Q.Examine the consistency of the following system of equations: $x + 2y = 2$ $2x + 3y = 3$

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 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

3. 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

4. AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
   Q.The number of solutions of the following system of linear homogenous equations $x - y + z = 0,\; x + 2y - z = 0,\; 2x + y + 3z = 0$ is ________ (A) 1 (B) 8 (C) Countable infinite (D) Uncountable

   ›Reveal solutionSolution

   A homogeneous linear system has only the trivial solution when its coefficient determinant is non-zero. Answer: (A).

   Concept and Intuition

   For a homogeneous system $A\mathbf{x}=\mathbf{0}$, if $\det(A)\neq0$, the only solution is $\mathbf{x}=\mathbf{0}$ (a unique, single solution). If $\det(A)=0$, infinitely many solutions exist.

   Step-by-Step Solution

   1. Coefficient matrix: $\begin{bmatrix}1&-1&1\\1&2&-1\\2&1&3\end{bmatrix}$.
   2. Expand along the first row: $1(2\cdot3-(-1)\cdot1) - (-1)(1\cdot3-(-1)\cdot2) + 1(1\cdot1-2\cdot2)$.
   3. $= 1(6+1) + 1(3+2) + 1(1-4) = 7+5-3 = 9$.
   4. Since $\det=9\neq0$, the system has only the trivial solution $x=y=z=0$ — exactly one solution.

   Common Mistakes

   • Assuming a homogeneous system automatically has infinitely many solutions (true only when the determinant is zero).

   ✓Final answer

   The correct option is (A) — 1.

   ANSWER: A

5. 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

6. 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).

7. 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

8. 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 inconsistent, giving no solution, for every positive $\mu$ and every $\lambda$.
   6. Check (B): setting $\mu=-1,\lambda=1$ does give infinitely many solutions, but it is not the only such pair (e.g. $\mu=-2$ with $(\lambda-1)^2=4/3$ also works), so the word "Only" makes (B) false.
   7. Check (C): infinite solutions require the specific relation $-\frac{4}{3}(\mu+1)=(\lambda-1)^2$, not any $\mu,\lambda$ — so (C) is false.
   8. (D) is the only universally true statement.

   Common Mistakes

   • Assuming a $3\times 3$-looking coefficient matrix automatically allows a unique solution — always check the actual rank; here rows 2 and 3 collapse to the same direction regardless of $\mu$.
   • Missing that row 3 constrains the same variable $z$ that row 2 already fixed, rather than introducing a genuinely new equation.

   ✓Final answer

   The correct option is (D) — For all positive values of $\mu$, $AX=B$ has no solution.

   ANSWER: D

9. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.Consider two systems of 3 linear equations in 3 unknowns $AX=B$ and $CX=D$. If $AX=B$ has unique solution $D$ and $CX=D$ has unique solution $B$, then the solution of $(A-C^{-1})X=O$ is (A) B (B) D (C) B+D (D) B-D

   ›Reveal solutionSolution

   This tests translating "$X=D$ solves $AX=B$" and "$X=B$ solves $CX=D$" into matrix equations and combining them algebraically.

   Concept and Intuition

   "$AX=B$ has unique solution $D$" is just a restatement that plugging $X=D$ into the system satisfies it: $AD=B$. Likewise "$CX=D$ has unique solution $B$" means $CB=D$. The question is which of the listed vectors satisfies the new homogeneous system $(A-C^{-1})X=O$; the trick is to express everything in terms of $D$ and $B$ and see what cancels.

   Step-by-Step Solution

   1. From "$AX=B$ has unique solution $D$": $AD = B$. — (i)
   2. From "$CX=D$ has unique solution $B$": $CB = D$. Multiply both sides by $C^{-1}$: $B = C^{-1}D$. — (ii)
   3. Substitute (ii) into (i): $AD = C^{-1}D$.
   4. Rearranging: $AD - C^{-1}D = O \Rightarrow (A-C^{-1})D = O$.
   5. This says exactly that $X=D$ satisfies $(A-C^{-1})X=O$.

   Common Mistakes

   • Confusing which vector solves which system (mixing up $B$ and $D$'s roles).
   • Trying to invert $(A-C^{-1})$ directly instead of substituting the two given relations.

   ✓Final answer

   The correct option is (B) — D.

   ANSWER: B

10. 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

11. AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
    Q.Let $A = \begin{bmatrix} 1 & 3 \\ 4 & -3 \end{bmatrix}$. Let $S = \left\{ \begin{bmatrix} x \\ y \end{bmatrix} \in R^2 \;/\; A\begin{bmatrix} x \\ y \end{bmatrix} = 3\begin{bmatrix} x \\ y \end{bmatrix} \right\}$ what is the cardinality of $S$? (A) 1 (B) Countably infinite (C) $|S| > 1$ but $S$ is finite (D) Uncountable

    ›Reveal solutionSolution

    $S$ is the eigenspace for eigenvalue $3$; since $\det(A-3I)=0$, it is a nontrivial line through the origin containing infinitely (uncountably) many vectors. Answer: (D).

    Concept and Intuition

    The condition $A\mathbf{v}=3\mathbf{v}$ means $\mathbf{v}$ is an eigenvector of $A$ for eigenvalue $3$ (or the zero vector). The solution set to $(A-3I)\mathbf{v}=0$ is either just $\{0\}$ (if $3$ is not an eigenvalue) or an entire subspace (a line, in 2D) if it is.

    Step-by-Step Solution

    1. $A - 3I = \begin{bmatrix}1-3 & 3\\ 4 & -3-3\end{bmatrix} = \begin{bmatrix}-2&3\\4&-6\end{bmatrix}$.
    2. $\det(A-3I) = (-2)(-6)-(3)(4) = 12-12=0$.
    3. Since the determinant is zero, $3$ IS an eigenvalue of $A$, so the null space is not just $\{0\}$ — it is a full 1-dimensional line through the origin in $\mathbb{R}^2$.
    4. A line in $\mathbb{R}^2$ (all real scalar multiples of an eigenvector) contains uncountably many points.

    Common Mistakes

    • Assuming a determinant of zero always gives "no solution" — for a homogeneous system it instead gives infinitely (here, uncountably) many solutions.

    ✓Final answer

    The correct option is (D) — Uncountable.

    ANSWER: D

12. 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