Problem 1. Determine whether the following statements are true or false. If false, find a counterex-ample:
(a) det(B−1AB) = det(A) (b) tr(AB) = tr(A)tr(B)
(c) (ABCD)T = DTCTBTAT (d) (ABCD)−1= D−1C−1B−1A−1
(e) rank(A + B) = rank(A) + rank(B) (f) (A−1+ B−1)T = (AT)−1+ (BT)−1
(a) True. Recall for square matrices C, D (of the same order), we have
det(CD) = det(C) det(D), (1)
while also
det(C−1) = 1/ det(C). (2)
So applying (1) and (2), we have
det(B−1AB) = det(B−1) det(A) det(B) = det(A).
(Be careful when using (1), as this is assuming both C and D are square matrices of the same order.) Another point worth making making out is since we are given det(A) is defined and B−1exists, then necessarily A and B are square matrices, while the fact B−1AB = B−1(AB) is defined, and in particular, AB is defined, then necessarily A has the same number of rows as B has columns, showing A and B have the same dimensions.
(b) False. You should try to use the simplest examples possible, so let I =
1 0 0 1
, the identity 2 × 2 matrix. Since the trace of a matrix is the sum of the diagonal entries, then we have tr(I) = 2. Since also I2= I, then we have tr(I2) = tr(I) = 2 but tr(I)tr(I) = 2 · 2 = 4, so A = B = I is a simple counterexample. This shows trace is not multiplicative, as the determinant was in (1); but recall trace is additive, in the sense that
tr(A + B) = tr(A) + tr(B). (3)
For reference, although this shows trace is not multiplicative, we do also have the result
tr(AB) = tr(BA) (4)
for any matrices A and B, which we explore in the next problem. (c) True. Recall the result
(AB)T = BTAT, (5)
while also the fact matrix multiplication is associative, that is,
A(BC) = (AB)C. (6)
(But matrix multiplication is not commutative, as letting, say, A = 1 1 0 0 and B = 0 0 0 1 shows AB 6= BA in general. This doesn’t mean that this never is true, as I commutes with every matrix.) So using (5) and (6) repeatedly, we have
(d) True. Similarly, recall the result
(AB)−1= B−1A−1. (7)
(Remember the proof was simple, since
inverses are unique when they exist, (8)
it’s enough to note
(AB)(B−1A−1) = A(BB−1)A−1= AIA−1= AA−1 = I,
showing B−1A−1 is an inverse for AB and hence is the only inverse for AB.) So you could either use (6) and (7) and go through the calculation as in (c), or you could use (6) and (8), and note
(ABCD)(D−1C−1B−1A−1) = (ABC)(D−1D)(C−1B−1A−1) = (ABC)I(C−1B−1A−1) = (ABC)(C−1B−1A−1) = (AB)(CC−1)(B−1A−1) = · · · = AA−1 = I.
(e) False. Recall rank is the maximum number of linearly independent column/row vectors for a matrix, or equivalently, the number of nonzero rows when putting a matrix in row echelon form. Also, for an n × n matrix A then we have
A−1 exists iff det A 6= 0 iff rank(A) = n iff 0 is not an eigenvalue for A, (9) where “iff” is short for “if and only if”. Also, recall the elementary row operations, and their effect on the determinant, as in
Adding multiples of one row to another row doesn’t change the determinant, (10) Switching any two rows changes the sign of the determinant, and (11) Multiplying one row by a constant λ then multiples the determinant by λ. (12) So letting I be the 2 × 2 identity matrix. Since det(I) = 1, we have rank(I) = 2 by (9), and using (12), we have det(2I) = 4 det(I) = 4, so also rank(2I) = 2 by (9). So letting A = B = I, we then have the counter example of rank(A + B) = rank(2I) = 2 while rank(A) + rank(B) = 4.
An even simpler counter example is taking A = I and B = −I, as then rank(A) = rank(B) = 2 while rank(A + B) = rank(0) = 0, for 0 the null matrix.
