Matrices & DeterminantshardFree

Matrices & Determinants: Vmatrix Vmatrix Equal

JEE Maths question with a full step-by-step solution.

Question
If Δr=∣2r−1x2n−12⋅3r−1y3n−14⋅5r−1z5n−1∣\Delta_r = \begin{vmatrix} 2^{r-1} & x & 2^n-1 \\ 2\cdot3^{r-1} & y & 3^n-1 \\ 4\cdot5^{r-1} & z & 5^n-1 \end{vmatrix}, then ∑r=1nΔr\displaystyle\sum_{r=1}^n \Delta_r is equal to:
A500500 only if n=15n = 15
B500500 only if x=y=z=0x = y = z = 0
C500500 only if x+y+z=0x + y + z = 0
DNone of thesecorrect
Solution
Step 1: Apply linearity in column 1 to sum over rr
∑r=1nΔr=∣∑r=1n2r−1x2n−1∑r=1n2⋅3r−1y3n−1∑r=1n4⋅5r−1z5n−1∣\sum_{r=1}^n \Delta_r = \begin{vmatrix} \sum_{r=1}^n 2^{r-1} & x & 2^n-1 \\ \sum_{r=1}^n 2\cdot3^{r-1} & y & 3^n-1 \\ \sum_{r=1}^n 4\cdot5^{r-1} & z & 5^n-1 \end{vmatrix}
 Step 2: Evaluate the geometric sums
∑r=1n2r−1=2n−1,∑r=1n2⋅3r−1=3n−1,∑r=1n4⋅5r−1=5n−1\sum_{r=1}^n 2^{r-1} = 2^n-1, \quad \sum_{r=1}^n 2\cdot3^{r-1} = 3^n-1, \quad \sum_{r=1}^n 4\cdot5^{r-1} = 5^n-1
 Step 3: Identify identical columns
=∣2n−1x2n−13n−1y3n−15n−1z5n−1∣=0= \begin{vmatrix} 2^n-1 & x & 2^n-1 \\ 3^n-1 & y & 3^n-1 \\ 5^n-1 & z & 5^n-1 \end{vmatrix} = 0
 Columns 1 and 3 are identical, so the sum equals 00 regardless of nn, xx, yy, zz. Answer: (4)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Matrices & Determinants · hard
Let AA be a 3×33\times3 matrix such that AT[101]=[522]A^T\begin{bmatrix}1\\0\\1\end{bmatrix}=\begin{bmatrix}5\\2\\2\end{bmatrix}, AT[001]=[311]A^T\begin{bmatrix}0\\0\\1\end{bmatrix}=\begin{bmatrix}3\\1\\1\end{bmatrix}, A[101]=[344]A\begin{bmatrix}1\\0\\1\end{bmatrix}=\begin{bmatrix}3\\4\\4\end{bmatrix} and A[001]=[131]A\begin{bmatrix}0\\0\\1\end{bmatrix}=\begin{bmatrix}1\\3\\1\end{bmatrix}. If det⁡(A)=1\det(A)=1, then det⁡(adj(A2+A))\det\big(\mathrm{adj}(A^2+A)\big) is equal to
Matrices & Determinants · medium
Let MM be a 3×33\times3 matrix such that M(100)=(123)M\begin{pmatrix}1\\0\\0\end{pmatrix}=\begin{pmatrix}1\\2\\3\end{pmatrix}, M(010)=(012)M\begin{pmatrix}0\\1\\0\end{pmatrix}=\begin{pmatrix}0\\1\\2\end{pmatrix}, and M(001)=(−111)M\begin{pmatrix}0\\0\\1\end{pmatrix}=\begin{pmatrix}-1\\1\\1\end{pmatrix}. If M(xyz)=(1711)M\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1\\7\\11\end{pmatrix}, then x+y+zx+y+z equals:
Matrices & Determinants · medium
If f:N→Zf:\mathbb{N} \to \mathbb{Z} is defined by f(n)=∣n−1−5−2n23(2k+1)2k+1−3n33k(2k+1)3k(k+2)+1∣,k∈Nf(n) = \begin{vmatrix} n & -1 & -5 \\ -2n^2 & 3(2k+1) & 2k+1 \\ -3n^3 & 3k(2k+1) & 3k(k+2)+1 \end{vmatrix}, \quad k \in \mathbb{N} and ∑n=1kf(n)=98\displaystyle\sum_{n=1}^k f(n) = 98, then kk is equal to:
Matrices & Determinants · hard
If f(x)=∣x5∣sin⁡x∣2x4tan⁡3x1sec⁡2xsin⁡3xx45∣f(x) = \begin{vmatrix} x^5 & |\sin x| & 2x^4 \\ \tan^3 x & 1 & \sec 2x \\ \sin^3 x & x^4 & 5 \end{vmatrix}, then ∫−π/2π/2f(x) dx\displaystyle\int_{-\pi/2}^{\pi/2} f(x)\,dx is equal to:

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath