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Previous year question hub

Matrices And Determinants - Algebra - Mathematics Previous Year Questions

Practice Matrices And Determinants - Algebra - Mathematics previous year questions organised from real papers, with year-wise coverage and clear topic navigation.

4Papers
4Years
8Questions
1Topics

Matrices And Determinants question pattern

Every graph below is calculated only from this selection.

Questions by year

Year-wise coverage for Matrices And Determinants. Each bar uses a separate theme-derived color.

Difficulty distribution

How the classified questions are distributed by difficulty.

Not classified 8 100%

Question type distribution

MCQ, numerical, multiple-select and other formats found in these papers.

Multiple Choices 8 100%

Subject weightage

Top subjects by unique question coverage.

Mathematics
8 Qs

Most asked topics

Top topics across the included previous year papers.

Algebra
8 Qs

Subtopic coverage

Top subtopics inside this exact selection.

Matrices And Determinants
8 Qs

Paper coverage

Question coverage for the most populated papers. Every active PYP paper remains listed below.

VITEEE 2024
2 Qs
VITEEE 2023
3 Qs
VITEEE 2022
1 Qs
VITEEE 2021
2 Qs

Included previous year papers

Newest papers appear first. Sort by year, question coverage or name.

PaperYear / sessionQuestions in this viewOpen
VITEEE 202420242View paper
VITEEE 202320233View paper
VITEEE 202220221View paper
VITEEE 202120212View paper

All Matrices And Determinants previous year questions

Practice every matching question in batches of 20, with every available option.

1
2021 · Mathematics · Algebra · Matrices And Determinants
VITEEE 2021

If \(A^{-1}=\left[\begin{array}{rr}5 & -2 \\ -7 & 3\end{array}\right]\) and \(B^{-1}=\frac{1}{2}\left[\begin{array}{rr}9 & -7 \\ -8 & 6\end{array}\right]\), then \((A B)^{-1}\) is equal to

A
\(\left[\begin{array}{rr}47 & -39 / 2 \\ -41 & 17\end{array}\right]\)
B
\(\left[\begin{array}{rr}94 & -82 \\ -39 & 34\end{array}\right]\)
C
\(\left[\begin{array}{rr}-47 & 46 \\ 39 / 2 & -17\end{array}\right]\)
D
\(\left[\begin{array}{rr}-47 & 39 / 2 \\ 46 & -17\end{array}\right]\)
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2
2021 · Mathematics · Algebra · Matrices And Determinants
VITEEE 2021

If \(A=\left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right]\), then \(\left(A-A^{\prime}\right)\) is equal to (where, \(A^{\prime}\) is transpose of matrix \(A\) )

A
null matrix
B
identity matrix
C
symmetric
D
skew-symmetric
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3
2022 · Mathematics · Algebra · Matrices And Determinants
VITEEE 2022

For all values of \(\lambda\), rank of matrix

$$A=\left[\begin{array}{ccc} { }^h C_0 & { }^4 C_3 & { }^5 C_4 \\ \[\lambda & 8 & 8 \lambda-6 \\\] 1+\lambda^2 & 8 \lambda+4 & 2 \lambda+21 \end{array}\right]$$

A
for \(\lambda=2, \rho(A)=1\)
B
for \(\lambda=-1, \rho(A)=1\)
C
for \(\lambda=2, \rho(A)=3\)
D
for \(\lambda=-1, \rho(A)=4\)
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4
2023 · Mathematics · Algebra · Matrices And Determinants
VITEEE 2023

If matrix \(A=\left[\begin{array}{ccc}0 & 2 b & -2 \\ 3 & 1 & 3 \\ 3 a & 3 & -1\end{array}\right]\) is given to be symmetric, then the value of \(a b\) is

A
1
B
0
C
\(-\)1
D
9/4
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5
2023 · Mathematics · Algebra · Matrices And Determinants
VITEEE 2023

Suppose, \(A=\left[\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right]\) is an adjoint of the matrix \(\left[\begin{array}{rrr}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4\end{array}\right]\). The value of \(\frac{a_1+b_2+c_3}{b_1 a_2}\) is

A
3
B
0
C
5
D
4
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6
2023 · Mathematics · Algebra · Matrices And Determinants
VITEEE 2023

The determinant of the matrix \(\left[\begin{array}{ccc}1 & 4 & 8 \\ 1 & 9 & 27 \\ 1 & 16 & 64\end{array}\right]\) is

A
13
B
208
C
52
D
104
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7
2024 · Mathematics · Algebra · Matrices And Determinants
VITEEE 2024

If $A, B$ are two square matrices, such that $A B=A, B A=B$, then $(A+B)^7$ equals

A
$A+B$
B
$2^7(A+B)$
C
$2^6(A+B)$
D
$2^8(A+B)$
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8
2024 · Mathematics · Algebra · Matrices And Determinants
VITEEE 2024

If $A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$ and $B=\left[\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right]$, then $\left(B B^T A\right)^5$ is equal to

A
$\left[\begin{array}{cc}2+\sqrt{3} & 1 \\ -1 & 2-\sqrt{3}\end{array}\right]$
B
$\frac{1}{2}\left[\begin{array}{ll}1 & 5 \\ 0 & 1\end{array}\right]$
C
$\left[\begin{array}{ll}1 & 5 \\ 0 & 1\end{array}\right]$
D
$\left[\begin{array}{ll}5 & 1 \\ 0 & 1\end{array}\right]$
Open complete paper