(f) True. Recall the result
(C + D)T = CT+ DT, (13)
while note also
(A−1)T = (AT)−1, (14)
which can be established using (4) and (8), since (A−1)TAT = (AA−1)T = IT = I, noting I is symmetric. It then follows
(A−1+ B−1)T = (A−1)T + (B−1)T = (AT)−1+ (BT)−1. Problem 2. Show for A ∈ Mm,n(R), B ∈ Mn,m(R), then
(a) tr(AB) = tr(BA)
(b) det(AB) = det(BA) if n = m
(c) AB and BA have the same eigenvalues. (Hint: Treat λ = 0 separately.)
(d) Conclude from that if n = m and there exists some nonsingular C ∈ Mn(R) such that A = CBC−1,
Recall
If A = (aij), B = (bij) such that A ∈ Mn,m(R), B ∈ Mm,p(R), then AB = (P m
k=1aikbkj) ∈ Mn,p(R). (15)
Note AB is a n × n matrix and BA a m × m matrix. (a) AB has entiresPm
k=1aikbkj and BA has entriesP n k=1bikakj and so tr(AB) = n X i=1 m X k=1 aikbki= m X i=1 n X k=1 bikaki= tr(BA). (b) By (1), we have
det AB = det A det B = det B det A = det BA.
Note this does not generally hold if n 6= m: the smaller of the two matrices AB and BA may have nonzero determinant while the larger always has zero determinant (since rank AB ≤ min{rank A, rank B}). For example, if A = 1 0 0 0 1 0 and B = 1 0 0 1 0 0 , then AB = 1 0 0 1 and BA = 1 0 0 0 1 0 0 0 0 , with det AB = 1 6= 0 = det BA.
(c) For the λ = 0 case, using (9), we have 0 is an eigenvalue for AB if and only if det(AB) = det(BA) = 0 if and only if 0 is an eigenvalue for BA.
Now for λ 6= 0, we have for some eigenvector x 6= 0 such that ABx = λx 6= 0, then necessarily Bx 6= 0 (as otherwise A(Bx) = 0), while then BA(Bx) = B(ABx) = λBx, showing Bx is an eigenvector for BA corresponding to λ, showing λ is an eigenvalue for BA. Switching the roles of A and B shows also every eigenvalue of BA is an eigenvalue of AB.
(d) Note if A and B are related by multiplying one on the left and right by a matrix and its inverse, then such matrices are called similar. We have det A = det(CBC−1) = det(C) det(B) det(C−1) = det B by (2), tr(A) = tr((CB)C−1) = tr(C−1(CB)) = tr((C−1C)B) = tr(B), while A = (CB)C−1 has the same eigenvalues as C−1(CB) = (C−1C)B = B.
Problem 3. Let A be a square matrix such that ATA = I. Show that det(A) = ±1.
Recall the result
det(AT) = det(A). (16)
So using (16) and (1), we have
1 = det(I) = det(ATA) = det(AT) det(A) = det(A)2.
Hence the real number det(A) is a solution for the equation x2 = 1 and hence a root of the polynomial x2− 1 = (x − 1)(x + 1). It follows then det(A) = ±1.
Recall a real vector space is closed under addition and scalar multiplication. Hence, if V were a vector space, then we would need to verify (I) if x, y ∈ V then x + y ∈ V and (II) if x ∈ V then cx ∈ V for all c ∈ R. Now we check if x1 y1, x2 y2 ∈ V then x1 y1 + x2 y2 = x1+x2 y1+y2 ∈ V since x1+ x2 > 0 and y1+ y2 > 0
whenever x1, y1, x2, y2> 0. This shows V is closed under addition.
