Q.Solve the system of equations $2x + 5y = 1$, $3x + 2y = 7$.

Showing the 12 most recent of 32 on this concept.

1. 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): $(1+3y-2z)+4y-3z=5 \Rightarrow 1+7y-5z=5 \Rightarrow 7y-5z=4$.
   6. Steps 4 and 5 both compute $7y-5z$ but give different values ($5$ vs $4$) — this is a direct contradiction ($5\ne4$), so no $(y,z)$ (hence no $(x,y,z)$) can satisfy the system simultaneously.
   7. Therefore the system has no solution (0 solutions), not infinitely many.

   Common Mistakes

   • Concluding "determinant $=0$ $\Rightarrow$ infinite solutions" without checking consistency — zero determinant only rules out a unique solution, it doesn't decide between 0 and $\infty$.
   • Arithmetic slips while eliminating $x$ — always redo the elimination from a second pair of equations as a check, as done here.

   ✓Final answer

   The correct option is (B) — 0.

   ANSWER: B

2. 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): $2x+5=-1 \Rightarrow 2x=-6 \Rightarrow x=-3$.
   6. So $\alpha=x=-3$ and $\beta=y=5$; then $\alpha+\beta = -3+5=2$.
   7. (Check: $z=x+y-6=-3+5-6=-4$; verify (2): $3(-3)+2(5)-(-4)=-9+10+4=5$ ✓.)

   Common Mistakes

   • Sign errors when rewriting $2x-y-2z+3=0$ as $2x-y-2z=-3$.
   • Mixing up $\alpha,\beta,\gamma$ with $x,y,z$ — the question asks for $\alpha+\beta$, i.e. $x+y$, not $x+y+z$.

   ✓Final answer

   The correct option is (B) — $2$.

   ANSWER: B

3. 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.
   8. Verify in (ii): 3(-1)-4+2(4)=1 checks. Verify in (iii): 2(-1)-2(4)+3(4)=2 checks.
   9. (alpha,beta,gamma)=(-1,4,4), sum of squares = 1+16+16=33.

   Common Mistakes

   • Sign errors during elimination.
   • Forgetting to verify the solution in all three equations.

   ✓Final answer

   The correct option is (A) — 33.

   ANSWER: A

4. 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.
   6. Since no values of $y,z$ (and hence no $x$) can satisfy the system simultaneously, the system has no solution.

   Common Mistakes

   • Seeing that the coefficient determinant is zero and jumping straight to "infinite solutions" — a zero determinant only rules out a unique solution; you still must check consistency to distinguish "infinite solutions" from "no solution."
   • Arithmetic slips while eliminating $x$ — it's worth double-checking both substitutions independently since the whole conclusion hinges on the two derived equations genuinely conflicting.

   ✓Final answer

   The correct option is (B) — no solution.

   ANSWER: B

5. 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): $2v=1\Rightarrow v=\dfrac12$.
   6. So $x=1/u=1=\alpha$, $y=1/v=2=\beta$, $z=1/w=3=\gamma$.
   7. $\alpha^2+\gamma^2=1^2+3^2=1+9=10$. Since $\beta=2$, $5\beta=10$ — matches.

   Common Mistakes

   • Arithmetic slips while eliminating variables in the linear system.
   • Forgetting to invert back from $u,v,w$ to $x,y,z=\alpha,\beta,\gamma$ at the end.

   ✓Final answer

   The correct option is (A) — $5\beta$.

   ANSWER: A

6. 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. $C = \dfrac{1}{5}(2A+3B) = \begin{bmatrix}2&1&6/5&7/5\\1&-7/5&2&3/5\end{bmatrix}$.
   5. This matches option (D) exactly (note the $-7/5$ in row 2, column 2).

   Common Mistakes

   • Sign slip on the $-7/5$ entry (options A and C show it as $+7/5$, and options B/C show a sign error on the leading $2$).

   ✓Final answer

   The correct option is (D).

   ANSWER: D

7. 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. $\Delta=\det(A)=4(1{\cdot}1-(-2){\cdot}1)-1(1{\cdot}1-(-2){\cdot}4)+(-7)(1{\cdot}1-1{\cdot}4)=4(3)-1(9)-7(-3)=12-9+21=24$.
   5. Given $\Delta_1=24$ (verified by direct expansion) $\Rightarrow x=\Delta_1/\Delta=24/24=1$.
   6. $\Delta_2$ (col 2 -> B): $\begin{vmatrix}4&-11&-7\\1&-4&-2\\4&5&1\end{vmatrix}=-24 \Rightarrow y=-24/24=-1$.
   7. Given $\Delta_3=48$ (verified) $\Rightarrow z=48/24=2$.
   8. $X=(x,y,z)=(1,-1,2)$.

