Let the system of equations : $x+5y-z=1$, $4x+3y-3z=7$ , $24x + \lambda y + \lambda z = \mu$ $\lambda, \mu \in R$, have infinitely many solutions. Then the number of the solutions of this system,if x, y, z are integers and satisfy $7\leq x+y+z\leq77$, is :
Determining Conditions for Infinite Solutions
The given system of linear equations is: $x+5y-z=1$ $4x+3y-3z=7$ $24x + \lambda y + \lambda z = \mu$ For this system to have infinitely many solutions, the determinant of the coefficient matrix must be zero, and the augmented matrix must be consistent. The coefficient matrix is: $ A = \begin{pmatrix} 1 & 5 & -1 \\ 4 & 3 & -3 \\ 24 & \lambda & \lambda \end{pmatrix} $ Calculate the determinant of $A$: $det(A) = 1 \times (3\lambda - (-3\lambda)) - 5 \times (4\lambda - (-3 \times 24)) + (-1) \times (4\lambda - 3 \times 24)$ $det(A) = 1(6\lambda) - 5(4\lambda + 72) - 1(4\lambda - 72)$ $det(A) = 6\lambda - 20\lambda - 360 - 4\lambda + 72$ $det(A) = -18\lambda - 288$ For infinite solutions, $det(A) = 0$. $-18\lambda - 288 = 0 \implies -18\lambda = 288 \implies \lambda = -16$.
Additionally, the system must be consistent. This requires the constant terms to follow the same linear dependency as the coefficients. Using row operations or matrix properties, the consistency condition is found to be $\mu = -3\lambda$. Substituting $\lambda = -16$, we get $\mu = -3(-16) = 48$. So, for infinitely many solutions, $\lambda = -16$ and $\mu = 48$.
Finding the Parametric Solution
Substitute $\lambda = -16$ and $\mu = 48$ into the system: $x+5y-z=1$ (1) $4x+3y-3z=7$ (2) $24x - 16y - 16z = 48 \implies 3x - 2y - 2z = 6$ (3) From equation (1), express $x$ in terms of $y$ and $z$: $x = 1 - 5y + z$. Substitute this into equation (2): $4(1 - 5y + z) + 3y - 3z = 7$ $4 - 20y + 4z + 3y - 3z = 7$ $-17y + z = 3 \implies z = 17y + 3$. Now substitute $z$ back into the expression for $x$: $x = 1 - 5y + (17y + 3) = 12y + 4$. Wait, $x = 1 - 5y + z = 1 - 5y + (17y+3) = 1 + 12y + 3 = 12y + 4$. Hmm let me recheck the previous step $x = 1 - 5y + z$. $x = 1 - 5y + (17y + 3) = 1 + 12y + 3 = 12y + 4$. Let's re-derive using row reduction on the consistent system. Augmented matrix with $\lambda=-16, \mu=48$: $ \begin{pmatrix} 1 & 5 & -1 & | & 1 \\ 4 & 3 & -3 & | & 7 \\ 24 & -16 & -16 & | & 48 \end{pmatrix} $ $R_2 \rightarrow R_2 - 4R_1$, $R_3 \rightarrow R_3 - 24R_1$: $ \begin{pmatrix} 1 & 5 & -1 & | & 1 \\ 0 & -17 & 1 & | & 3 \\ 0 & -136 & -4 & | & 24 \end{pmatrix} $ $R_3 \rightarrow R_3 - 8R_2$: $ \begin{pmatrix} 1 & 5 & -1 & | & 1 \\ 0 & -17 & 1 & | & 3 \\ 0 & 0 & -12 & | & 0 \end{pmatrix} $ The last row $0=0$ confirms consistency. From $R_2$: $-17y + z = 3 \implies z = 17y + 3$. From $R_1$: $x + 5y - z = 1 \implies x = 1 - 5y + z = 1 - 5y + (17y + 3) = 12y + 4$. Let the free variable $y = t$. The parametric solution is: $x = 12t + 4$ $y = t$ $z = 17t + 3$
Applying Integer and Sum Constraints
We are looking for integer solutions, which means $t$ must be an integer. The constraint on the sum of variables is $7 \leq x+y+z \leq 77$. Calculate the sum $x+y+z$: $x+y+z = (12t + 4) + t + (17t + 3) = 30t + 7$. Apply the constraint to the sum: $7 \leq 30t + 7 \leq 77$ Subtract 7 from all parts: $0 \leq 30t \leq 70$ Divide by 30: $0 \leq t \leq 70/30$ $0 \leq t \leq 7/3$ $0 \leq t \leq 2.333...$
Counting the Number of Integer Solutions
Since $t$ must be an integer, the possible values for $t$ in the interval $[0, 2.333...]$ are $0, 1, 2$. Each integer value of $t$ gives a unique integer solution $(x, y, z)$.
- For $t=0$: $(x, y, z) = (4, 0, 3)$. Sum = 7.
- For $t=1$: $(x, y, z) = (16, 1, 20)$. Sum = 37.
- For $t=2$: $(x, y, z) = (28, 2, 37)$. Sum = 67.
There are 3 possible integer values for $t$, resulting in 3 distinct integer solutions that satisfy all the conditions.
Similar Questions
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Important Questions from Algebra
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