Mathematics

Linear Algebra

510 Questions

Linear algebra involves the study of matrices, vectors, and linear transformations. Common topics include finding the rank of a matrix, calculating eigenvalues and eigenvectors, and performing LU decomposition. These advanced mathematical concepts are frequently tested in engineering, statistics, and civil service examinations.

Matrix rank calculationLU decompositionEigenvalues and eigenvectorsMatrix invertibilityDeterminant properties

Linear Algebra Questions

Multiple choice maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

If the matrices $A, B, (A+B)$ are non singular then ${[A{(A+B)}^{-1}B]}^{-1}$ is equal to-

  1. $A+B$
  2. ${A}^{-1}+{B}^{-1}$
  3. $A{(A+B)}^{-1}$

  4. None
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$ (A(A+B)^{-1}B)^{-1}$

$ = [(A(A^{-1}+B^{-1}))B]^{-1}$

$ = [(AA^{-1}+AB^{-1})B]^{-1}$

$ = ((I+AB^{-1})B)^{-1} = (B+AB^{-1}B)^{-1}$

$ = (B+A)^{-1} = A^{-1}+B^{-1}$
Multiple choice maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

If $A$ is an invertible matrix of order $2$, then $det({A}^{-1})$ is equal to

  1. $det(A)$
  2. $\cfrac{1}{det(A)}$
  3. $1$
  4. $0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We know that 

$AA^{-1}=I$
Taking determinant both sides
$|AA^{-1}|=|I|$
$|A||A^{-1}|=|I|$       $[\because |AB|=|A||B|]$
$|A||A^{-1}|=1$         $[\because |I|=1]$
$|A^{-1}|=\dfrac{1}{|A|}$
Since $|A|\neq0$
Hence, $|A^{-1}|=\dfrac{1}{|A|}$

Multiple choice maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

Let $A,B$ and $C$ be square matrices of order $3\ \times 3$. If $A$ invertible $(A-B)C=BA^{-1}$, then

  1. $C\ (A-B)=A^{-1}B$
  2. $C\ (A-B)=BA^{-1}$
  3. $(A-B)C=A^{-1}B$
  4. $All\ the\ above$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given (A-B)C = BA^-1, we can multiply both sides by A on the right to get (A-B)CA = B. Rearranging or manipulating the equation leads to C(A-B) = A^-1 B.

Multiple choice maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

A square non-singular matrix A satisfies $\displaystyle A^{2}-A+2I=0$, then $\displaystyle A^{-1}=$

  1. $\displaystyle I-A$
  2. $\displaystyle \frac{1}{2}\left ( I-A \right )$
  3. $\displaystyle I+A$
  4. $\displaystyle \frac{1}{2}\left ( I+A \right )$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, $\displaystyle A^{2}-A+2I=0$


$\Rightarrow A^{2}A^{-1}-AA^{-1}+2IA^{-1}=0$

$\Rightarrow A-I+2A^{-1}=0$

$\Rightarrow 2A^{-1}=I-A$

$\Rightarrow A^{-1}=\displaystyle \frac{1}{2}(I-A)$

Multiple choice maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

Let p be a nonsingular matrix, and $I + p + p^2 + ..... + p^n = 0$, then find $p^{-1}$.

  1. $I$
  2. $p^{n+1}$
  3. $p^n$
  4. $\left( p^{n+1} - I\right) \left( p-I\right)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We have $I + p + p^2 + ..... + p^n = O$  ----------$(1)$
Since p is a nonsingular matrix, p is invertible.
Multiplying both sides of (1) by $p^{-1}$, we get
$p^{-1} + I + Ip + ..... + p^{n - 1} I = O. p^{-1}$
or $ p^{-1} + I (1 + p + ...... + p^{n-1}) = O$
or $ p^{-1} = - I (I + p + p^2 + ..... + p^{n - 1}) = - I(-p^n) = p^n$

Multiple choice maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

Matrices A and B satisfy $AB = B^{-1}$, where $ B\quad =\quad \begin{bmatrix} 2 & -1 \ 2 & 0 \end{bmatrix}$, then find without finding $A^{-1}$, the matrix X satisfying $A^{-1}XA = ?$

  1. $B$
  2. $B^2$
  3. $A$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given, $A^{-1} XA = B$


$AA^{-1} XA = AB$

$IXA =AB$

$XAB =AB^2$ 

$XAB =I$ since $\left[AB=B^{-1}\Rightarrow AB^2=I\right]$

$XAB^2 = B$

$XI = B$

$\therefore X = B$

Hence option $'A'$ is the answer.

