Q.Examine the consistency of the following system of equations: $5x - y + 4z = 5$ $2x + 3y + 5z = 2$ $5x - 2y + 6z = -1$

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1. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.The system of linear equations $x+2y+z=-3$, $3x+3y-2z=-1$, $2x+7y+7z=-4$ has (A) infinite number of solutions (B) no solution (C) unique solution (D) finite number of solutions

   ›Reveal solutionSolution

   Eliminating one variable reduces the system to two equations that contradict each other outright, which is the signature of an inconsistent linear system: the coefficient determinant is zero, but the system does not have infinitely many solutions — it has none.

   Concept and Intuition

   When the determinant of the coefficient matrix of a $3\times3$ linear system is zero, the system is not guaranteed a unique solution — but that alone doesn't tell you whether it has infinitely many solutions or none at all. You have to check consistency by actually trying to solve (or by comparing ranks of the coefficient matrix and the augmented matrix). A quick, reliable way for a 3-variable system is to eliminate one variable using two different pairs of equations and see if the resulting two-variable equations agree or contradict.

   Step-by-Step Solution

   1. Equations: (1) $x+2y+z=-3$, (2) $3x+3y-2z=-1$, (3) $2x+7y+7z=-4$.
   2. From (1): $x=-3-2y-z$.
   3. Substitute into (2): $3(-3-2y-z)+3y-2z=-1 \Rightarrow -9-6y-3z+3y-2z=-1 \Rightarrow -9-3y-5z=-1 \Rightarrow 3y+5z=-8$. Call this (A).
   4. Substitute into (3): $2(-3-2y-z)+7y+7z=-4 \Rightarrow -6-4y-2z+7y+7z=-4 \Rightarrow -6+3y+5z=-4 \Rightarrow 3y+5z=2$. Call this (B).
   5. Compare (A) and (B): both say "$3y+5z=$ something," but (A) requires it to equal $-8$ while (B) requires it to equal $2$. These cannot both be true — a direct contradiction. …

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2. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.The number of solutions of the system of equations $2x+y-z=7$, $x-3y+2z=1$, $x+4y-3z=5$ is (A) 1 (B) 0 (C) Infinite (D) 2

   ›Reveal solutionSolution

   The coefficient determinant is $0$, and substitution shows the equations are mutually contradictory — the system has no solution. Answer: (B).

   Concept and Intuition

   When the determinant of the coefficient matrix of a $3\times3$ linear system is zero, the system is NOT guaranteed a unique solution — it is either inconsistent (no solution) or has infinitely many solutions, depending on whether the equations are compatible. The way to tell them apart is to actually eliminate variables and see whether you reach a contradiction (like $0=$ nonzero) or a genuine identity ($0=0$).

   Step-by-Step Solution

   1. System: (1) $2x+y-z=7$; (2) $x-3y+2z=1$; (3) $x+4y-3z=5$.
   2. Coefficient determinant: $\begin{vmatrix}2&1&-1\\1&-3&2\\1&4&-3\end{vmatrix} = 2[(-3)(-3)-2(4)] - 1[1(-3)-2(1)] + (-1)[1(4)-(-3)(1)]$ $= 2(9-8)-1(-3-2)-1(4+3) = 2(1)-1(-5)-1(7) = 2+5-7=0$.
   3. Since the determinant is $0$, solve by substitution to check consistency. From (2): $x=1+3y-2z$.
   4. Substitute into (1): $2(1+3y-2z)+y-z=7 \Rightarrow 2+6y-4z+y-z=7 \Rightarrow 7y-5z=5$.
   5. Substitute into (3): $(1+3y-2z)+4y-3z=5 \Rightarrow 1+7y-5z=5 \Rightarrow 7y-5z=4$. …

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3. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.If the solution of the system of simultaneous linear equations $x+y-z=6$, $3x+2y-z=5$ and $2x-y-2z+3=0$ is $x=\alpha, y=\beta, z=\gamma$, then $\alpha+\beta=$ (A) $-7$ (B) $2$ (C) $1$ (D) $-2$

   ›Reveal solutionSolution

   Solving the 3×3 linear system directly gives $x=-3,\ y=5,\ z=8$, so $\alpha+\beta=x+y=2$.

   Concept and Intuition

   A straightforward system of 3 linear equations in 3 unknowns — eliminate one variable at a time using simple linear combinations, rather than invoking full Cramer's rule, since the numbers are small.

