Solutions to Linear Algebra Practice Problems. Determine which of the following augmented matrices are in row echelon from, row reduced echelon form or neither. Also determine which variables are free if it s in row echelon form or row reduced echelon form. 2 2 0 0 is neither row-echelon from nor reduced row-echelon 0 0 form (because the leading in the third row is not to the right of the leading in the second row). 2 0 is in row-echelon from. Free variables is x 3. 0 0 0 2 2 0 0 is neither row-echelon from nor reduced row-echelon 0 0 0 form (because the leading coefficient in first row is not.) 2 0 0 0 0 0 is in row reduced echelon from. Free variables is x 2. 0 0 0 2. Determine if the following systems are consistent and if so give all solutions in parametric vector form. (a) x 2x 2 = 3 2x 7x 2 = 0 5x +8x 2 = 5 Solution: The augmented matrix is 2 3 2 7 0 (new row 2:=-2 row + row 2) 5 8 5 2 3 0 3 6 5 8 5
2 3 (new row 3:=5row +row 3) 0 3 6 0 2 20 2 3 (new row 2:= row 2) 3 0 2 0 2 20 2 3 (new row 3:=2 row 2 + row 3) 0 2. The last row implies 0 0 24 that 0 = 24 which is impossible. So this system is inconsistent. (b) x +2x 2 3x 3 +x 4 = x 2x 2 +4x 3 x 4 = 6 2x 4x 2 +7x 3 x 4 = 2 3 The augmented matrix is 2 4 6 (new row 2:=row 2 4 7 2 3 +row 2) 0 0 0 7 (new row 3:=row +2row 3) 2 4 7 2 3 0 0 0 7 0 0 3 2 3 (new row 3:=(-)row 2+row 3) 0 0 0 7 (new 0 0 0 4 2 3 0 5 row :=(-)row 3+row ) 0 0 0 7 (new row :=3row 0 0 0 4 2
2+row ) 2 0 0 26 0 0 0 7 0 0 0 4. So x 2 is free. The solution is x = 26 2x 2, x 3 = 7, x 4 = 47. Its parametric vector form is 26 2t t 7 4 = 26 0 7 4 + t 2 0. 0 (c) x +2x 2 3x 3 +x 4 = x 2x 2 +4x 3 x 4 = 6 2x 4x 2 +7x 3 2x 4 = 2 3 The augmented matrix is 2 4 6 2 4 7 2-2 * Row 2 + Row 3 := new Row 3 2 3 2 4 6 0 0 0 * Row + Row 2 := new Row 2 2 3 0 0 0 7 0 0 0 * Row 2 + Row 3 := new Row 3 2 3 0 0 0 7 0 0 0 0 4 x x 2 x 4 = x 4 The last row means that 0 = 4 which is impossible. So this system is inconsistent. 3
3 4 7 3. Let A = 2 6 5. 3 9 4 5 (a) Find all the solutions of the non-homogeneous system Ax = b, and write them in parametric form, where b = 2. Solution: 3 3 4 7 Consider the augmented matrix [A b] = 2 6 5 2. 3 9 4 5 3 Now we perform row operations on the augmented matrix. 3 4 7 2 6 5 2 newrow2 := ( 2)row + row2, newrow3 := ( 3)row + row3 3 9 4 5 3 3 4 7 0 0 3 3 0 0 0 6 6 0 newrow2 := row2, newrow3 := row3, newrow3 := row3 row2 3 6 3 4 7 0 0 0 0 0 0 0 0 newrow := 4row2 + row So the solution is So 3 0 3 0 0 0 0 0 0 0 0 x + 3x 2 + 3x 4 = x 3 x 4 = 0 x 2 and x 4 are free. x = 3x 2 3x 4 x 2 is free x 3 = x 4 x 4 is free. 4 () (2)
