Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Matrices and Determinants question

2022 · 28 Jul · Shift 1 · Q42
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Matrices and Determinants
  5. /2022 · 28 Jul · Shift 1 · Q42

Matrices and Determinants question

2022 · 28 Jul · Shift 1 · Q42

JEE MainMathematicsMatrices and DeterminantsNumerical+4 / −1
Let A=[1−12α]A=\left[\begin{array}{cc}1 & -1 \\ 2 & \alpha\end{array}\right]A=[12​−1α​] and B=[β110],α,β∈RB=\left[\begin{array}{cc}\beta & 1 \\ 1 & 0\end{array}\right], \alpha, \beta \in \mathbf{R}B=[β1​10​],α,β∈R. Let α1\alpha_{1}α1​ be the value of α\alphaα which satisfies (A+B)2=A2+[2222](\mathrm{A}+\mathrm{B})^{2}=\mathrm{A}^{2}+\left[\begin{array}{ll}2 & 2 \\ 2 & 2\end{array}\right](A+B)2=A2+[22​22​] and α2\alpha_{2}α2​ be the value of α\alphaα which satisfies (A+B)2=B2(\mathrm{A}+\mathrm{B})^{2}=\mathrm{B}^{2}(A+B)2=B2. Then ∣α1−α2∣\left|\alpha_{1}-\alpha_{2}\right|∣α1​−α2​∣ is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 2

  1. Given matrices

A=[1−12α],B=[β110]A=\begin{bmatrix}1&-1\\2&\alpha\end{bmatrix},\qquad B=\begin{bmatrix}\beta&1\\1&0\end{bmatrix}A=[12​−1α​],B=[β1​10​]

We need:

  • α1\alpha_1α1​ such that (A+B)2=A2+[2222](A+B)^2=A^2+\begin{bmatrix}2&2\\2&2\end{bmatrix}(A+B)2=A2+[22​22​]
  • α2\alpha_2α2​ such that (A+B)2=B2(A+B)^2=B^2(A+B)2=B2

Then compute ∣α1−α2∣|\alpha_1-\alpha_2|∣α1​−α2​∣.

  1. Compute useful matrices

First, A+B=[1+β03α]A+B=\begin{bmatrix}1+\beta&0\\3&\alpha\end{bmatrix}A+B=[1+β3​0α​]

So,

=\begin{bmatrix}(1+\beta)^2&0\\3(1+\beta)+3\alpha&\alpha^2\end{bmatrix} =\begin{bmatrix}(1+\beta)^2&0\\3(1+\beta+\alpha)&\alpha^2\end{bmatrix}$$ Now compute $A^2$: $$A^2=\begin{bmatrix}1&-1\\2&\alpha\end{bmatrix}\begin{bmatrix}1&-1\\2&\alpha\end{bmatrix} =\begin{bmatrix}-1&-1-\alpha\\2+2\alpha&\alpha^2-2\end{bmatrix}$$ Hence, $$A^2+\begin{bmatrix}2&2\\2&2\end{bmatrix} =\begin{bmatrix}1&1-\alpha\\4+2\alpha&\alpha^2\end{bmatrix}$$ Also compute $B^2$: $$B^2=\begin{bmatrix}\beta&1\\1&0\end{bmatrix}\begin{bmatrix}\beta&1\\1&0\end{bmatrix} =\begin{bmatrix}\beta^2+1&\beta\\\beta&1\end{bmatrix}$$ --- 3. **Find $\alpha_1$ from** $$ (A+B)^2=A^2+\begin{bmatrix}2&2\\2&2\end{bmatrix} $$ Equating entries: $$\begin{bmatrix}(1+\beta)^2&0\\3(1+\beta+\alpha)&\alpha^2\end{bmatrix} = \begin{bmatrix}1&1-\alpha\\4+2\alpha&\alpha^2\end{bmatrix}$$ From the $(1,2)$ entry: $$0=1-\alpha \implies \alpha=1$$ So, $$\alpha_1=1$$ (Checking consistency: with $\alpha=1$, from $(1,1)$ entry $(1+\beta)^2=1\Rightarrow \beta=0$ or $-2$, and from $(2,1)$ entry $3(2+\beta)=6\Rightarrow \beta=0$, so it is consistent.) --- 4. **Find $\alpha_2$ from** $$ (A+B)^2=B^2 $$ Equating entries: $$\begin{bmatrix}(1+\beta)^2&0\\3(1+\beta+\alpha)&\alpha^2\end{bmatrix} = \begin{bmatrix}\beta^2+1&\beta\\\beta&1\end{bmatrix}$$ From the $(1,2)$ entry: $$0=\beta \implies \beta=0$$ From the $(2,2)$ entry: $$\alpha^2=1 \implies \alpha=\pm 1$$ From the $(2,1)$ entry: $$3(1+0+\alpha)=0 \implies 1+\alpha=0 \implies \alpha=-1$$ Therefore, $$\alpha_2=-1$$ --- 5. **Compute required value** $$|\alpha_1-\alpha_2|=|1-(-1)|=2$$ --- 6. **Comparison with stored answer** Derived answer is $2$, which matches the stored correct answer.
PreviousNext

More from Matrices and Determinants

  • Let A and B be any two 3×3 symmetric and skew symmetric matrices respectively. Then which of the following is NOT true?2022 · MCQ
  • If the system of linear equations 2x+3y−z=−2x+y+z=4x−y+∣λ∣z=4λ−4 where, λ∈ R, has no solution, then2022 · MCQ
  • If the system of linear equations 2x−3y=γ+5, αx+5y=β+1, where α, β, γ∈ R has infinitely many solutions then the value of | 9 α + 3 β + 5 γ | is equal to ​…2022 · Numerical
  • Let A=(1+i−i​10​) where i=−1​. Then, the number of elements in the set { n }∈ {1, 2, ......, 100} : An = A } is ​.2022 · Numerical
  • Let A and B be two 3×3 non-zero real matrices such that AB is a zero matrix. Then2022 · MCQ
  • Let p and p + 2 be prime numbers and let Δ=​p!(p+1)!(p+2)!​(p+1)!(p+2)!(p+3)!​(p+2)!(p+3)!(p+4)!​​…2022 · Numerical
  • Which of the following matrices can NOT be obtained from the matrix [−11​2−1​] by a single elementary row operation ?2022 · MCQ
  • If the system of equations ​x+y+z=62x+5y+αz=βx+2y+3z=14​ has infinitely many solutions, then α+β is equal to2022 · MCQ