Q.The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹ 60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is ₹ 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is ₹ 70. Find cost of each item per kg by matrix method.

Showing the 12 most recent of 32 on this concept.

1. 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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2. 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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3. 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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4. 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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5. 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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6. 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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7. 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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8. 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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9. 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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10. 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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11. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
    Q.If $\begin{bmatrix} x & 4 & -1 \end{bmatrix} \begin{bmatrix} 2 & 1 & 0 \\ 1 & 0 & 2 \\ 0 & 2 & 4 \end{bmatrix} \begin{bmatrix} x \\ 4 \\ -1 \end{bmatrix} = 0$, then $x =$ (A) $-1+\sqrt{6}$ (B) $8 \pm \sqrt{5}$ (C) $-2 \pm \sqrt{10}$ (D) $3 \pm \sqrt{6}$

    ›Reveal solutionSolution

    Expanding the quadratic form gives the quadratic equation x^2+4x-6=0, whose roots are -2 +/- sqrt(10).

    Concept and Intuition

    A quadratic form v^T M v expands into a scalar quadratic expression; here only x is unknown, so the result reduces to an ordinary quadratic equation in x.

    Step-by-Step Solution

    1. Compute $M[x,4,-1]^T$: Row1: $2x+4$; Row2: $x-2$; Row3: $8-4=4$.
    2. Dot with $[x,4,-1]$: $x(2x+4)+4(x-2)+(-1)(4) = 2x^2+4x+4x-8-4$.
    3. Simplify: $2x^2+8x-12=0$.
    4. Divide by 2: $x^2+4x-6=0$. …

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12. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
    Q.If $A = \begin{pmatrix}3 & 4\\5 & 6\end{pmatrix}$ and $B = \begin{pmatrix}x & 0\\0 & y\end{pmatrix}$, $x,y\in N$, then (A) There is exactly one such matrix B such that AB = I (B) There is no matrix B such that AB = BA (C) There exist only a finite number of matrices B such that AB = BA (D) There exist infinite number of matrices B such that AB = BA

    ›Reveal solutionSolution

    Multiplying out $AB$ and $BA$ shows they're equal exactly when $x=y$, and since $x,y$ can be any of the infinitely many natural numbers with $x=y$, there are infinitely many commuting diagonal matrices B.

    Concept and Intuition

    For a diagonal matrix $B=\text{diag}(x,y)$ multiplying a general matrix $A$ on the left vs. right scales A's rows vs. columns differently — $AB$ scales A's columns by $x,y$ respectively, and $BA$ scales A's rows by $x,y$ respectively. So $AB=BA$ becomes a condition relating how each off-diagonal entry of $A$ gets scaled from each side, and typically forces the diagonal entries of B to be equal whenever the off-diagonal entries of A are both non-zero (as they are here).

    Step-by-Step Solution

    1. Compute $AB=\begin{pmatrix}3&4\\5&6\end{pmatrix}\begin{pmatrix}x&0\\0&y\end{pmatrix}=\begin{pmatrix}3x&4y\\5x&6y\end{pmatrix}$ (this scales each column of A by x then y).
    2. Compute $BA=\begin{pmatrix}x&0\\0&y\end{pmatrix}\begin{pmatrix}3&4\\5&6\end{pmatrix}=\begin{pmatrix}3x&4x\\5y&6y\end{pmatrix}$ (this scales each row of A by x then y).
    3. Set $AB=BA$ entrywise: $(1,1)$: $3x=3x$ (always true); $(1,2)$: $4y=4x\Rightarrow x=y$; $(2,1)$: $5x=5y\Rightarrow x=y$ (same condition); $(2,2)$: $6y=6y$ (always true).
    4. So the only requirement is $x=y$, with $x,y\in\mathbb{N}$. Since $x$ can be $1,2,3,\dots$ (infinitely many choices, each giving a valid B with $x=y$), there are infinitely many such matrices B. …

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