   Step-by-Step Solution

   1. The equations are:

   $x+y-z=6 \quad (1)$

   $3x+2y-z=5 \quad (2)$

   $2x-y-2z=-3 \quad (3)\ \text{(rewriting } 2x-y-2z+3=0\text{)}$

   2. Subtract (1) from (2): $(3x+2y-z)-(x+y-z) = 5-6 \Rightarrow 2x+y=-1$ ... (4)
   3. From (1): $z = x+y-6$.
   4. Substitute into (3): $2x - y - 2(x+y-6) = -3 \Rightarrow 2x-y-2x-2y+12=-3 \Rightarrow -3y+12=-3 \Rightarrow -3y=-15 \Rightarrow y=5$.
   5. Substitute $y=5$ into (4): $2x+5=-1 \Rightarrow 2x=-6 \Rightarrow x=-3$. …

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4. AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQ
   Q.The system of equations $2x + 6y = -11,\ 6x + 20y - 6z = -3,\ 6y - 18z = -1$ are (A) Inconsistent (B) Consistent with unique solution (C) Consistent with countable infinite many solutions (D) Consistent with infinitely many solutions

   ›Reveal solutionSolution

   Eliminating $x$ and $z$ from the three equations produces two contradictory conditions on $y-3z$ ($15$ vs $-1/6$), so the system is inconsistent (no solution).

   Concept and Intuition

   A linear system is inconsistent when the equations, after elimination, reduce to a statement that is never true (like $0=$ nonzero number). This happens when the coefficient matrix is singular (determinant zero) but the augmented matrix has higher rank — i.e., the planes described by the equations don't share a common point.

   Step-by-Step Solution

   1. Write the equations: (1) $2x+6y=-11$; (2) $6x+20y-6z=-3$; (3) $6y-18z=-1$.
   2. Eliminate $x$ between (1) and (2): multiply (1) by 3: $6x+18y=-33$. Subtract from (2): $(6x+20y-6z)-(6x+18y) = -3-(-33)$ $\Rightarrow 2y - 6z = 30 \Rightarrow y - 3z = 15$. Call this Equation A.
   3. Now look at equation (3) directly: $6y - 18z = -1$. Divide by 6: $y - 3z = -\dfrac16$. Call this Equation B.
   4. Equations A and B both express $y-3z$ but give different values: $15$ from A, and $-\tfrac16$ from B. Since $15 \ne -\tfrac16$, no values of $y,z$ can satisfy both simultaneously.
   5. This contradiction means the original system of three equations has no solution — it is inconsistent. …

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5. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
   Q.If the solution for the system of equations $x+2y-z=3,\ 3x-y+2z=1$ and $2x-2y+3z=2$ is $(\alpha,\beta,\gamma)$, then $\alpha^2+\beta^2+\gamma^2 =$ (A) 33 (B) 5 (C) 17 (D) 14

   ›Reveal solutionSolution

   Solving the system gives (x,y,z) = (-1, 4, 4), so the sum of squares is 33.

   Concept and Intuition

   A system of three linear equations in three unknowns can be solved by systematic elimination.

   Step-by-Step Solution

   1. Equations: (i) x+2y-z=3; (ii) 3x-y+2z=1; (iii) 2x-2y+3z=2.
   2. From (i): x = 3-2y+z.
   3. Substitute into (ii): 3(3-2y+z)-y+2z=1 -> -7y+5z=-8 -> 7y-5z=8. (iv)
   4. Substitute into (iii): 2(3-2y+z)-2y+3z=2 -> -6y+5z=-4 -> 6y-5z=4. (v)
   5. (iv)-(v): y=4.
   6. From (v): 6(4)-5z=4 -> z=4.
   7. x = 3-2(4)+4 = -1. …

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6. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
   Q.If $A = \begin{pmatrix}1 & 5 & 3\\ 2 & 4 & 0\\ 3 & -1 & -5\end{pmatrix}$, $B = \begin{pmatrix}-1\\ -2\\ 4\end{pmatrix}$ and $[x\ y\ z]A^{T} = B^{T}$, then $x+y+z=$ (A) $4$ (B) $-2$ (C) $6$ (D) $3$

   ›Reveal solutionSolution

   Transposing the matrix equation converts it into the ordinary linear system A·v = B, which solves to x=6, y=−7/2, z=7/2, giving x+y+z=6.

   Concept and Intuition

   "[x y z]Aᵀ = Bᵀ" is a row-vector equation. Taking the transpose of both sides converts it to the more familiar column form: (v Aᵀ)ᵀ = A vᵀ, and (Bᵀ)ᵀ = B. So the equation is equivalent to A·(x,y,z)ᵀ = B, a standard system of 3 linear equations.

