Q.If $A = \begin{bmatrix} 3 & 5 \end{bmatrix}$, $B = \begin{bmatrix} 7 & 3 \end{bmatrix}$, then find a non-zero matrix $C$ such that $AC = BC$.

Showing the 12 most recent of 32 on this concept.

1. 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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2. 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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3. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
   Q.If $A = \begin{bmatrix} 1 & 0 & 0 \\ a & -1 & 0 \\ b & c & 1 \end{bmatrix}$ is such that $A^2 = I$, then (A) $b = \dfrac{ac}{2}$ (B) $b = -\dfrac{ac}{2}$ (C) $b = \dfrac{a+c}{2}$ (D) $b = \sqrt{ac}$

   ›Reveal solutionSolution

   Multiplying the lower-triangular matrix $A$ by itself and forcing the result to equal $I$ gives a single condition on $a,b,c$: $b=-\dfrac{ac}{2}$.

   Concept and Intuition

   $A$ is lower triangular with $1$'s on the diagonal (except the $-1$ in the middle). Squaring a triangular matrix keeps it triangular, and the diagonal entries of $A^2$ are just the squares of the diagonal entries of $A$ ($1^2=1$, $(-1)^2=1$, $1^2=1$), which already match $I$'s diagonal automatically. The real content of the condition $A^2=I$ is in the off-diagonal (lower-triangular) entries.

   Step-by-Step Solution

   1. Write $A=\begin{bmatrix}1&0&0\\a&-1&0\\b&c&1\end{bmatrix}$.
   2. Compute $A^2 = A\cdot A$ row by row.
      • Row 1: $[1,0,0]\cdot A = [1,0,0]$ (matches $I$ automatically).
      • Row 2: $[a,-1,0]\cdot A$: first entry $= a(1)+(-1)(a)+0(b) = 0$; second entry $= a(0)+(-1)(-1)+0(c)=1$; third entry $=0$. So row 2 $=[0,1,0]$ (matches $I$ automatically, for any $a$).
      • Row 3: $[b,c,1]\cdot A$: first entry $= b(1)+c(a)+1(b) = 2b+ac$; second entry $= b(0)+c(-1)+1(c) = 0$; third entry $= 0+0+1=1$. So row 3 $=[2b+ac,\ 0,\ 1]$. …

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4. 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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5. 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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6. 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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7. 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:
      • $(1,1)$: $-3\alpha+\beta = 2$
      • $(1,2)$: $-8\alpha+2\beta = 0$ …

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8. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.If $A = \begin{bmatrix} 2 & 3 & 3 \\ 3 & 2 & 3 \\ 3 & 3 & 2 \end{bmatrix}$, then $A^2 - 8I =$ (A) $0$ (B) $8A$ (C) $7A$ (D) $5A$

   ›Reveal solutionSolution

   The key idea is to compute $A^2$ directly and then subtract $8I$; the result simplifies to $5A$, so the correct option is (D).

   We start with the matrix

   $A = \begin{bmatrix} 2 & 3 & 3 \\ 3 & 2 & 3 \\ 3 & 3 & 2 \end{bmatrix}.$

   The problem asks for $A^2 - 8I$. Instead of guessing, we can compute $A^2$ by matrix multiplication and then subtract $8$ times the identity. This is a straightforward algebraic check — no need for eigenvalues or characteristic polynomials here, though those would also work.

   1. Compute $A^2$ Multiply $A$ by itself. The $(i,j)$ entry of $A^2$ is the dot product of row $i$ of $A$ with column $j$ of $A$.
      • For diagonal entries (e.g., $(1,1)$):

   $(2)(2) + (3)(3) + (3)(3) = 4 + 9 + 9 = 22.$

    By symmetry, all diagonal entries are $22$.
   

   • For off-diagonal entries (e.g., $(1,2)$):

   $(2)(3) + (3)(2) + (3)(3) = 6 + 6 + 9 = 21.$

    Again by symmetry, every off-diagonal entry is $21$.  
   

   So

   $A^2 = \begin{bmatrix} 22 & 21 & 21 \\ 21 & 22 & 21 \\ 21 & 21 & 22 \end{bmatrix}.$

   2. Subtract $8I$ The identity matrix $I$ has 1’s on the diagonal and 0’s elsewhere. Thus

   $A^2 - 8I = \begin{bmatrix} 22-8 & 21 & 21 \\ 21 & 22-8 & 21 \\ 21 & 21 & 22-8 \end{bmatrix} = \begin{bmatrix} 14 & 21 & 21 \\ 21 & 14 & 21 \\ 21 & 21 & 14 \end{bmatrix}.$

   3. Compare with multiples of $A$ Notice that $A$ is

   $A = \begin{bmatrix} 2 & 3 & 3 \\ 3 & 2 & 3 \\ 3 & 3 & 2 \end{bmatrix}.$

   Multiply $A$ by 5:

   $$$$

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9. 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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10. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
    Q.If $A = \begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 8 & 0 \\ 7 & 1 \end{bmatrix}$ and $A^3 = B$, then $x =$ (A) $-2$ or $3$ (B) $-2$ (C) $2$ or $-3$ (D) $2$

    ›Reveal solutionSolution

    Directly cubing the lower-triangular matrix $A$ and matching both entries to $B$ pins down $x=2$ uniquely. The answer is (D).

    Concept and Intuition

    $A$ is lower triangular, so its powers stay lower triangular, and the diagonal entries of $A^n$ are just the diagonal entries of $A$ raised to the $n$-th power — this makes computing $A^3$ by hand straightforward via repeated matrix multiplication.

    Step-by-Step Solution

    1. $A=\begin{bmatrix}x&0\\1&1\end{bmatrix}$.
    2. $A^2 = A\cdot A = \begin{bmatrix}x^2&0\\x+1&1\end{bmatrix}$ (bottom-left entry: $1\cdot x+1\cdot1=x+1$).
    3. $A^3 = A^2\cdot A = \begin{bmatrix}x^3&0\\(x+1)x+1&1\end{bmatrix} = \begin{bmatrix}x^3&0\\x^2+x+1&1\end{bmatrix}$.
    4. Set $A^3=B=\begin{bmatrix}8&0\\7&1\end{bmatrix}$: from the $(1,1)$ entry, $x^3=8\Rightarrow x=2$ (the unique real cube root). …

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11. 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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12. 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. …

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