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 :
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 $\mu = 48$.
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: $x = 12t + 4$ $y = t$ $z = 17t + 3$
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: $0 \leq t \leq 70/30$ $0 \leq t \leq 7/3$ $0 \leq t \leq 2.333...$
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 $(x, y, z)$.
There are 3 possible integer values for $t$, resulting in 3 distinct integer solutions that satisfy all the conditions.
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If $\int_{b-a}^{b+a}[x^2] dx=p-\sqrt{q}-\sqrt{r}$, $p,q,r\in N$, $gcd(p,q,r) = 1$, where $[.]$ is the greatest integer function, then $p + q + r$ is equal to
Let A be a matrix of order $3 \times 3$ and $|A| = 5$. If $|2\text{adj} (3A \text{adj} (2A))| = 2^\alpha \cdot 3^\beta \cdot 5^\gamma$, $\alpha, \beta, \gamma \in N$, then $\alpha + \beta + \gamma$ is equal to
Let $a_1, a_2, a_3,....$ be a G.P. of increasing positive numbers. If $a_3a_5 = 729$ and $a_2 + a_4 = \frac{111}{4}$, then $24 (a_1 + a_2 + a_3)$ is equal to
All five letter words are made using all the letters A, B, C, D, E and arranged as in an English dictionary with serial numbers. Let the word at serial number $n$ be denoted by $W_n$. Let the probability $P(W_n)$ of choosing the word $W_n$ satisfy $P(W_n) = 2P(W_{n-1})$, $n > 1$.
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Let $a \in \mathbf{R}$ and $A$ be a matrix of order $3 \times 3$ such that $\det (A) = -4$ and $A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix}$, where $I$ is the identity matrix of order $3 \times 3$. If $\det ((a+1)\text{adj}((a-1)A))$ is $2^m 3^n$, $m, n \in \{0, 1, 2, \dots, 20\}$, then $m+n$ is equal to :
Let A = {-3, -2, -1, 0, 1, 2, 3}. Let R be a relation on A defined by xRy if and only if $0\le x^2+2y\le 4$. Let $l$ be the number of elements in R and $m$ be the minimum number of elements required to be added in R to make it a reflexive relation. Then $l + m$ is equal to
Let $\alpha$ and $\beta$ be the roots of $x^2 + \sqrt{3}x-16=0$, and $\gamma$ and $\delta$ be the roots of $x^2 + 3x - 1 = 0$. If $P_n = \alpha^n + \beta^n$ and $Q_n = \gamma^n + \delta^n$, then $\frac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} + \frac{Q_{25}-Q_{23}}{Q_{24}}$ is equal to
Let the domain of the function $f (x) = \log_2 \log_4 \log_6(3 + 4x -x^2)$ be $(a, b)$.
If $\int_{b-a}^{b+a}[x^2] dx=p-\sqrt{q}-\sqrt{r}$, $p,q,r\in N$, $gcd(p,q,r) = 1$, where $[.]$ is the greatest integer function, then $p + q + r$ is equal to
Let A be a matrix of order $3 \times 3$ and $|A| = 5$. If $|2\text{adj} (3A \text{adj} (2A))| = 2^\alpha \cdot 3^\beta \cdot 5^\gamma$, $\alpha, \beta, \gamma \in N$, then $\alpha + \beta + \gamma$ is equal to
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