   Step-by-Step Solution

   1. A = [[1,5,3],[2,4,0],[3,−1,−5]], B = (−1,−2,4)ᵀ.
   2. Write the system A(x,y,z)ᵀ = B:
      • x + 5y + 3z = −1
      • 2x + 4y = −2
      • 3x − y − 5z = 4
   3. From equation 2: 2x+4y=−2 ⟹ x+2y=−1 ⟹ x = −1−2y.
   4. Substitute into equation 1: (−1−2y)+5y+3z = −1 ⟹ 3y+3z=0 ⟹ z=−y.
   5. Substitute x and z into equation 3: 3(−1−2y) − y − 5(−y) = 4 ⟹ −3−6y−y+5y = 4 ⟹ −3−2y=4 ⟹ y=−7/2.
   6. Then x = −1−2(−7/2) = 6, and z = −y = 7/2. …

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7. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
   Q.What are the values of $(x,y,z,t)$ where $3\begin{bmatrix} x & y \\ z & t \end{bmatrix} = \begin{bmatrix} x & 6 \\ -1 & 2t \end{bmatrix} + \begin{bmatrix} 4 & x+y \\ z+t & 3 \end{bmatrix} = ?$ (A) $(2,4,3,1)$ (B) $(2,4,1,3)$ (C) $(1,3,2,4)$ (D) $(1,3,4,2)$

   ›Reveal solutionSolution

   Equating corresponding entries on both sides of the matrix equation and solving the resulting simple linear equations gives $(x,y,z,t)=(2,4,1,3)$.

   Concept and Intuition

   Two matrices are equal exactly when every corresponding entry is equal. Setting up the entry-wise equations turns a matrix equation into a small system of linear equations, which can usually be solved one variable at a time by picking the simplest equation first.

   Step-by-Step Solution

   1. Left side: $3\begin{bmatrix}x&y\\z&t\end{bmatrix} = \begin{bmatrix}3x&3y\\3z&3t\end{bmatrix}$.
   2. Right side (sum of the two given matrices): $\begin{bmatrix}x&6\\-1&2t\end{bmatrix}+\begin{bmatrix}4&x+y\\z+t&3\end{bmatrix} = \begin{bmatrix}x+4 & 6+x+y\\ -1+z+t & 2t+3\end{bmatrix}$.
   3. Equate the $(1,1)$ entries: $3x = x+4 \Rightarrow 2x=4 \Rightarrow x=2$.
   4. Equate the $(2,2)$ entries: $3t = 2t+3 \Rightarrow t=3$.
   5. Equate the $(1,2)$ entries: $3y = 6+x+y \Rightarrow 2y = 6+x = 6+2=8 \Rightarrow y=4$. …

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8. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
   Q.If the solution of the system of simultaneous equations $\frac{1}{x}+\frac{2}{y}-\frac{3}{z}-1=0$, $\frac{2}{x}-\frac{4}{y}+\frac{3}{z}-1=0$ and $\frac{3}{x}+\frac{6}{y}-\frac{6}{z}-4=0$ is $x=\alpha, y=\beta, z=\gamma$ then $\alpha^2+\gamma^2 =$ (A) $5\beta$ (B) $\beta^2$ (C) $3\beta$ (D) $2\beta^2$

   ›Reveal solutionSolution

   Substitute $u=1/x,\,v=1/y,\,w=1/z$ to reduce the system to a linear system, solve for $x,y,z$, then evaluate $\alpha^2+\gamma^2$ against $\beta$.

   Concept and Intuition

   The equations are linear in $1/x,1/y,1/z$ even though they look nonlinear in $x,y,z$. Substituting turns this into a standard $3\times3$ linear system, easily solved by elimination.

   Step-by-Step Solution

   1. Let $u=1/x,\ v=1/y,\ w=1/z$. The system becomes: $u+2v-3w=1$ ... (1); $2u-4v+3w=1$ ... (2); $3u+6v-6w=4$ ... (3).
   2. Add (1)+(2): $3u-2v=2$ ... (i).
   3. Compute (3)$-3\times$(1): $(3u+6v-6w)-3(u+2v-3w)=4-3\Rightarrow3w=1\Rightarrow w=\dfrac13$.
   4. Substitute $w=1/3$ into (1): $u+2v-1=1\Rightarrow u+2v=2$ ... (ii).
   5. Add (i)+(ii): $4u=4\Rightarrow u=1$; then from (ii): $2v=1\Rightarrow v=\dfrac12$. …

