Let A = ( \beginmatrix [x + 1] & [x + 2] & [x + 3] [x] & [x + 3] & [x + 3] [x] & [x + 2] & [x + 4] \endmatrix ) , where [t] denotes the greatest integer less than or equal to t. If det(A) = 192, then the set of values of x is the interval :

Matrices and Determinants PYQ — JEE Main 2021 · 27 Aug · Shift 2 · Q24 with Solution

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Correct answer: B

Let $n=[x].$ Since adding an integer inside the floor just adds that integer, $[x+k]=[x]+k=n+k \quad (k\in \mathbb Z).$

So the matrix becomes

[x+1] & [x+2] & [x+3]\\ [x] & [x+3] & [x+3]\\ [x] & [x+2] & [x+4] \end{pmatrix} = \begin{pmatrix} n+1 & n+2 & n+3\\ n & n+3 & n+3\\ n & n+2 & n+4 \end{pmatrix}.$$ 2. **Compute the determinant** We evaluate $$\det(A)= \begin{vmatrix} n+1 & n+2 & n+3\\ n & n+3 & n+3\\ n & n+2 & n+4 \end{vmatrix}.$$ Apply row operations that do not change the determinant: $$R_2\to R_2-R_1,\qquad R_3\to R_3-R_1.$$ Then $$\det(A)= \begin{vmatrix} n+1 & n+2 & n+3\\ -1 & 1 & 0\\ -1 & 0 & 1 \end{vmatrix}.$$ Now expand along the first row: $$\det(A)=(n+1) \begin{vmatrix} 1 & 0\\ 0 & 1 \end{vmatrix} -(n+2) \begin{vmatrix} -1 & 0\\ -1 & 1 \end{vmatrix} +(n+3) \begin{vmatrix} -1 & 1\\ -1 & 0 \end{vmatrix}.$$ Compute minors: $$\begin{vmatrix}1&0\\0&1\end{vmatrix}=1,$$ $$\begin{vmatrix}-1&0\\-1&1\end{vmatrix}=(-1)(1)-0(-1)=-1,$$ $$\begin{vmatrix}-1&1\\-1&0\end{vmatrix}=(-1)(0)-1(-1)=1.$$ Hence $$\det(A)=(n+1)- (n+2)(-1) + (n+3)(1).$$ So, $$\det(A)= (n+1)+(n+2)+(n+3)=3n+6.$$ 3. **Use the given determinant value** Given $$\det(A)=192,$$ so $$3n+6=192.$$ Thus, $$3n=186 \implies n=62.$$ Since $n=[x]$, we get $$[x]=62.$$ Therefore, $$x\in[62,63).$$ 4. **Match with the options** The correct interval is: $$[62,63).$$ So the correct option is **B**.

View written solutionFree

Correct answer: B

Let $n=[x].$ Since adding an integer inside the floor just adds that integer, $[x+k]=[x]+k=n+k \quad (k\in \mathbb Z).$

So the matrix becomes

[x+1] & [x+2] & [x+3]\\ [x] & [x+3] & [x+3]\\ [x] & [x+2] & [x+4] \end{pmatrix} = \begin{pmatrix} n+1 & n+2 & n+3\\ n & n+3 & n+3\\ n & n+2 & n+4 \end{pmatrix}.$$ 2. **Compute the determinant** We evaluate $$\det(A)= \begin{vmatrix} n+1 & n+2 & n+3\\ n & n+3 & n+3\\ n & n+2 & n+4 \end{vmatrix}.$$ Apply row operations that do not change the determinant: $$R_2\to R_2-R_1,\qquad R_3\to R_3-R_1.$$ Then $$\det(A)= \begin{vmatrix} n+1 & n+2 & n+3\\ -1 & 1 & 0\\ -1 & 0 & 1 \end{vmatrix}.$$ Now expand along the first row: $$\det(A)=(n+1) \begin{vmatrix} 1 & 0\\ 0 & 1 \end{vmatrix} -(n+2) \begin{vmatrix} -1 & 0\\ -1 & 1 \end{vmatrix} +(n+3) \begin{vmatrix} -1 & 1\\ -1 & 0 \end{vmatrix}.$$ Compute minors: $$\begin{vmatrix}1&0\\0&1\end{vmatrix}=1,$$ $$\begin{vmatrix}-1&0\\-1&1\end{vmatrix}=(-1)(1)-0(-1)=-1,$$ $$\begin{vmatrix}-1&1\\-1&0\end{vmatrix}=(-1)(0)-1(-1)=1.$$ Hence $$\det(A)=(n+1)- (n+2)(-1) + (n+3)(1).$$ So, $$\det(A)= (n+1)+(n+2)+(n+3)=3n+6.$$ 3. **Use the given determinant value** Given $$\det(A)=192,$$ so $$3n+6=192.$$ Thus, $$3n=186 \implies n=62.$$ Since $n=[x]$, we get $$[x]=62.$$ Therefore, $$x\in[62,63).$$ 4. **Match with the options** The correct interval is: $$[62,63).$$ So the correct option is **B**.