Q.Find the values of $a$, $b$, $c$ and $d$, if $3\begin{bmatrix}a & b\\ c & d\end{bmatrix} = \begin{bmatrix}a & 6\\ -1 & 2d\end{bmatrix} + \begin{bmatrix}4 & a+b\\ c+d & 3\end{bmatrix}$.

Showing the 12 most recent of 32 on this concept.

1. 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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2. 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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3. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.The augmented matrix of a nonhomogeneous system of equations $AX = B$, $[A\ B]$ is reduced to the following form after applying a series of elementary row transformations $\begin{bmatrix} 1 & 1 & 1 & 5 \\ 0 & 0 & -3 & 4 \\ 0 & 0 & \mu+1 & \lambda^2 - 2\lambda + 1 \end{bmatrix}$, Then (A) Only for $\mu \neq -1$, $AX = B$ has unique solution (B) Only for $\mu = -1$ and $\lambda = 1$, $AX = B$ has infinite number of solutions (C) For any $\mu$ and for any $\lambda$, $AX = B$ has infinite number of solutions (D) For all positive values of $\mu$, $AX = B$ has no solution

   ›Reveal solutionSolution

   The coefficient matrix always has rank 2 (rows 2 and 3 are parallel), so a unique

   solution is never possible; and for any positive $\mu$, row 3 always contradicts the

   $z$-value fixed by row 2, forcing "no solution". Answer: (D).

   Concept and Intuition

   For $AX=B$ with augmented matrix reduced to

   $\begin{bmatrix} 1 & 1 & 1 & 5 \\ 0 & 0 & -3 & 4 \\ 0 & 0 & \mu+1 & \lambda^2-2\lambda+1 \end{bmatrix}$,

   read off the coefficient rows (ignore the last column) for $x,y,z$: $(1,1,1)$,

   $(0,0,-3)$, $(0,0,\mu+1)$. The second and third rows are both scalar multiples of $(0,0,1)$ for every value of $\mu$ — they can never be linearly independent of each

   other. So $\text{rank}(A) \le 2$ always (it equals 2, from row 1 and row 2, since row

   2 is never the zero vector). Since there are 3 unknowns but rank never reaches 3, a

   unique solution is structurally impossible no matter what $\mu,\lambda$ are.

   Row 2 directly gives $-3z=4 \Rightarrow z=-4/3$, a fixed value. Row 3 imposes a second

   constraint on the same variable $z$: $(\mu+1)z=\lambda^2-2\lambda+1=(\lambda-1)^2$.

   Substituting $z=-4/3$: $-\frac{4}{3}(\mu+1)=(\lambda-1)^2$. For consistency, this

   equality must hold; since the right side $(\lambda-1)^2\ge 0$, we need $-\frac{4}{3} (\mu+1)\ge 0$, i.e. $\mu+1\le 0$, i.e. $\mu \le -1$.

   Step-by-Step Solution

   1. From row 2: $-3z=4 \Rightarrow z=-\dfrac{4}{3}$.
   2. From row 3: $(\mu+1)z=\lambda^2-2\lambda+1=(\lambda-1)^2$.
   3. Substitute: $(\mu+1)\left(-\dfrac{4}{3}\right)=(\lambda-1)^2$.
   4. Check option (A): coefficient-matrix rank is always 2 (rows 2, 3 parallel) — a unique solution needs rank 3, which never happens. So (A) is false.
   5. Check option (D): if $\mu>0$, then $\mu+1>0$, so the left side $-\frac{4}{3}(\mu+1)$ is strictly negative, but the right side $(\lambda-1)^2$ is always $\ge 0$. A negative number can never equal a non-negative number, so the equation in step 3 is never satisfied for any positive $\mu$ — the system is …

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4. 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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5. 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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6. 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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7. 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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8. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.While solving a system of linear equations $AX = B$ using Cramer's rule with the usual notation, if $\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 2 & -1 & 2 \\ -1 & 1 & 5 \end{vmatrix}$; $\Delta_1 = \begin{vmatrix} 5 & 1 & 1 \\ 4 & -1 & 2 \\ 11 & 1 & 5 \end{vmatrix}$ and $X = \begin{bmatrix} \alpha \\ 2 \\ \beta \end{bmatrix}$, then $\alpha^2 + \beta^2 =$ (A) 9 (B) 13 (C) 5 (D) 25

   ›Reveal solutionSolution

   This tests Cramer's rule bookkeeping: computing $\alpha$ directly from $\Delta_1/\Delta$, then recovering $\beta$ by reconstructing the right-hand-side vector $B$ implicit in $\Delta_1$ and using the known value of $y=2$.

