Q.The sum of three numbers is 6. If we multiply third number by 3 and add second number to it, we get 11. By adding first and third numbers, we get double of the second number. Represent it algebraically and find the numbers using matrix method.

Showing the 12 most recent of 32 on this concept.

1. 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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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): $2x+5=-1 \Rightarrow 2x=-6 \Rightarrow x=-3$. …

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3. 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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4. 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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5. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
   Q.If $A+2B=\begin{pmatrix}1 & 2 & 0\\ 6 & -3 & 3\\ -5 & 3 & 1\end{pmatrix}$ and $2A-B=\begin{pmatrix}2 & -1 & 5\\ 2 & -1 & 6\\ 0 & 1 & 2\end{pmatrix}$, then $\mathrm{Tr}[A]-\mathrm{Tr}[B]=$ (A) $1$ (B) $2$ (C) $3$ (D) $4$

   ›Reveal solutionSolution

   Trace is linear, so instead of solving for $A$ and $B$ as full matrices, take the trace of each given equation and solve the resulting $2\times2$ linear system.

   Concept and Intuition

   Trace (sum of diagonal entries) is a linear functional: $\mathrm{Tr}[A+2B]=\mathrm{Tr}[A]+2\mathrm{Tr}[B]$. So applying trace to both matrix equations converts a matrix problem into a scalar linear system in $a=\mathrm{Tr}[A]$ and $b=\mathrm{Tr}[B]$.

   Step-by-Step Solution

   1. Trace of first matrix: $1+(-3)+1=-1$. So $a+2b=-1$.
   2. Trace of second matrix: $2+(-1)+2=3$. So $2a-b=3$.
   3. From equation 1: $a=-1-2b$.
   4. Substitute into equation 2: $2(-1-2b)-b=3 \Rightarrow -2-4b-b=3 \Rightarrow -5b=5 \Rightarrow b=-1$.
   5. Then $a=-1-2(-1)=1$. …

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6. 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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7. 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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8. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
   Q.If $A = \begin{bmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{bmatrix}$, then $A + A^3 + A^4 + A^5 + 3I =$ (A) $\begin{bmatrix} 4 & 2 & 1 \\ 2 & 5 & 6 \\ -3 & 2 & 3 \end{bmatrix}$ (B) $\begin{bmatrix} 4 & 1 & 3 \\ 5 & 5 & 6 \\ -2 & -1 & 0 \end{bmatrix}$ (C) $\begin{bmatrix} 3 & 1 & 4 \\ 3 & 1 & -2 \\ -1 & 2 & -1 \end{bmatrix}$ (D) $\begin{bmatrix} 4 & 1 & 3 \\ 2 & 3 & 5 \\ -3 & -2 & -3 \end{bmatrix}$

   ›Reveal solutionSolution

   A's characteristic polynomial is lambda^3=0, so by Cayley-Hamilton A^3=0 (hence A^4=A^5=0 too), collapsing the sum to A+3I, which matches option B.

   Concept and Intuition

   Rather than computing $A^3,A^4,A^5$ by repeated multiplication, use the Cayley-Hamilton theorem: every square matrix satisfies its own characteristic equation. If the characteristic polynomial turns out to be $\lambda^3=0$, we immediately get $A^3=0$, collapsing the higher powers.

   Step-by-Step Solution

   1. $\mathrm{tr}(A) = 1+2+(-3) = 0$.
   2. $\det(A) = 1[(2)(-3)-(6)(-1)] - 1[(5)(-3)-(6)(-2)] + 3[(5)(-1)-(2)(-2)] = 1(0) - 1(-3) + 3(-1) = 0+3-3=0$.
   3. Sum of principal 2x2 minors: $M_{11}=\begin{vmatrix}2&6\\-1&-3\end{vmatrix}=0$; $M_{22}=\begin{vmatrix}1&3\\-2&-3\end{vmatrix}=3$; $M_{33}=\begin{vmatrix}1&1\\5&2\end{vmatrix}=-3$. Sum $=0$. …

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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): $AD = C^{-1}D$. …

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10. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
    Q.Let $A=\begin{pmatrix}0\\-6\\8\end{pmatrix}$, $B=\begin{pmatrix}3 & 5 & -7\\0 & -1 & 8\\6 & -1 & 0\end{pmatrix}$ and $X=\begin{pmatrix}x\\y\\z\end{pmatrix}$. If $D=[\alpha\ \beta\ \gamma]^T$ is the solution of $X^TB^T=A^T$, then $D^TA=$ (A) 0 (B) 4 (C) $-2$ (D) 6

    ›Reveal solutionSolution

    This tests matrix-equation manipulation via transposes: $X^TB^T=A^T$ is just $(BX)^T=A^T$, i.e. $BX=A$. Solve the resulting linear system, then take the dot product $D^TA$. Answer: 4.

    Concept and Intuition

    The transpose of a product reverses order: $(BX)^T = X^TB^T$. So the given equation $X^TB^T = A^T$ is really $(BX)^T = A^T$, and taking the transpose of both sides gives $BX = A$ — an ordinary linear system for the unknown column $X=(x,y,z)^T$. Once $X=D$ is found, $D^TA$ is just the plain dot product of two column vectors.

    Step-by-Step Solution

    1. Transpose: $X^TB^T=A^T \Rightarrow (BX)^T=A^T \Rightarrow BX=A$.
    2. Write $B=\begin{pmatrix}3&5&-7\\0&-1&8\\6&-1&0\end{pmatrix}$, $A=\begin{pmatrix}0\\-6\\8\end{pmatrix}$. The system is: $3x+5y-7z=0$ $-y+8z=-6$ $6x-y=8$
    3. From the second equation: $y=8z+6$.
    4. Substitute into the third: $6x-(8z+6)=8 \Rightarrow 6x-8z=14 \Rightarrow 3x-4z=7 \Rightarrow x=\dfrac{7+4z}{3}$. …

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11. 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: $1(1)+1(2)+1(\beta) = 5 \Rightarrow 3+\beta=5 \Rightarrow \beta=2$. …

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12. 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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