   Common Mistakes

   • Not realizing $\Delta_2$ (and even $\Delta$ itself) must be reconstructed from the shared structure of $\Delta_1,\Delta_3$ rather than assumed given.
   • Mixing up which column of $\Delta_1/\Delta_3$ corresponds to $B$ versus the original matrix's columns.

   ✓Final answer

   The correct option is (C) — $\begin{pmatrix}1\\-1\\2\end{pmatrix}$.

   ANSWER: C

8. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.While solving a system of linear equations $AX = B$ using Cramer's rule with the usual notation, if $\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 2 & -1 & 2 \\ -1 & 1 & 5 \end{vmatrix}$; $\Delta_1 = \begin{vmatrix} 5 & 1 & 1 \\ 4 & -1 & 2 \\ 11 & 1 & 5 \end{vmatrix}$ and $X = \begin{bmatrix} \alpha \\ 2 \\ \beta \end{bmatrix}$, then $\alpha^2 + \beta^2 =$ (A) 9 (B) 13 (C) 5 (D) 25

   ›Reveal solutionSolution

   This tests Cramer's rule bookkeeping: computing $\alpha$ directly from $\Delta_1/\Delta$, then recovering $\beta$ by reconstructing the right-hand-side vector $B$ implicit in $\Delta_1$ and using the known value of $y=2$.

   Concept and Intuition

   In Cramer's rule for $AX=B$ with $A$ a $3\times3$ coefficient matrix and $X=(\alpha, y, \beta)^T$: $\Delta = \det(A)$, and $\Delta_1$ is the determinant formed by replacing the first column of $A$ (the column of coefficients of $\alpha$) with $B$. So $\alpha = \Delta_1/\Delta$ directly. Moreover, since $\Delta_1$'s 2nd and 3rd columns are unchanged from $A$'s, we can read the vector $B$ straight off $\Delta_1$'s first column. Once $B$ is known, and given $y=2$ is already provided, we can plug into the original equations $AX=B$ (using $A$'s actual rows) to solve for the remaining unknown $\beta$.

   Step-by-Step Solution

   1. Compute $\Delta = \begin{vmatrix}1&1&1\\2&-1&2\\-1&1&5\end{vmatrix}$. Expanding along row 1: $1[(-1)(5)-(2)(1)] - 1[(2)(5)-(2)(-1)] + 1[(2)(1)-(-1)(-1)] = 1(-7) - 1(12) + 1(1) = -7-12+1 = -18$.
   2. Compute $\Delta_1 = \begin{vmatrix}5&1&1\\4&-1&2\\11&1&5\end{vmatrix}$. Expanding along row 1: $5[(-1)(5)-(2)(1)] - 1[(4)(5)-(2)(11)] + 1[(4)(1)-(-1)(11)] = 5(-7) - 1(-2) + 1(15) = -35+2+15=-18$.
   3. By Cramer's rule, $\alpha = \Delta_1/\Delta = (-18)/(-18) = 1$.
   4. Since $\Delta_1$'s columns 2 and 3 match $A$'s columns 2 and 3 exactly (compare: $A$'s column 2 is $(1,-1,1)^T$ and column 3 is $(1,2,5)^T$, matching $\Delta_1$), $\Delta_1$'s column 1, $(5,4,11)^T$, must be the RHS vector $B$.
   5. Now use $A$'s actual rows with $X=(\alpha,y,\beta)=(1,2,\beta)$ and $B=(5,4,11)$: Row 1: $1(1)+1(2)+1(\beta) = 5 \Rightarrow 3+\beta=5 \Rightarrow \beta=2$.
   6. Verify with row 2: $2(1)+(-1)(2)+2(\beta) = 2-2+2(2) = 4$ ✓ (matches $B_2=4$). Verify with row 3: $-1(1)+1(2)+5(\beta) = -1+2+10=11$ ✓ (matches $B_3=11$). Both check out, confirming $\beta=2$.
   7. Finally, $\alpha^2+\beta^2 = 1^2+2^2 = 1+4=5$.

   Common Mistakes

   • Forgetting that $\Delta_1$ replaces the column corresponding to the first unknown ($\alpha$), not the second ($y$) — this determines which column of $\Delta_1$ actually represents $B$.
   • Arithmetic slips in the $3\times3$ determinant expansions (sign errors are especially common).
   • Trying to find $\beta$ without first identifying the actual right-hand-side vector $B$ from $\Delta_1$ — this reconstruction step is the crux of the problem.

   ✓Final answer

   The correct option is (C) — 5.

