Q.If $X$ and $Y$ are $2 \times 2$ matrices, then solve the following matrix equations for $X$ and $Y$: $2X + 3Y = \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix}$, $3X + 2Y = \begin{bmatrix} -2 & 2 \\ 1 & -5 \end{bmatrix}$.

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 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 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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4. 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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5. 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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6. 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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7. 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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8. 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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9. 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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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 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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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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