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9. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.In solving a system of linear equations $AX=B$ by Cramer's rule, in the usual notation, if $\Delta_1=\begin{vmatrix}-11 & 1 & -7\\ -4 & 1 & -2\\ 5 & 1 & 1\end{vmatrix}$ and $\Delta_3=\begin{vmatrix}4 & 1 & -11\\ 1 & 1 & -4\\ 4 & 1 & 5\end{vmatrix}$, then $X=$ (A) $\begin{pmatrix}-1\\ 1\\ 2\end{pmatrix}$ (B) $\begin{pmatrix}2\\ 1\\ -1\end{pmatrix}$ (C) $\begin{pmatrix}1\\ -1\\ 2\end{pmatrix}$ (D) $\begin{pmatrix}1\\ 2\\ -1\end{pmatrix}$

   ›Reveal solutionSolution

   Reconstructing the coefficient matrix and RHS vector from the shared columns of $\Delta_1,\Delta_3$ and computing $\Delta,\Delta_1,\Delta_2,\Delta_3$ gives $X=(1,-1,2)$.

   Concept and Intuition

   In Cramer's rule for $AX=B$, $\Delta_i$ is formed by replacing the $i$-th column of $A$ with $B$, keeping the other columns unchanged. So $\Delta_1$'s columns 2,3 and $\Delta_3$'s columns 1,2 are literally the ORIGINAL matrix $A$'s corresponding columns — comparing the two given determinants lets us recover the full system.

   Step-by-Step Solution

   1. $\Delta_1=\begin{vmatrix}-11&1&-7\\-4&1&-2\\5&1&1\end{vmatrix}$: column 2 = original $A$'s column 2 $=(1,1,1)^T$; column 3 = original $A$'s column 3 $=(-7,-2,1)^T$; column 1 $=B=(-11,-4,5)^T$.
   2. $\Delta_3=\begin{vmatrix}4&1&-11\\1&1&-4\\4&1&5\end{vmatrix}$: column 1 = original $A$'s column 1 $=(4,1,4)^T$; column 2 = same $(1,1,1)^T$ (consistent); column 3 $=B=(-11,-4,5)^T$ (matches $\Delta_1$'s column 1, confirming $B$).
   3. Reconstructed system: $A=\begin{pmatrix}4&1&-7\\1&1&-2\\4&1&1\end{pmatrix}$, $B=\begin{pmatrix}-11\\-4\\5\end{pmatrix}$.
   4. $\Delta=\det(A)=4(1{\cdot}1-(-2){\cdot}1)-1(1{\cdot}1-(-2){\cdot}4)+(-7)(1{\cdot}1-1{\cdot}4)=4(3)-1(9)-7(-3)=12-9+21=24$. …

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10. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
    Q.If $A=\begin{bmatrix}2 & 1 & 3 & -1\\1 & -2 & 2 & -3\end{bmatrix}, B=\begin{bmatrix}2 & 1 & 0 & 3\\1 & -1 & 2 & 3\end{bmatrix}$ and $2A+3B-5C=O$, then $C=$ (A) $\begin{bmatrix}2 & 1 & 6/5 & 7/5\\1 & 7/5 & 2 & 3/5\end{bmatrix}$ (B) $\begin{bmatrix}-2 & 1 & 6/5 & 7/5\\1 & -7/5 & 2 & 3/5\end{bmatrix}$ (C) $\begin{bmatrix}-2 & 1 & 6/5 & 7/5\\1 & 7/5 & 2 & 3/5\end{bmatrix}$ (D) $\begin{bmatrix}2 & 1 & 6/5 & 7/5\\1 & -7/5 & 2 & 3/5\end{bmatrix}$

    ›Reveal solutionSolution

    Direct matrix arithmetic: $C=(2A+3B)/5$, computed entrywise, gives $\begin{bmatrix}2&1&6/5&7/5\\1&-7/5&2&3/5\end{bmatrix}$.

    Concept and Intuition

    This is pure entrywise matrix algebra — scale each matrix, add, then divide by 5 (since $5C=2A+3B$).