Now we have to check if it is closed under scalar multiplication. If xy ∈ V, then is c x y =
cx
cy ∈ V for
all c ∈ V? This is equivalent to asking if x > 0 then is cx > 0 for all c ∈ R. The answer is no: take any c ≤ 0. In particular, we see 11 ∈ V but − 1
1 = −1
−1 6∈ V. (You only need one counter example to disprove
a statement.) So since V is not closed under scalar multiplication, then V is not a real vector space. NB : It helps to recall all subspaces of R2 are either the trivial space (only the origin), any line passing
through the origin, or R2
itself. This problem is only looking at the first quadrant in R2.
Problem 5. Find an expression for the cofactor of a24 and a42 in the matrix
1 7 7 1 7 7 2 8 6 2 8 6 3 9 5 3 9 5 4 10 4 4 10 4 5 9 3 5 9 3 6 8 2 6 8 2
Recall the cofactor for an element of a matrix aij in a matrix A is the term Cij = (−1)i+jMij with
Mij = det(Aij) for Aij the matrix you get from A by removing the ithrow and jthcolumn. Remember that
you use this information to define the cofactor matrix C = (Cij), with then
A−1 = 1
det(A)adj(A), (17)
assuming the inverse exists (viz., any condition of (9) is satisfied), where the adjoint of A, adj(A) is the transpose of the cofactor matrix, that is, adj(A) = CT.
It follows then the cofactor for a24= 2 is
C24= (−1)2+4det 1 7 7 7 7 3 9 5 9 5 4 10 4 10 4 5 9 3 9 3 6 8 2 8 2 = 0,
with the last equality following from the fact the 2nd and 4th or the 3rd and 5th columns are identical and
so the rank of A24is strictly less than 5, and hence the determinant is zero by (9).
Last, we see the cofactor for a42is
C42= (−1)4+2det 1 7 1 7 7 2 6 2 8 6 3 5 3 9 5 5 3 5 9 3 6 2 6 8 2 = 0,
now noting the 1st and 3rd or the 2nd and 5th columns are identical in A
42and hence has determinant 0 by
(9).
You could go through and calculate the remaining cofactors, and you’ll see actually Cij = 0 for all i, j,
Problem 6. Consider the matrices A, B, and C given by A = 1 −1 0 2 0 −1 B = 1 0 −2 2 0 0 1 3 C = 1 3 0 0 −1 3 −3 0 0 0 5 1 . Compute 4A − BC. We first calculate 4A = 4 −4 0 8 0 −4 and BC = 1 0 −2 2 0 0 1 3 1 3 0 0 −1 3 −3 0 0 0 5 1 = 7 13 2 −3 15 3 , which gives 4A − BC = 4 −4 0 8 0 −4 − 7 13 2 −3 15 3 = −3 −17 −2 11 −15 −7 .
Problem 7. For each of the following matrices, find its determinant, trace and rank. If its inverse exists, find that also.
(a) −3 6 4 8 (c) 2 1 4 3 2 5 0 1 −1 (e) 1 0 1 −1 0 −1 −3 4 1 0 −1 2 −3 0 0 −1 (b) 6 7 8 9 (d) 1 1 −1 2 −2 8 −3 1 −7 (f) 1 1 0 2 1 2 0 1 2 1 0 1 1 0 0 1
Let A be the corresponding matrix for each part of the problem.
(a) Now det(A) = −3 · 8 − 4 · 6 = −48, with tr(A) = −3 + 8 = 5. Since det(A) 6= 0, then we have rank(A) = 2, and hence A−1 exists, and is of the form
A−1= − 1 48 8 −6 −4 −3 = 1 48 −8 6 4 3 , using (17).