Thus the solution of Ax = 2 is 3 x 3s 3t 3 3 x = x 2 x 3 = s t = 0 0 + s 0 + t 0. where x 4 t 0 0 s and t are any numbers. (b) Find all the solutions of the homogeneous system Ax = 0, and write them in parametric form. 3 4 7 0 Solution: Consider the augmented matrix [A 0] = 2 6 5 0. 3 9 4 5 0 3 0 3 From previous example, it can be row reduced to 0 0 0 0 0 0 0 0 So the solution is So x + 3x 2 + 3x 4 = 0 x 3 x 4 = 0 x 2 and x 4 are free. x = 0 3x 2 3x 4 x 2 is free x 3 = x 4 x 4 is free. So the solution of Ax = 0 is of the form 3 3 x = s 0 + t 0 0 where s and t are any numbers. (3) (4) 5
{ 4. Let S = Span 2 3, 0 2, 3 2 3, 0 4 0 0 Solution: Let A = 2 3 3 2. In part (a), we need to determine if Ax = v is consistent or not. In part (b), we need to determine 2 3 4 if Ax = w is consistent or not. We can just consider the augmented matrix [A v w]. 0 0 2 3 3 3 3 2 2 2 2 3 4 2 * Row + Row 2 new Row 2 0 0 0 3 2 2 2 2 3 4-3 * Row + Row 3 new Row 3 0 0 0 5 0 5 2 3 4 - * Row + Row 4 new Row 4 0 0 0 5 0 2 2 4 2 0 - * Row 2 + Row 3 new Row 3 0 0 0 5 0 0 0 0 0 0 0 2 2 4 2 0 }. 6
2 * Row 2 + Row 4 new Row 4 0 0 0 5 0 0 0 0 0 0 0 0 0 2 4 0 swap Row 3 with Row 4 0 0 0 5 0 0 0 2 4 0 0 0 0 0 0 0 -/2 * Row 3 new Row 3 0 0 0 5 0 0 0 2 5 0 0 0 0 0 0 - * Row 3 + Row 2 new Row 2 0 0 0 0 3 0 0 0 0 2 5 0 0 0 0 0 0 This implies the system Ax = v is consistent and v is a linear combination of the vectors in S Similarly, the system Ax = w is inconsistent and w is a not linear combination of the vectors in S 5. Consider a linear system whose augmented matrix is of the form 2 2 b 2 a (a) For what values of a will the system have a unique solution? What is the solution?(your answer may involve a Solution: 2 2 b 2 a 7
(-)row + row 2 row 2 2 0 0 2 + b 2 a row + row 3 row 3 2 0 0 2 + b 0 3 a + 3-3 * row 2 + row 3 row 3 2 0 0 2 + b 0 0 a + 9 + 3b From this, we know that if a+ 0 then this system has a unique solution. We can continue the row reduction. row 3 new row 3 a+ 2 0 0 2 + b 9+3b 0 0 a+ (-) row 3 + row new row 0 2 9+3b a+ 0 0 2 + b 0 0 9+3b a+ (-) row 2 + row new row 0 0 4 b 9+3b a+ 0 0 2 + b 0 0 9+3b a+ Hence if a then this system has a unique solution x = 4 b 9+3b, x a+ 2 = 2 + b and x 3 = 9+3b. a+ 8
(b) For what values of a and b will the system have infinitely many solutions? Solution: From previous problem, we know that is row reduced to 2 2 b 2 a 2 0 0 2 + b 0 0 a + 9 + 3b we know that if a + = 0 and 9 + 3b = 0 then this system has infinitely many solutions. (c) For what values of a and b will the system be inconsistent? Solution: From previous problem, we know that 2 2 b 2 a is row reduced to 2 0 0 2 + b 0 0 a + 9 + 3b we know that if a + = 0 and 9 + 3b 0 then this system is inconsistent. 6. (a) Find the inverses of the following matrices if they exist. [ ] 2 3 4 7 2 A =, B = and C = 5 6 7. 4 0 0 8 9 [ 0 ] 7 2 0 Solution: Consider the augmented matrix [A I 2 ]. 4 0 2 * Row 2 + Row new Row 9,,
[ 0 ] 2 4 0 - * Row new Row [ ] 0 2 4 0 4 * Row + Row 2 new Row 2 [ ] [ ] 0 2 2 So A 0 4 7 =. 4 7 Consider the augmented matrix [B I 3 ]. - * Row + Row 2 new Row 2 0 0 0 2 0 0 0 0 0 0 - * Row + Row 3 new Row 3 0 0 0 2 0 0 0 0 /2 * Row 2 new Row 2 0 0 0 0 /2 /2 0 0 0 - * Row 2 + Row 3 new Row 3 0 0 0 0 /2 /2 0 0 0 /2 /2 - * Row 3 new Row 3 0 0 0 0 /2 /2 0 0 0 /2 /2 - * Row 3 + Row new Row 0 0 0 0 0 0 0 0 0
0 /2 /2 0 0 /2 /2 0 0 0 /2 /2 * Row 2 + Row new Row 0 0 0 0 0 0 0 0 /2 /2 0 So B = /2 /2 0 0 0 /2 /2 /2 /2 2 3 4 0 0 Now consider [C I 3 ]. 5 6 7 0 0 8 9 0 0 0-2 * Row + Row 2 new Row 2 2 3 4 0 0 0 2 0 8 9 0 0 0 Row swap Row 2 0 2 0 2 3 4 0 0 8 9 0 0 0-2 * Row + Row 2 new Row 2 0 2 0 0 3 6 5 2 0 8 9 0 0 0-8 * Row + Row 3 new Row 3 0 2 0 0 3 6 5 2 0 0 9 8 6 8 /3 * Row 2 new Row 2 0 2 0 0 2 5/3 2/3 0 0 9 8 6 8-9 * Row 2 + Row 3 new Row 3 0 2 0 0 2 5/3 2/3 0 0 0 0 2
Thus C is not invertible. (b) What s (B T )? 0 0 0 /2 /2 Solution: (B T ) = (B ) T = ( /2 /2 0 ) T = 0 /2 /2. /2 /2 0 [ ] 7. Let A = and f(x) = x 2 2x + 2. Show that f(a) = 0 2. Here 0 2 is the 2 2 zero matrix. [ ] [ ] [ ] 0 2 Solution: First, we compute A 2 = =. So [ ] [ ] [ ] 2 0 0 2 0 f(a) = A 2 2A + 2I 2 = 2 + 2 [ ] 2 [ 0 ] 0 0 2 + 2 2 + 2 + 0 0 0 = =. 2 2 + 0 0 2 + 2 0 0 8. Let A be an 3 3 matrix. Suppose A 3 + 2A 2 4A + I 3 = 0. Is A invertible? Express A in terms of A if possible. Solution: From A 3 + 2A 2 4A + I 3 = 0, we have A 3 2A 2 + 4A = I 3 and A( A 2 2A + 4I 3 ) = I 3. This implies that A is invertible and A = A 2 2A + 4I 3. 9. Express the following matrices as a product of elementary matrices and a matrix in reduced row-echelon form. [ ] A = 2 3, B =. 4 5 6 Solution: (a) Step. A = - * Row + Row 2 new Row 2 0 0 0 0 0 Repeat the same operation on the identity matrix 0 0 0 0 2
- * Row + Row 2 new Row 2 0 0 E = 0 0 0 Step 2. * Row + Row 3 new Row 3 0 0 0 0 0 2 0 0 Repeat the same operation on the identity matrix 0 0 0 0 * Row + Row 3 new Row 3 0 0 E 2 = 0 0 0 Step 3. Row 2 Row 3 0 0 2. 0 0 0 0 0 Repeat the same operation on the identity matrix 0 0 0 0 Row 2 Row 3 0 0 E 3 = 0 0 0 0 Step 4. /2 * Row 2 new Row 2 0 0 0 0 0 3
0 0 Repeat the same operation on the identity matrix 0 0 0 0 /2 * Row 2 new Row 2 0 0 E 4 = 0 /2 0 0 0 Step 5. - * Row 2 + Row new Row 0 0 0 0 0 0 0 0 Repeat the same operation on the identity matrix 0 0 0 0 - * Row 2 + Row new Row 0 E 5 = 0 0. 0 0 Finally 0 A = E 5 E 4 E 3 E 2 E 0 0. 0 0 0 (b) step. B = [ 2 ] 3 4 5 6-4 * Row + Row 2 new Row 2 [ ] 2 3 0 3 6 Repeat the same operation on the identity matrix I 2 [ 0 0-4 * Row + Row 2 new Row 2 ] 4
E = step 2. [ ] 0 4 -/3 * Row 2 new Row 2 [ ] 2 3 0 2 Repeat the same operation on the identity matrix I 2 [ 0 0 -/3 * Row 2 new Row 2 [ ] 0 E 2 = 0 /3 step 3. -2 * Row 2 + Row new Row [ ] 0 0 2 Repeat the same operation on the identity matrix I 2 [ 0 0-2 * Row 2 + Row new Row [ ] 2 E 3 = 0 [ ] 0 So B = E 3 E 2 E. 0 2 ] ] 5