Find a row-reduced echelon matrix R which is row-equivalent to A and an invertible 3×3 matrix P such that R=PA.
solution: We perform row operations on A′=[AY] with Y=[y1,y2,y3]T, thus A′=⎣⎡1−1120−2131051y1y2y3⎦⎤→⎣⎡10022−4140051y1y2+y1y3−y1⎦⎤→⎣⎡1002201480511y1y2+y12y2+y3+y1⎦⎤→⎣⎡100210121025811y121(y2+y1)81(2y2+y3+y1)⎦⎤→⎣⎡100210001−811−4181187y1−41y2−81y341(y1−y3)81(2y2+y3+y1)⎦⎤→⎣⎡100010001−87−4181183y1−41y2+83y341(y1−y3)81(2y2+y3+y1)⎦⎤ thus R=⎣⎡100010001−87−41811⎦⎤, and the matrix P s.t. R=PA is P=⎣⎡3/81/41/8−1/401/43/8−1/41/8⎦⎤
2. Do Exercise 1, but with
A=⎣⎡21i0−31i−i1⎦⎤ solution: We perform row operations on A′=[AY] with Y=[y1,y2,y3]T, thus A′=⎣⎡21i0−31i−i1y1y2y3⎦⎤→⎣⎡1000−312i−23i232y1y2−2y1y3−2iy1⎦⎤→⎣⎡1000102i2329−3i2y1y3−2iy13y3−23i+1y1+y2⎦⎤→⎣⎡1000102i2312y1y3−2iy1153+i(3y3−23i+1y1+y2)⎦⎤ It’s easy to see R=I and to calculate P, we see A′→[IY′], in which Y′=⎣⎢⎢⎢⎢⎢⎢⎡2y1−15i2(3+i)(3y3−23i+1y1+y2)y3−2iy1−452(3+i)(3y3−23i+1y1+y2)153+i(3y3−23i+1y1+y2)⎦⎥⎥⎥⎥⎥⎥⎤=⎣⎢⎢⎢⎢⎡21+30i2(3+i)(3i+1)−2i+45(3+i)(3i+1)−30(3+i)(3i+1)−15i2(3+i)−452(3+i)153+i−5i2(3+i)1−152(3+i)53+i⎦⎥⎥⎥⎥⎤⎣⎡y1y2y3⎦⎤ thus P is the matrix on the left.
3. For each of the two matrices
⎣⎡2465−14−121⎦⎤,⎣⎡130−12124−2⎦⎤
use elementary row operations to discover whether it is invertible, and to find the inverse in case it is.
solution: As ⎣⎡2465−14−121100010001⎦⎤→⎣⎡2005−11−11−1441−2−3010001⎦⎤→⎣⎡2005−110−1401−2−101−1001⎦⎤ the first matrix is not invertible. As ⎣⎡130−12124−2100010001⎦⎤→⎣⎡100−1512−2−21−30010001⎦⎤→⎣⎡100−1102−2810−300101−5⎦⎤→⎣⎡100−1100017/4−3/4−3/8−1/41/41/85/4−1/4−5/8⎦⎤→⎣⎡1000100011−3/4−3/801/41/81−1/4−5/8⎦⎤ the second matrix is invertible and its inverse is ⎣⎡1−3/4−3/801/41/81−1/4−5/8⎦⎤
4. Let
A=⎣⎡510051005⎦⎤
For which X does there exist a scalar c such that AX=cX?
solution: If X=0, then AX=cX for any scalar c. If X=0, then the system (A−cI)X=0 has non trivial solutions, if c=5 we have ⎣⎡5−c1005−c1005−c⎦⎤→⎣⎡10005−c1005−c⎦⎤→⎣⎡111⎦⎤ thus the system (A−cI)X=0 has only trivial solutions, a contradiction. If c=5, then as long as X=(0,0,k),k=0, we have AX=5X. In conclusion we can say for X=(0,0,k),k∈R does there exists a scalar c s.t. AX=cX.