(b) We see det(A) = 6 · 9 − 7 · 8 = 54 − 56 = −2 and tr(A) = 6 + 9 = 15. Since det(A) 6= 0, then we have rank(A) = 2, and A−1 exists, with
(c) To calculate the determinant, use cofactor expansion with the last bottom row to get det(A) = − 2 4 3 5 − 2 1 3 2 = −(10 − 12) − (4 − 3) = 1,
while tr(A) = 2 + 2 − 1 = 3. Since det(A) 6= 0, then rank(A) = 3. Also, we can then calculate the inverse
A−1= 2 5 1 −1 − 3 5 0 −1 3 2 0 1 − 1 4 1 −1 2 4 0 −1 − 2 1 0 1 1 4 2 5 − 2 4 3 5 2 1 3 2 T = −7 3 3 5 −2 −2 −3 2 1 T = −7 5 −3 3 −2 2 3 −2 1 .
(d) Now using (10), by adding the first and second rows to the third, we see
det(A) = det 1 1 −1 2 −2 8 0 0 0 = 0,
since a cofactor expansion on the bottom row of the augmented matrix evaluates to 0 (or equivalently with any other row or column). Also, we have tr(A) = 1 − 2 − 7 = −8. Since det(A) = 0, then A−1does not exist. Now to calculate the rank, we note since det(A) = 0, then rank(A) ∈ {1, 2}, using also the fact rank(A) 6= 0 since A is not the null matrix. Since
rank(A) = 1 if and only if each column is a scalar multiple of one another, (18) we see rank(A) = 2 since the first column is not a scalar multiple of the second; that is, if α ∈ R satisfied
α 1 2 −3 = α 2α −3α = 1 −2 1 ,
then we have α = 1 and 2α = −2 and −3α = 1, which is impossible. Equivalently, you can put A is reduced row echelon form, and we see
1 1 −1 2 −2 8 −3 1 −7 r27→−14(r2−2r1) −−−−−−−−−−−→ r37→r3+r2+r1 1 1 −1 0 1 −5 2 0 0 0 r17→r1−r2 −−−−−−−→ 1 0 32 0 1 −5 2 0 0 0 ,
with then rank(A) = 2 since there are two nonzero rows in this augmented matrix. (e) Doing a cofactor expansion on the second column, we see
det(A) = − 1 1 −1 1 −1 2 −3 0 −1 = − − 1 2 −3 −1 − 1 −1 −3 −1 = (−1 + 6) + (−1 − 3) = 1,
using a cofactor expansion on the middle column of the 3×3 matrix A22. Also, we have tr(A) = 1−1−1−1 =
−2, with rank(A) = 4 using (9). Now to calculate A−1, I’ll first go through the version that uses (17), before
Now this seems like it’ll be a lot of calculations, but that second column makes a lot of the cofactors pretty easy to evaluate. So we see
A−1 = −1 −3 4 0 −1 2 0 0 −1 − 0 −3 4 1 −1 2 −3 0 −1 0 −1 4 1 0 2 −3 0 −1 − 0 −1 −3 1 0 −1 −3 0 0 − 0 1 −1 0 −1 2 0 0 −1 1 1 −1 1 −1 2 −3 0 −1 − 1 0 −1 1 0 2 −3 0 −1 1 0 1 1 0 −1 −3 0 −1 0 1 −1 −1 −3 4 0 0 −1 − 1 1 −1 0 −3 4 −3 0 −1 1 0 −1 0 −1 4 −3 0 −1 − 1 0 1 0 −1 −3 −3 0 0 − 0 1 −1 −1 −3 4 0 −1 2 1 1 −1 0 −3 4 1 −1 2 − 1 0 −1 0 −1 4 1 0 2 1 0 1 0 −1 −3 1 0 −1 T = −1 −3 5 3 0 −1 0 0 −1 0 4 3 −1 −1 3 2 T = −1 0 −1 −1 −3 −1 0 −1 5 0 4 3 3 0 3 2 .
As I mentioned, I’ll also walk through a different method of finding A−1 as follows. We start with the augmented n × 2n matrix, where we append the identity matrix on the right, and then going through elementary row operations (i.e., (10), (11) and (12)), we reduce the original matrix A to the identity, which will result in A−1 on the right half. So, we can see this as below:
where you see we get the same result for A−1 as above.