   Concept and Intuition

   In Cramer's rule for $AX=B$ with $A$ a $3\times3$ coefficient matrix and $X=(\alpha, y, \beta)^T$: $\Delta = \det(A)$, and $\Delta_1$ is the determinant formed by replacing the first column of $A$ (the column of coefficients of $\alpha$) with $B$. So $\alpha = \Delta_1/\Delta$ directly. Moreover, since $\Delta_1$'s 2nd and 3rd columns are unchanged from $A$'s, we can read the vector $B$ straight off $\Delta_1$'s first column. Once $B$ is known, and given $y=2$ is already provided, we can plug into the original equations $AX=B$ (using $A$'s actual rows) to solve for the remaining unknown $\beta$.

   Step-by-Step Solution

   1. Compute $\Delta = \begin{vmatrix}1&1&1\\2&-1&2\\-1&1&5\end{vmatrix}$. Expanding along row 1: $1[(-1)(5)-(2)(1)] - 1[(2)(5)-(2)(-1)] + 1[(2)(1)-(-1)(-1)] = 1(-7) - 1(12) + 1(1) = -7-12+1 = -18$.
   2. Compute $\Delta_1 = \begin{vmatrix}5&1&1\\4&-1&2\\11&1&5\end{vmatrix}$. Expanding along row 1: $5[(-1)(5)-(2)(1)] - 1[(4)(5)-(2)(11)] + 1[(4)(1)-(-1)(11)] = 5(-7) - 1(-2) + 1(15) = -35+2+15=-18$.
   3. By Cramer's rule, $\alpha = \Delta_1/\Delta = (-18)/(-18) = 1$.
   4. Since $\Delta_1$'s columns 2 and 3 match $A$'s columns 2 and 3 exactly (compare: $A$'s column 2 is $(1,-1,1)^T$ and column 3 is $(1,2,5)^T$, matching $\Delta_1$), $\Delta_1$'s column 1, $(5,4,11)^T$, must be the RHS vector $B$.
   5. Now use $A$'s actual rows with $X=(\alpha,y,\beta)=(1,2,\beta)$ and $B=(5,4,11)$: Row 1: $1(1)+1(2)+1(\beta) = 5 \Rightarrow 3+\beta=5 \Rightarrow \beta=2$. …

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9. 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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10. 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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11. 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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12. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
    Q.The system of linear equations $x+2y+z=-3$, $3x+3y-2z=-1$, $2x+7y+7z=-4$ has (A) infinite number of solutions (B) no solution (C) unique solution (D) finite number of solutions

    ›Reveal solutionSolution

    Eliminating one variable reduces the system to two equations that contradict each other outright, which is the signature of an inconsistent linear system: the coefficient determinant is zero, but the system does not have infinitely many solutions — it has none.

    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 — but that alone doesn't tell you whether it has infinitely many solutions or none at all. You have to check consistency by actually trying to solve (or by comparing ranks of the coefficient matrix and the augmented matrix). A quick, reliable way for a 3-variable system is to eliminate one variable using two different pairs of equations and see if the resulting two-variable equations agree or contradict.

    Step-by-Step Solution

    1. Equations: (1) $x+2y+z=-3$, (2) $3x+3y-2z=-1$, (3) $2x+7y+7z=-4$.
    2. From (1): $x=-3-2y-z$.
    3. Substitute into (2): $3(-3-2y-z)+3y-2z=-1 \Rightarrow -9-6y-3z+3y-2z=-1 \Rightarrow -9-3y-5z=-1 \Rightarrow 3y+5z=-8$. Call this (A).
    4. Substitute into (3): $2(-3-2y-z)+7y+7z=-4 \Rightarrow -6-4y-2z+7y+7z=-4 \Rightarrow -6+3y+5z=-4 \Rightarrow 3y+5z=2$. Call this (B).
    5. Compare (A) and (B): both say "$3y+5z=$ something," but (A) requires it to equal $-8$ while (B) requires it to equal $2$. These cannot both be true — a direct contradiction. …

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