4. Given the system of linear equations 2.1 - X2 + 23 = -1 8x1 - 4x2 - 23 = -4 21 - 22 = -1 determine if a solution exists. If it does, what is the solution? A) *Let t be a variable such that t goes from negative infinity to positive infinity. The solution set is (t - 5, t, 0). B) Let t be a variable such that t goes from 0 to positive infinity. The solution set is (?t, 3t-?, t) C) Let t be a variable such that t goes from negative to positive infinity. The solution set is (t, 3t - 3, t) D) Let t be a variable such that t goes from negative to positive infinity. The solution set is( 3 t + 1, - 5t, t). E) None of the above.Consider the following function: 2': "f: k. h,(:1:) = 3. 1 I; > 35133 ikamZO and find the value of k: that makes Mm) continuous on the interval (0, 00) A) -3 B) U C) D) *3 E) None of the above. Step 1: Formulate the Augmented Matrix From the system of equations: 1. 2x1 - 22 + 23 = -1 2. 8x1 - 4x2 - 23 = -4 3. 2x1 - 2 = -1 we can form the augmented matrix as follows: N 1 - 1 8 - 4 Concept Used: An augmented matrix combines the coefficients of the system of linear equations, with the constants moved to the right of a bar (or a line), effectively consolidating the system into a matrix for simplified handling.Row Operations for RREF: 1. Operation: Divide the first row (R1) by 2 to get a leading 1. Resulting Matrix: 1 0.5 0.5 |0.5 8 4 1 |4 2 1 0 |1 2. Operation: Replace the second row (R2) with R2 8R1 and the third row (R3) with R3 - 2R1 to make the rest of the first column zero. Resulting Matrix: 1 0.5 0.5 |0.5 0 0 5 |0 0 0 1 l0 3. Operation: Divide the third row (R3) by -1 to get a leading 1 in the third row. Resulting Matrix: 1 0.5 0.5 |0.5 0 0 5 |0 0 0 1 |0 4. Operation: Replace the second row (R2) with R2 + 5*R3 to eliminate the -5 in the third column of the second row. Resulting Matrix: 1 0.5 0.5 |0.5 0 0 0 |0 0 0 1 |0 Resulting Matrix (RREF): 1 0.5 0 |0.5 0 0 0 |0 0 0 1 |0 Interpreting the RREF: From the RREF, we can see that the system of equations is consistent and has an infinite number of solutions because there's a row with all zeros in the coefficient matrix and the constant matrix (row 2). Also, we see that $2 is a free variable because its column doesn't have a pivot (a leading 1 in RREF). Given 3:2 is free, we let m2 = t where t is a parameter that can take any value from negative to positive infinity. The RREF gives us the following relations: 1. 1:1 0.5m : 0.50ra:1 : 0.52132 0.5 2. CB3 : 0 Given that 3:2 is a free variable, we can set 3:2 2 t (where t can be any real number) and solve for $1 and $3 in terms oft. From the RREF, we have the equations: 1 _ 1 _1 1 1. $1$2or$1$2 2.33320 Substituting 332 : t, we get: 75 l, t, 0), where t can be any real NIrt So, the solution set in parametric form is (:31, 5132, $3) = ( number from negative infinity to positive infinity