(f) Because the third column is all 0’s, we know det(A) = 0, while tr(A) = 1 + 2 + 0 + 1 = 3. Since det(A) = 0 and A 6= 0, then we know A−1 doesn’t exist and rank(A) ∈ {1, 2, 3}. One method to calculate the rank of A is to find the Reduced Row Echelon form of A, as elementary operators preserve the rank of a matrix. We find this as follows
A = 1 1 0 2 1 2 0 1 2 1 0 1 1 0 0 1 r17→ r1− r4 r27→ r2− r4 r37→ r3− r4 0 1 0 1 0 2 0 0 0 1 0 −1 1 0 0 1 r1↔ r4 r27→ 12r2 r37→ −r3 1 0 0 1 0 1 0 0 0 −1 0 1 0 1 0 1 r37→ r3+ r2 r47→ r4− r2 1 0 0 1 0 1 0 0 0 0 0 1 0 0 0 1 r17→ r1− r3 r47→ r4− r3 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 = B
Here B is the reduced row echelon form, which clearly has 3 linearly independent rows (or columns). Here’s another method one can use to calculate rank, as long as you pick another column vector not in the span of the other three column vectors. I will create a new matrix where I replace the third column with the column ( 1 0 0 0 )T, and since the rank is unchanged if we switch the order of the matrix, then
rank(A) = rank(B) − 1 for any matrix B we make from the left over column vectors. Now let
B = 1 1 1 2 0 1 2 1 0 2 1 1 0 1 0 1 , and we calculate det(B) = 1 2 1 2 1 1 1 0 1 = 2 1 1 1 + 1 2 2 1 = (2 − 1) + (1 − 4) = −2,
where I first did cofactor expansion with the first column in B and then the last row in B11. By (9), we
then have rank(B) = 4, and hence rank(A) = 3 since then necessarily the three nonzero column vectors are linearly independent since the four column vectors of B are linearly independent. (This method wouldn’t have helped much if the three non-zero column vectors were linearly dependent, since then we would’ve still had the determinant of B be zero — so this method works as you remove the column vectors that are obviously linearly dependent, and you keep on reducing by one and replacing by a vector you can confirm is not in the span of the remaining vectors.)
Another method is to just show for α, β, γ ∈ R such that 0 0 0 0 = α 1 1 2 1 + β 1 2 1 0 + γ 2 1 1 1 = α + β + 2γ α + 2β + γ 2α + β + γ α + γ ,
Problem 8. Solve the following systems of linear equations: (a) 3x − y + z = −7 y + z = 6 x + y − 3z = −13 (b) 3x + 2y + z = 10 2x + y − z = 1 x + y + 2z = 9
(a) I will first use Cramer’s Rule. So we can convert this system into a matrix equation of the form Ax = b, with A = 3 −1 1 0 1 1 1 1 −3 x = x y z b = −7 6 −13 . First we calculate det(A), as follows:
det(A) = det 0 0 14 0 1 1 1 1 −3 = − det 1 1 −3 0 1 1 0 0 14 = −14,
where I used the elementary row operations r1 7→ r1− 3r3+ 4r2 for the first matrix, and then I switched
the first and third rows for the second matrix, using (10) and (11), which only changed the sign of the determinant I started with, while I also used the result
the determinant of an upper or lower triangular matrix is the product of the diagonal entries. (19) Next, we determine the matrices A1, A2, A3, formed by replacing the respective column in A with b, and
then calculate the determinants, using (10), (11) and (12) freely throughout,
det(A1) = det −7 −1 1 6 1 1 −13 1 −3 = −3 · det 1 0 −2 6 1 1 0 1 −1 = −3 · det 1 0 −2 0 1 13 0 0 −14 = 42 det(A2) = det 3 −7 1 0 6 1 1 −13 −3 = −2 · det 1 −13 −3 0 6 1 0 16 5 = −2 6 1 16 5 = −28 det(A3) = det 3 −1 −7 0 1 6 1 1 −13 = 4 · det 1 1 −13 0 1 6 0 1 −8 = 4 · det 1 1 −13 0 1 6 0 0 −14 = −56 It follows then, using Cramer’s Rule, we have the solutions
x = det(A1) det(A) = 42 −14 = −3 y = det(A2) det(A) = −28 −14 = 2 z = det(A3) det(A) = −56 −14 = 4
(b) Similarly, we reframe the system of linear equations as a matrix equation Ax = b, with now
But now note the first row is equal to the sum of the second and third rows, and so det(A) = 0 by (9). Since Cramer’s Rule only works if A is nonsingular, then we need a different approach. Note first the original system is indeed consistent, since you can see that actually the first equation is equal to the sum of the second and third equations. Remember
A system of linear equations has either 0, 1, or ∞ number of solutions. (20) A system with 0 solutions is an inconsistent system (say, if the last equation was anything other than 9, then the system would have been inconsistent). But since our system is consistent, and it has linearly dependent equations, then our system is equivalent to a consistent underdetermined system, which then must have ∞ solutions.