Multiple choice maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

If $A$ satisfies the equation $x^3-5x^2+4x+kI=0,$ then $A^{-1}$ exists if

  1. $k\neq -1$
  2. $k\neq 0$
  3. $k\neq 1$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since A satisfies the given equation, therefore ${ A }^{ 3 }-5{ A }^{ 2 }+4A+kI=0$

${ A }^{ -1 }$ exits if $k\neq 0$ since if $k=0$ then the above equation gives $A=0$ and in that case ${ A }^{ -1 }$ wont exist.

Multiple choice maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

If $A$ and $B$ are symmetric matrices and $AB=BA$, then ${ A }^{ -1 }B$ is a

  1. Symmetric matrix

  2. Skew-symmetric matrix

  3. Identity matrix

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We have $AB=BA=B'A'=\left( AB \right) '$

$\Rightarrow AB$ is symmetric.
Also, $AB{ A }^{ -1 }=BA{ A }^{ -1 }=B\quad \quad \left( \because AB=BA \right) $
$\Rightarrow { A }^{ -1 }AB{ A }^{ -1 }={ A }^{ -1 }B\Rightarrow B{ A }^{ -1 }={ A }^{ -1 }B$
Therefore, $\left( { A }^{ -1 }B \right) '=\left( B{ A }^{ -1 } \right) =\left( { A }^{ -1 } \right) 'B'=A'B$   $[\because { A }^{ -1 }$ and $B$ are symmetric $]$
Thus, the matrix ${ A }^{ -1 }B$ is symmetric.

Multiple choice maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

IF $A,B,C$ are non-singular $n\times n$ matrices, then $(ABC)^{-1}$ = ____________.

  1. $A^{-1}C^{-1}B^{-1}$
  2. $C^{-1}B^{-1}A^{-1}$
  3. $C^{-1}A^{-1}B^{-1}$
  4. $B^{-1}C^{-1}A^{-1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

using the property $(AB)^{-1}=B^{-1}A^{-1}$


$(ABC)^{-1}=(BC)^{-1}A^{-1}=C^{-1}B^{-1}A^{-1}$

Hence, option B.

Multiple choice maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

If $A^{-1}=\begin{bmatrix} 1 & -2 \ -2 & 2 \end{bmatrix}$, then what is $det(A)$ equal to ?

  1. $2$
  2. $-2$
  3. $1/2$
  4. $-1/2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$det{\left( {A}^{-1} \right)} = \left( 1 \times 2 \right) - \left( -2 \right) \times \left( -2 \right) = 2 - 4 = -2$

As we know that,
$det{\left( {A}^{-1} \right)} = \cfrac{1}{det{\left( A \right)}}$
$\Rightarrow det{\left( A \right)} = \cfrac{1}{det{\left( {A}^{-1} \right)}} = \cfrac{1}{-2} = - \cfrac{1}{2}$

Multiple choice maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

A square, non-singular matrix $A$ satifies $A^2 - A + 2I = 0$, then $A^{-1} = $

  1. $I - A$
  2. $\dfrac {(I - A) }{2}$
  3. $I + A$
  4. $\dfrac {(I + A)}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, $A^{2}-A+2I = 0$

$\Rightarrow 2I = A-A^{2}$

$\Rightarrow 2A^{-1}I = A^{-1}A-A^{-1}A^{2}$

$\therefore A^{-1} = \displaystyle\frac{I-A}{2}$

Multiple choice maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

If matrix $A=\left| \begin{matrix} sin\theta  & cosec\theta  & 1 \ cosec\theta  & 1 & sin\theta  \ 1 & sin\theta  & cosec\theta  \end{matrix} \right| $ a non invertible matrix. then possible value of $\theta$ is-

  1. $n\pi+(-1)^n\dfrac{\pi}{4}$
  2. $n\pi+(-1)^n\dfrac{\pi}{3}$
  3. $n\pi+(-1)^n\dfrac{\pi}{6}$
  4. $2n\pi+\dfrac{\pi}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The matrix is non-invertible if its determinant is zero. Calculating the determinant of the given matrix and setting it to zero leads to a trigonometric equation that yields the specified value for theta.