Q.If $\begin{bmatrix}xy & 4\\ z+6 & x+y\end{bmatrix} = \begin{bmatrix}8 & w\\ 0 & 6\end{bmatrix}$, then find values of $x$, $y$, $z$ and $w$.

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 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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3. 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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4. 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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5. 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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6. 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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7. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
   Q.$A = \begin{bmatrix} a^{2} & 15 & 31 \\ 12 & b^{2} & 41 \\ 35 & 61 & c^{2} \end{bmatrix}$ and $B = \begin{bmatrix} 2a & 3 & 5 \\ 2 & 2b & 8 \\ 1 & 4 & 2c-3 \end{bmatrix}$ are two matrices such that the sum of the principal diagonal elements of both $A$ and $B$ are equal, then the product of the principal diagonal elements of $B$ is ________ (A) 4 (B) 0 (C) $-4$ (D) $-12$

   ›Reveal solutionSolution

   Equating the traces of A and B forces $(a-1)^2+(b-1)^2+(c-1)^2=0$, so $a=b=c=1$; substituting into B's diagonal product gives $2\times2\times(-1)=-4$.

   Concept and Intuition

   When an equation reduces to a sum of squares equal to zero, each square must individually be zero (since squares of real numbers are never negative) — this is a very common and powerful trick to pin down exact values of multiple unknowns from a single scalar equation.

   Step-by-Step Solution

   1. Trace (sum of principal diagonal elements) of $A$: $a^2+b^2+c^2$.
   2. Trace of $B$: $2a+2b+(2c-3)$.
   3. Given these are equal: $a^2+b^2+c^2 = 2a+2b+2c-3$.
   4. Rearrange: $a^2-2a + b^2-2b + c^2-2c+3 = 0$.
   5. Complete the square for each variable: $(a^2-2a+1) + (b^2-2b+1) + (c^2-2c+1) = 0$, i.e., $(a-1)^2+(b-1)^2+(c-1)^2 = 0$.
   6. Since each square term is $\ge 0$ and they sum to zero, each must be exactly zero: $a=1,\ b=1,\ c=1$. …

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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 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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10. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
    Q.If $x^a y^b=e^m$, $x^c y^d=e^n$, $\Delta_1=\begin{vmatrix}m & b\\ n & d\end{vmatrix}$, $\Delta_2=\begin{vmatrix}a & m\\ c & n\end{vmatrix}$, $\Delta_3=\begin{vmatrix}a & b\\ c & d\end{vmatrix}$, then the values of $x$ and $y$ are respectively ($e$ is the base of natural logarithm) (A) $\dfrac{\Delta_1}{\Delta_3}$ and $\dfrac{\Delta_2}{\Delta_3}$ (B) $\dfrac{\Delta_2}{\Delta_1}$ and $\dfrac{\Delta_3}{\Delta_1}$ (C) $\log\left(\dfrac{\Delta_1}{\Delta_3}\right)$ and $\log\left(\dfrac{\Delta_2}{\Delta_3}\right)$ (D) $e^{\Delta_1/\Delta_3}$ and $e^{\Delta_2/\Delta_3}$

    ›Reveal solutionSolution

    Taking natural logs turns the exponential system into a linear system in $u=\log x,\ v=\log y$, which Cramer's rule solves directly in terms of the given determinants. Exponentiating back gives $x,y$ in terms of $\Delta_1,\Delta_2,\Delta_3$.

    Concept and Intuition

    Whenever variables appear as exponents (here $x^a y^b = e^m$), taking logarithms converts a nasty exponential relation into an ordinary linear system — a standard trick. Once linear, Cramer's rule (using $2\times2$ determinants) gives a clean closed-form solution, and we just need to recognise that the given $\Delta_1,\Delta_2,\Delta_3$ are exactly the Cramer's-rule determinants for this linear system.

    Step-by-Step Solution

    1. Take $\log$ of both given equations:
       • $x^ay^b=e^m \Rightarrow a\log x + b\log y = m$
       • $x^cy^d=e^n \Rightarrow c\log x + d\log y = n$
    2. Substitute $u=\log x,\ v=\log y$: the system becomes

    $au+bv=m,\qquad cu+dv=n.$

    3. By Cramer's rule, with coefficient determinant $\Delta_3=\begin{vmatrix}a&b\\c&d\end{vmatrix}$ (nonzero, assumed):

    $u=\dfrac{\begin{vmatrix}m&b\\n&d\end{vmatrix}}{\Delta_3},\qquad v=\dfrac{\begin{vmatrix}a&m\\c&n\end{vmatrix}}{\Delta_3}.$

    4. Compare with the given $\Delta_1=\begin{vmatrix}m&b\\n&d\end{vmatrix}$ and $\Delta_2=\begin{vmatrix}a&m\\c&n\end{vmatrix}$ — these are exactly the numerator determinants above. …

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11. 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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12. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
    Q.$A=\begin{bmatrix}0 & 1 & 2\\ 2 & 3 & 0\\ 4 & 0 & 3\end{bmatrix}$ and B is a matrix such that $AB=BA$. If AB is not an identity matrix, then the matrix that can be taken as B is (A) $\begin{bmatrix}-9 & -3 & 6\\ -6 & 8 & -4\\ 12 & -4 & -2\end{bmatrix}$ (B) $\begin{bmatrix}9 & -3 & 6\\ -6 & 8 & -4\\ -12 & -4 & 2\end{bmatrix}$ (C) $\begin{bmatrix}9 & -3 & -6\\ -6 & 8 & -4\\ -12 & 4 & -2\end{bmatrix}$ (D) $\begin{bmatrix}9 & -3 & -6\\ -6 & -8 & 4\\ -12 & 4 & -2\end{bmatrix}$

    ›Reveal solutionSolution

    Direct computation is the only reliable way here — multiplying $A$ by each candidate $B$ shows option (D) is the unique one where $AB=BA$ (in fact $AB=BA=-30I$, a non-identity scalar matrix), so it satisfies every condition in the question.

    Concept and Intuition

    With no special structure making $B$ an obvious polynomial in $A$, the only certain way to check "does $B$ commute with $A$" among four numerically given candidates is to actually multiply both products, $AB$ and $BA$, and compare entry by entry. The question's extra clause — "if $AB$ is not an identity matrix" — is there to rule out a trivial commuting case ($B=A^{-1}$ scaled so $AB=I$), so once we find the candidate that truly commutes, we should double check its product isn't simply $I$.

    Step-by-Step Solution

    1. $A=\begin{bmatrix}0&1&2\\2&3&0\\4&0&3\end{bmatrix}$.
    2. Testing option (D), $B=\begin{bmatrix}9&-3&-6\\-6&-8&4\\-12&4&-2\end{bmatrix}$:
    3. Compute $AB$ row by row: row 1 of $A$ is $(0,1,2)$, so $(AB)_{1j}=0\cdot B_{1j}+1\cdot B_{2j}+2\cdot B_{3j}$, giving $(AB)_{1,\cdot}=(0-6-24,\,0-8+8,\,0+4-4)=(-30,0,0)$. Similarly rows 2 and 3 give $(0,-30,0)$ and $(0,0,-30)$.
    4. So $AB=\begin{bmatrix}-30&0&0\\0&-30&0\\0&0&-30\end{bmatrix}=-30I$.
    5. Compute $BA$ the same way (row of $B$ dotted into columns of $A$): it also comes out to $-30I$. …

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