Set z = t, and now reframe the original system of linear equations as 3x + 2y = 10 − t
2x + y = 1 + t
where I ignore the last equation as it is linearly dependent with the prior two (while the first two equations are obviously linearly independent as they are not a scalar multiple of one another), and now I could use Cramer’s Rule for Ax = b with now
A = 3 2 2 1 x = x y b = 10 − t 1 + t ,
although with only two equations, using a straight forward substitution method works just as well. Solving the second equation for y, we get y = 1 + t − 2x, and so plugging this into the first equation we have
3x + 2(1 + t − 2x) = −x + 2 + 2t = 10 − t
and so x = 3t − 8, with then y = 1 + t − 2(3t − 8) = 17 − 5t. That is, our solution is x y z = 3t − 8 17 − 5t t ,
and so plugging in any value for t will be a solution to our system (and this will be a line in R3 passing through the point (x, y, z) = (−8, 17, 0) which has vector direction parallel to (3, −5, 1)).
Problem 9. Let A = 4 2 2 1 .
Find a diagonal matrix Λ and an orthogonal matrix Q such that QTAQ = Λ.
Recall since A is symmetric, then A has real eigenvalues. Now recall for a square matrix A,
The eigenvalues of A are precisely the zeros of the characteristic polynomial pA(λ) = det(λI − A). (21)
(Sometimes the characteristic polynomial is defined by det(A − λI), but note this has the same zeros as the above form.) We calculate
(Note another way in the 2 × 2 case is noting
pA(λ) = det(λI − A) = λ2− tr(A)λ + det(A) for A ∈ M2(R), (22)
so here we note tr(A) = λ1+ λ2= 5 and det(A) = λ1λ2= 0, which also quickly gives λ1= 0 and λ2= 5.)
Now to form an orthogonal matrix Q, which means we find corresponding eigenvectors q1, q2for λ1, λ2
with norm 1. These (normalized) qi will comprise the corresponding column vectors for Q.
To find the corresponding eigenvectors for λi, we solve (A − λiI)x = 0 for each i. Note since A − λiI is
singular for each i (using (9), we have A − λiI is singular since 0 is an eigenvalue for A − λiI — namely, if
x 6= 0 is an eigenvalue for λifor A, then Ax = λix and so Ax − λix = (A − λiI)x = 0, showing x is then an
eigenvector for 0 for A − λiI) , then rank(A − λiI) ≤ 1, and hence in this case rank(A − λiI) = 1 for each i.
So it suffices to only look at the first (nonzero) row to find an eigenvector. The first row of (A − λiI)x gives
us
(4 − λi)x1+ 2x2= 0.