5. Discover whether
A=⎣⎢⎢⎡1000220033304444⎦⎥⎥⎤
is invertible, and find A−1 if it exists.
solution: A is invertible. we have ⎣⎢⎢⎡10002200333044441000010000100001⎦⎥⎥⎤→⎣⎢⎢⎡1000220033300004100001000010−1−1−11⎦⎥⎥⎤→⎣⎢⎢⎡100022000030000410000100−1−110−1−1−11⎦⎥⎥⎤→⎣⎢⎢⎡10000200003000041000−1100−1−110−1−1−11⎦⎥⎥⎤→⎣⎢⎢⎡10000100001000011000−11/200−1−1/21/30−1−1/2−1/31/4⎦⎥⎥⎤ thus A−1=⎣⎢⎢⎡1000−11/200−1−1/21/30−1−1/2−1/31/4⎦⎥⎥⎤
6. Suppose A is a 2×1 matrix and that B is a 1×2 matrix. Prove that C=AB is not invertible.
solution: Suppose A=[ab],B=[cd], then C=AB=[acbcadbd], use elementary row operation one can eliminate one of C’s rows if a=0, and the first row of C vanishes if a=0, thus C can’t be row equivalent to the identity matrix, and C is not invertible.
7. Let A be an n×n (square) matrix. Prove the following two statements:
(a) If A is invertible and AB=0 for some n×n matrix B, then B=0.
(b) If A is not invertible, then there exists an n×n matrix B such that AB=0 but B=0.
solution: (a) We have PA=I for some P, thus B=IB=PAB=P0=0. (b) The system AX=0 has non trivial solutions, let x=0 be a solution and let B=[x0…0], then AB=0.
8. Let
A=[acbd]
Prove, using elementary row operations, that A is invertible if and only if (ad−bc)=0.
solution: If c=0, then to make A invertible if and only if (a=0)∧(d=0), which is equivalent to ad=ad−bc=0. If c=0, then use elementary row operations, we have A=[acbd]→[acacbcad]→[ac0bcad−bc]→[a0bad−bc],a=0A=[acbd]→[cadb]=[c0db],a=0 thus if A is invertible, we shall have ad−bc=0. Conversely if ad−bc=0, then in the case a=0 we directly have A is invertible since [a0bad−bc] is row equivalent to I2, in the case a=0, notice then bc=0 and thus b=0, so [c0db] is row equivalent to I2, in either case, A is invertible.
9. An n×n matrix A is called upper-triangular if Aij=0,i>j, that is, if every entry below the main diagonal is 0. Prove that an upper-triangular (square) matrix is invertible if and only if every entry on its main diagonal is different from 0.
solution: If every entry on its main diagonal is not 0, then we use row n to eliminate the last column to en, and next use row n−1 to eliminate column n−1 to en−1, continuing we make A to In, thus A is invertible. Conversely, suppose A is invertible and assume some entry on its main diagonal is 0, let k be the largest integer s.t. Akk=0, then if k=n then row n is all 0, if not then repeating the same steps in the previous paragragh n−k times, we can get row k to be all 0, then A is not invertible, as the matrix can’t be row-equivalent to In.
10. Prove the following generalization of Exercise 6. If A is an m×n matrix, B is an n×m matrix and n<m, then AB is not invertible.
solution: Since n<m, if row-reduce A to an row-deduced echelon form, the last row must be all 0, i.e. there’s an m×m invertible matrix P s.t. PA=C, where the last row of C is all 0, now assume AB is invertible, as it’s square we can have D be it’s right inverse, so ABD=I, thus PABD=CBD=P, now the last row of C is 0, means the last row of CBD ,or P, is 0, this contradicts P being invertible.
11. Let A be an m×n matrix. Show that by means of a finite number of elementary row and/or column operations one can pass from A to a matrix R which is both ‘row-reduced echelon’ and ‘column-reduced echelon’, i.e. Rij=0 if i=j, Rii=1,1≤i≤r,Rii=0 if i>r. Show that R=PAQ, where P is an invertible m×m matrix and Q is an invertible n×n matrix.
solution: We’ve proved use elementary row operations we can get a row-reduced echelon matrix R′ such that R′=PA, where P is an invertible m×m matrix. Starting from R′ we can use elementary column operations to get a column-reduced echelon matrix R, and an invertible n×n matrix Q s.t. R=R′Q, thus R=PAQ.
12. The result of Example 16 suggests that perhaps the matrix
is invertible and A−1 has integer entries. Can you prove that?
solution: This matrix is called the Hilbert matrix, it’s invertible and its inverse is given by B=(Bij), where Bij=(−1)i+j(i+j+1)(ii+j)(ji+j)(n−jn+i+1)(n−in+j+1) One can direct compute the product AB and verify that AB=I.