\ 1 & 1 & 0 \end{bmatrix}
- \begin{bmatrix} 5 & 1 & 3 \\ 4 & 1 & 1 \\ 4 & 5 & 1 \end{bmatrix}
- \begin{bmatrix} 3 & 2 & 5 & 7 \\ 0 & 0 & 2 & 0 \\ 0 & 4 & 5 & 0 \\ 2 & 1 & 2 & 4 \end{bmatrix}
- \begin{bmatrix} 0 & 2 & 1 & 0 \\ 1 & 2 & -3 & 4 \\ 5 & 6 & -7 & 8 \\ 1 & 2 & 3 & -2 \end{bmatrix}
- Answer
-
- -2
- 8
- 0
- -6
- -3
- 28
- 16
- -24
For which x are the following matrices singular (not invertible).
- \begin{bmatrix} 1 & 3 \\ 1 & x \end{bmatrix}
- \begin{bmatrix} 3 & x \\ 1 & 3 \end{bmatrix}
- \begin{bmatrix} x & 3 \\ 3 & x \end{bmatrix}
- \begin{bmatrix} x & 1 & 0 \\ 1 & 4 & 0 \\ 1 & 6 & 2 \end{bmatrix}
- Answer
-
- 3
- 9
- 3
- \frac{1}{4}
Compute \det \left( \begin{bmatrix} 3 & 4 & 7 & 12 \\ 0 & -1 & 9 & -8 \\ 0 & 0 & -2 & 4 \\ 0 & 0 & 0 & 2 \end{bmatrix}^{-1} \right) \nonumber without computing the inverse.
- Answer
-
12
Find all the x that make the matrix inverse \begin{bmatrix} 1 & 2 \\ 1 & x \end{bmatrix}^{-1} \nonumber have only integer entries (no fractions). Note that there are two answers.
- Answer
-
1 and 3