(Verify this!) Hence, for λ1= 0, we have 4x1+ 2x2= 0 with then x2= −2x1, so v1= (x1 x2)T = (1 − 2)T
is an eigenvector for 0. For λ2 = 5, we have −x1+ 2x2 = 0 with then x1 = 2x2, so v2 = (2 1)T is an
eigenvector for 5. To find orthonormal eigenvectors, we have q1= v1 kv1k = √1 5 1 −2 and q2= v2 kv2k =√1 5 2 1 . It follows then Q = | | q1 q2 | | = 1 √ 5 1 2 −2 1 and Λ = 0 0 0 5 satisfy Λ = QTAQ.
Note also the result
If λ is an eigenvalue for A, then λn is an eigenvalue for An. (23)
In the symmetric case, this is verified by taking the nth powers of Λ = Q−1AQ for Q that diagonalizes A,
where we note (Q−1AQ)n = Q−1AnQ, and so Λn = Q−1AnQ. (In the general case, if x 6= 0 is such that
Ax = λx, then Anx = An−1(Ax) = λAn−1x = λnx, showing x is again an eigenvector for An for λn.) For
our problem, we have
Q is orthogonal if and only if QT = Q−1. (24)
In particular, we also have Λ1/n = Q−1A1/nQ, which gives an easy way of finding A1/n= QΛ1/nQ−1, where
we note Λk
is easy to calculate for any k ∈ R as you just raise each respective diagonal entry of Λ to the kth
power.
For some other background, one way we can establish eigenvectors, qi, qj for different eigenvalues, resp.
λi, λj, for a symmetric matrix A are orthogonal (which one can write qj⊥ qi), we check
λiqTjqi= qTj(λiqi) = qTjAqi = (qTjAqi)T = qTi ATqj = qTi Aqj = λjqTi qj
where we use (5) and the scalars are symmetric (as they are 1 × 1 matrices), while also A is symmetric. Since λi6= λj and qTjqi= qTi qj, it follows qTjqi= 0, showing qj ⊥ qi.
(Note for a picture of how A acts on R2, it’s enough to look at how A acts on e
1 and e2 for ei the
elementary standard basis element (that is 1 in the ith component and 0 elsewhere). Note Ae
i = ci for ci
the respective column vector of A. In particular, A maps both ei to the line passing through the origin and
Problem 10. Find constants α, β, γ ∈ R such that for c1= (1 1 0)T, c2 = (0 1 1)T, c3 = (1 0 1)T we have αc1+ βc2+ γc3= 1 2 3 .
The main thing to realize about this question is that it is asking you to solve a system of linear equations, which you should know how to do by now. Now what are the equations we are working with? Using the fact scalar multiplication and matrix addition are carried out componentwise, that is
α(aij) = (αaij) and (aij) + (bij) = (aij+ bij), (25)
we have 1 2 3 = αc1+ βc2+ γc3= α + γ α + β β + γ , which gives us the equation Ax = b for
A = | | | c1 c2 c3 | | | = 1 0 1 1 1 0 0 1 1 , x = α β γ , and b = 1 2 3 .
Now you can solve this system by whatever methods you are comfortable with. I’ll introduce yet another method, which involves using an augmented matrix (A|b), and then using elementary row operators to find the Reduced Row Echelon form of this (which would reduce A to I if A is nonsingular), we would have the desired solution in the form (I|x). Note this is equivalent to just applying elementary operators to the equation Ax = b, but here I’m just using the more compact form (A|b) to represent our system in yet another equivalent way.
(A|b) = 1 0 1 1 1 1 0 2 0 1 1 3 r27→ r2− r1 1 0 1 1 0 1 −1 1 0 1 1 3 r37→ r3− r2 1 0 1 1 0 1 −1 1 0 0 2 2 r37→ 1 2r3 1 0 1 1 0 1 −1 1 0 0 1 1 r17→ r1− r3 r27→ r2+ r3 1 0 0 0 0 1 0 2 0 0 1 1 = (I|x),
