Q.Find the values of $x$ and $y$ from the following equation: $2\begin{bmatrix} x & 5 \\ 7 & y-3 \end{bmatrix} + \begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix}$

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: $3y = 6+x+y \Rightarrow 2y = 6+x = 6+2=8 \Rightarrow y=4$. …

   🔒 Start your 14-day free trial to see the full solution →
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: $x^2+4x-6=0$. …

   🔒 Start your 14-day free trial to see the full solution →
3. 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 $a=-1-2(-1)=1$. …

   🔒 Start your 14-day free trial to see the full solution →
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. …

   🔒 Start your 14-day free trial to see the full solution →
5. 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. $C = \dfrac{1}{5}(2A+3B) = \begin{bmatrix}2&1&6/5&7/5\\1&-7/5&2&3/5\end{bmatrix}$. …

   🔒 Start your 14-day free trial to see the full solution →
6. 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. …

   🔒 Start your 14-day free trial to see the full solution →
7. 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): $2x+5=-1 \Rightarrow 2x=-6 \Rightarrow x=-3$. …

   🔒 Start your 14-day free trial to see the full solution →
8. 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 ($x+ky+z=-8$):

   $2+ky+(y-4)=-8 \Rightarrow y(k+1)=-6 \Rightarrow y=\dfrac{-6}{k+1}.$ …

   🔒 Start your 14-day free trial to see the full solution →
9. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.If $A = \begin{bmatrix} c & -a & b \\ a & b & -c \\ -b & c & a \end{bmatrix}$, $B = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 2 \\ 1 & 2 & 0 \end{bmatrix}$ and $AB = \begin{bmatrix} 4 & 4 & -2 \\ 1 & 1 & 10 \\ -1 & 5 & -4 \end{bmatrix}$, then $a^2 + b^2 + c^2 =$ (A) $14$ (B) $17$ (C) $11$ (D) $19$

   ›Reveal solutionSolution

   The key idea is to interpret the given matrix equation $AB = C$ as a system of linear equations in $a, b, c$ by comparing entries. Solving yields $a=2, b=1, c=3$, so $a^2+b^2+c^2 = 14$, which corresponds to option (A).

   We are given two matrices $A$ and $B$, and their product $AB$. The matrix $A$ contains the unknowns $a, b, c$. Instead of trying to invert $B$ directly (which is possible but messy), we can use the fact that matrix multiplication gives us explicit equations for each entry of the product. Since $B$ is known, each entry of $AB$ is a linear combination of $a, b, c$. By picking the right rows and columns, we can solve for $a, b, c$ efficiently.

   ---

   1. Write the product $AB$ in terms of $a, b, c$.

      The $(i,j)$ entry of $AB$ is the dot product of the $i$-th row of $A$ with the $j$-th column of $B$.

      Let’s compute a few key entries.

      • Row 1 of $A$: $\begin{bmatrix} c & -a & b \end{bmatrix}$ Column 1 of $B$: $\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}$ Entry $(1,1)$: $c\cdot 1 + (-a)\cdot 0 + b\cdot 1 = c + b$ Given $AB(1,1) = 4$, so

   $b + c = 4. \qquad(1)$

   • Row 1, Column 2: Column 2 of $B$ is $\begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}$ Entry $(1,2)$: $c\cdot 0 + (-a)\cdot 1 + b\cdot 2 = -a + 2b$ Given $AB(1,2) = 4$, so

   $-a + 2b = 4. \qquad(2)$

   • Row 1, Column 3: Column 3 of $B$ is $\begin{bmatrix} 2 \\ 2 \\ 0 \end{bmatrix}$ Entry $(1,3)$: $c\cdot 2 + (-a)\cdot 2 + b\cdot 0 = 2c - 2a$ Given $AB(1,3) = -2$, so

   $2c - 2a = -2 \quad \Rightarrow \quad c - a = -1. \qquad(3)$

   2. Solve the system from row 1.

      From (1): $b = 4 - c$.

      From (3): $a = c + 1$.

      Substitute into (2): $-(c+1) + 2(4-c) = 4$

      $\Rightarrow -c - 1 + 8 - 2c = 4$

      $\Rightarrow -3c + 7 = 4$

      $\Rightarrow -3c = -3$

      $\Rightarrow c = 1$.

      Then $a = c+1 = 2$, and $b = 4 - c = 3$.

   3. Verify with another row to be safe. …

   🔒 Start your 14-day free trial to see the full solution →
10. 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: $a=1,\ b=1,\ c=1$. …

    🔒 Start your 14-day free trial to see the full solution →
11. 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. …

    🔒 Start your 14-day free trial to see the full solution →
12. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
    Q.If $A = \begin{bmatrix} 1 & 2 \\ -2 & -5 \end{bmatrix}$ and $\alpha A^2 + \beta A = 2I$ for some $\alpha, \beta \in \mathbb{R}$ then $\alpha + \beta =$ (A) 7 (B) 10 (C) 12 (D) 5

    ›Reveal solutionSolution

    Computing $A^2$ and matching the matrix equation $\alpha A^2+\beta A=2I$ entry-by-entry gives $\alpha=2,\ \beta=8$, so $\alpha+\beta=10$ — option (B).

    Concept and Intuition

    Given a $2\times2$ matrix satisfying a polynomial relation like $\alpha A^2+\beta A = 2I$, the cleanest approach is to compute $A^2$ directly, then equate corresponding entries of both sides of the matrix equation — this converts a single matrix equation into a small system of linear equations in the unknown scalars $\alpha,\beta$ (using just two independent entries is enough, and the remaining entries serve as a consistency check, since the relation must hold for the whole matrix, not just isolated numbers — this consistency is itself guaranteed for genuine matrix polynomial identities via Cayley–Hamilton-type reasoning, but verifying arithmetic keeps you safe under exam conditions).

    Step-by-Step Solution

    1. Compute $A^2$: $A^2=\begin{bmatrix}1&2\\-2&-5\end{bmatrix}\begin{bmatrix}1&2\\-2&-5\end{bmatrix} = \begin{bmatrix}1(1)+2(-2) & 1(2)+2(-5)\\-2(1)+(-5)(-2) & -2(2)+(-5)(-5)\end{bmatrix} = \begin{bmatrix}-3&-8\\8&21\end{bmatrix}$.
    2. Write $\alpha A^2+\beta A = 2I$ entry-wise:
       • $(1,1)$: $-3\alpha+\beta = 2$
       • $(1,2)$: $-8\alpha+2\beta = 0$ …

    🔒 Start your 14-day free trial to see the full solution →