$x + 5y + 6z = 4,$
$2x + 3y + 4z = 7,$
$x + 6y + az = b$
has infinitely many solutions, then the point (a, b) lies on the line
To find the condition for the system of equations to have infinitely many solutions, we can use the method of row reduction on the augmented matrix.
Augmented Matrix Row Reduction
The given system of equations is:
$x + 5y + 6z = 4$
$2x + 3y + 4z = 7$
$x + 6y + az = b$
The augmented matrix is:
Perform row operations to simplify:
- $R_2 \rightarrow R_2 - 2R_1$
- $R_3 \rightarrow R_3 - R_1$
The matrix becomes:
Swap $R_2$ and $R_3$ for easier calculation:
Perform the next row operation:
- $R_3 \rightarrow R_3 + 7R_2$
The matrix becomes:
Simplify the last row:
Coefficient of $z$: $-8 + 7a - 42 = 7a - 50$
Constant term: $-1 + 7b - 28 = 7b - 29$
The final row-reduced matrix is:
Condition for Infinite Solutions
For the system to have infinitely many solutions, the last row must represent the equation $0z = 0$. This requires both the coefficient of $z$ and the constant term in the last row to be zero.
$7a - 50 = 0 \implies a = \frac{50}{7}$
$7b - 29 = 0 \implies b = \frac{29}{7}$
Therefore, the point $(a, b)$ is $(\frac{50}{7}, \frac{29}{7})$.
Identifying the Line
Now, we check which of the given lines the point $(\frac{50}{7}, \frac{29}{7})$ lies on.
- Option 1: $y - x = 3$ $\implies \frac{29}{7} - \frac{50}{7} = -\frac{21}{7} = -3 \neq 3$.
- Option 2: $x - y = 3$ $\implies \frac{50}{7} - \frac{29}{7} = \frac{21}{7} = 3$. This matches.
- Option 3: $x + y = 11$ $\implies \frac{50}{7} + \frac{29}{7} = \frac{79}{7} \neq 11$.
- Option 4: $x + y = 12$ $\implies \frac{50}{7} + \frac{29}{7} = \frac{79}{7} \neq 12$.
The point $(a, b)$ lies on the line $x - y = 3$.
Similar Questions
- Let A be a $3 \times 3$ matrix such that $A + A^T = O$. If $A\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$, $A^2\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$ and $\det(adj(2 \ adj(A + I))) = (2)^\alpha \cdot (3)^\beta \cdot (11)^\gamma$, $\alpha, \beta, \gamma$ are non-negative integers, then $\alpha + \beta + \gamma$ is equal to _________
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Important Questions from Algebra
- Let A be a $3 \times 3$ matrix such that $A + A^T = O$. If $A\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$, $A^2\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$ and $\det(adj(2 \ adj(A + I))) = (2)^\alpha \cdot (3)^\beta \cdot (11)^\gamma$, $\alpha, \beta, \gamma$ are non-negative integers, then $\alpha + \beta + \gamma$ is equal to _________
- Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
- Let ABC be a triangle. Consider four points $p_1, p_2, p_3, p_4$ on the side AB, five points $p_5, p_6, p_7, p_8, p_9$ on the side BC, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points $p_1, p_2, ..., p_{13}$, is _________
- If $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is a solution of the system of equations $AX = B$, where $\text{adj } A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$, then $|x + y + z|$ is equal to :
- Let $C_r$ denote the coefficient of $x^r$ in the binomial expansion of $(1 + x)^n$, $n \in \mathbb{N}, 0 \leq r \leq n$. If $P_n = C_0 - C_1 + \frac{2^2}{3} C_2 - \frac{2^3}{4} C_3 + \dots + \frac{(-2)^n}{n+1} C_n$, then the value of $\sum_{n=1}^{25} \frac{1}{P_{2n}}$ equals.