   ANSWER: C

9. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.If $A = \begin{bmatrix} 1 & 2 \\ -2 & -5 \end{bmatrix}$ and $\alpha A^2 + \beta A = 2I$ for some $\alpha, \beta \in \mathbb{R}$ then $\alpha + \beta =$ (A) 7 (B) 10 (C) 12 (D) 5

   ›Reveal solutionSolution

   Computing $A^2$ and matching the matrix equation $\alpha A^2+\beta A=2I$ entry-by-entry gives $\alpha=2,\ \beta=8$, so $\alpha+\beta=10$ — option (B).

   Concept and Intuition

   Given a $2\times2$ matrix satisfying a polynomial relation like $\alpha A^2+\beta A = 2I$, the cleanest approach is to compute $A^2$ directly, then equate corresponding entries of both sides of the matrix equation — this converts a single matrix equation into a small system of linear equations in the unknown scalars $\alpha,\beta$ (using just two independent entries is enough, and the remaining entries serve as a consistency check, since the relation must hold for the whole matrix, not just isolated numbers — this consistency is itself guaranteed for genuine matrix polynomial identities via Cayley–Hamilton-type reasoning, but verifying arithmetic keeps you safe under exam conditions).

   Step-by-Step Solution

   1. Compute $A^2$: $A^2=\begin{bmatrix}1&2\\-2&-5\end{bmatrix}\begin{bmatrix}1&2\\-2&-5\end{bmatrix} = \begin{bmatrix}1(1)+2(-2) & 1(2)+2(-5)\\-2(1)+(-5)(-2) & -2(2)+(-5)(-5)\end{bmatrix} = \begin{bmatrix}-3&-8\\8&21\end{bmatrix}$.
   2. Write $\alpha A^2+\beta A = 2I$ entry-wise:
      • $(1,1)$: $-3\alpha+\beta = 2$
      • $(1,2)$: $-8\alpha+2\beta = 0$
      • $(2,1)$: $8\alpha-2\beta=0$ (same equation as above)
      • $(2,2)$: $21\alpha-5\beta=2$
   3. From $(1,2)$: $-8\alpha+2\beta=0 \Rightarrow \beta = 4\alpha$.
   4. Substitute into $(1,1)$: $-3\alpha+4\alpha=2 \Rightarrow \alpha=2$, hence $\beta=4(2)=8$.
   5. Verify with $(2,2)$: $21(2)-5(8)=42-40=2$ ✓ — consistent.
   6. $\alpha+\beta=2+8=10$.

   Common Mistakes

   • Sign errors multiplying the negative entries when computing $A^2$.
   • Forgetting to verify with a third equation, risking an arithmetic slip going unnoticed.

   ✓Final answer

   The correct option is (B) — $\alpha+\beta=10$.

   ANSWER: B

10. 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: $3y = 6+x+y \Rightarrow 2y = 6+x = 6+2=8 \Rightarrow y=4$.
    6. Equate the $(2,1)$ entries: $3z = -1+z+t \Rightarrow 2z = -1+t = -1+3=2 \Rightarrow z=1$.
    7. So $(x,y,z,t) = (2,4,1,3)$.

    Common Mistakes

    • Solving for $z$ or $y$ before pinning down $x$ and $t$ first — since $y$'s and $z$'s equations both depend on $x$ and $t$ respectively, tackling $x$ and $t$ first (the two equations that are self-contained) is the efficient order.
    • Mixing up which entry is which after adding the two matrices on the right.

    ✓Final answer

    The correct option is (B) — $(2,4,1,3)$.

    ANSWER: B

11. 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.
    6. (This can be cross-checked: the coefficient matrix has determinant $0$, since row (3) is proportional to the "$y,z$-part" derived from eliminating (1),(2) — but the constants don't match up, which is exactly the signature of an inconsistent system rather than infinitely many solutions.)

    Common Mistakes

    • Stopping after finding determinant of the coefficient matrix is zero and concluding "infinitely many solutions" — a zero determinant only means either no solution or infinitely many; one must check whether the constants are consistent (as done above) to distinguish the two cases.

    ✓Final answer

    The correct option is (A) — Inconsistent.

    ANSWER: A

12. 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 ($x+ky+z=-8$):

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

    A unique solution needs the coefficient determinant $-4(k+1)\neq 0$, i.e. $k\neq -1$. For the solution $(x,y,z)$ to be an integer triple (consistent with the given integer $x=2$), $y=\dfrac{-6}{k+1}$ must be an integer, and the admissible values are

    • $k=2:\ (x,y,z)=(2,-2,-6)$ ✓
    • $k=-3:\ (x,y,z)=(2,3,-1)$ ✓

    Both values $k=2,\,-3$ are the roots of

    $(x-2)(x+3)=x^2+x-6=0.$

    ✓Final answer

    $k$ satisfies $x^2+x-6=0$ (roots $2$ and $-3$) — option (B).