    Step-by-Step Solution

    1. $2A = \begin{bmatrix}4&2&6&-2\\2&-4&4&-6\end{bmatrix}$.
    2. $3B = \begin{bmatrix}6&3&0&9\\3&-3&6&9\end{bmatrix}$.
    3. $2A+3B = \begin{bmatrix}10&5&6&7\\5&-7&10&3\end{bmatrix}$.
    4. $C = \dfrac{1}{5}(2A+3B) = \begin{bmatrix}2&1&6/5&7/5\\1&-7/5&2&3/5\end{bmatrix}$. …

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11. AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
    Q.If the system of simultaneous linear equations $x+y-z=6$, $3x-y+z=2$ and $x+ky+z=-8$ has a unique solution $x=2, y=\beta, z=\gamma$ then the value of k satisfies the following quadratic equation (A) $x^2-5x+6=0$ (B) $x^2+x-6=0$ (C) $x^2-x-6=0$ (D) $x^2+x-2=0$

    ›Reveal solutionSolution

    Adding the first two equations forces $x=2$; requiring an integer solution triple gives $k=2$ or $k=-3$, so $k$ satisfies $x^2+x-6=0$.

    Adding the first two equations:

    $(x+y-z)+(3x-y+z)=6+2 \Rightarrow 4x=8 \Rightarrow x=2.$

    From equation 1 with $x=2$: $y-z=4$, i.e. $z=y-4$.

    Substitute into equation 3 ($x+ky+z=-8$):

    $2+ky+(y-4)=-8 \Rightarrow y(k+1)=-6 \Rightarrow y=\dfrac{-6}{k+1}.$ …

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12. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
    Q.The augmented matrix of a nonhomogeneous system of equations $AX = B$, $[A\ B]$ is reduced to the following form after applying a series of elementary row transformations $\begin{bmatrix} 1 & 1 & 1 & 5 \\ 0 & 0 & -3 & 4 \\ 0 & 0 & \mu+1 & \lambda^2 - 2\lambda + 1 \end{bmatrix}$, Then (A) Only for $\mu \neq -1$, $AX = B$ has unique solution (B) Only for $\mu = -1$ and $\lambda = 1$, $AX = B$ has infinite number of solutions (C) For any $\mu$ and for any $\lambda$, $AX = B$ has infinite number of solutions (D) For all positive values of $\mu$, $AX = B$ has no solution

    ›Reveal solutionSolution

    The coefficient matrix always has rank 2 (rows 2 and 3 are parallel), so a unique

    solution is never possible; and for any positive $\mu$, row 3 always contradicts the

    $z$-value fixed by row 2, forcing "no solution". Answer: (D).

    Concept and Intuition

    For $AX=B$ with augmented matrix reduced to

    $\begin{bmatrix} 1 & 1 & 1 & 5 \\ 0 & 0 & -3 & 4 \\ 0 & 0 & \mu+1 & \lambda^2-2\lambda+1 \end{bmatrix}$,

    read off the coefficient rows (ignore the last column) for $x,y,z$: $(1,1,1)$,

    $(0,0,-3)$, $(0,0,\mu+1)$. The second and third rows are both scalar multiples of $(0,0,1)$ for every value of $\mu$ — they can never be linearly independent of each

    other. So $\text{rank}(A) \le 2$ always (it equals 2, from row 1 and row 2, since row

    2 is never the zero vector). Since there are 3 unknowns but rank never reaches 3, a

    unique solution is structurally impossible no matter what $\mu,\lambda$ are.

    Row 2 directly gives $-3z=4 \Rightarrow z=-4/3$, a fixed value. Row 3 imposes a second

    constraint on the same variable $z$: $(\mu+1)z=\lambda^2-2\lambda+1=(\lambda-1)^2$.

    Substituting $z=-4/3$: $-\frac{4}{3}(\mu+1)=(\lambda-1)^2$. For consistency, this

    equality must hold; since the right side $(\lambda-1)^2\ge 0$, we need $-\frac{4}{3} (\mu+1)\ge 0$, i.e. $\mu+1\le 0$, i.e. $\mu \le -1$.

    Step-by-Step Solution

    1. From row 2: $-3z=4 \Rightarrow z=-\dfrac{4}{3}$.
    2. From row 3: $(\mu+1)z=\lambda^2-2\lambda+1=(\lambda-1)^2$.
    3. Substitute: $(\mu+1)\left(-\dfrac{4}{3}\right)=(\lambda-1)^2$.
    4. Check option (A): coefficient-matrix rank is always 2 (rows 2, 3 parallel) — a unique solution needs rank 3, which never happens. So (A) is false.
    5. Check option (D): if $\mu>0$, then $\mu+1>0$, so the left side $-\frac{4}{3}(\mu+1)$ is strictly negative, but the right side $(\lambda-1)^2$ is always $\ge 0$. A negative number can never equal a non-negative number, so the equation in step 3 is never satisfied for any positive $\mu$ — the system is …

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