Q.Solve the equation for $x, y, z$ and $t$, if $2 \begin{bmatrix} x & z \\ y & t \end{bmatrix} + 3 \begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix} = 3 \begin{bmatrix} 3 & 5 \\ 4 & 6 \end{bmatrix}$.

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-2023-05-18-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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3. 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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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 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 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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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. …

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8. 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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9. 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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10. 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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11. 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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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): $2v=1\Rightarrow v=\dfrac12$. …

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