Tag: properties of multiplication of matrix

Questions Related to properties of multiplication of matrix

If $A^{2}-A+I=0$, then inverse of $A$ is

business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

  1. $A^{-2}$

  2. $A+I$

  3. $I-A$

  4. $A-I$

Show answer
Correct Option: C
Explanation:
$A^2-A+I=0$

Multiplying$=A^{-1}(0)$

$A^{-1}(A^2-A=I)=A^{-1}(0)$

$A-I+A^{-1}=0$

$A^{-1}=-(A-I)$

$A^{-1}=I-A$.

The matrices $\begin{bmatrix} \cos { \theta  }  & -\sin { \theta  }  \ \sin { \theta  }  & \cos { \theta  }  \end{bmatrix}$ and $\begin{bmatrix} a & 0 \ 0 & b \end{bmatrix}$ commute under multiplication

business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

  1. if $a=b$ or $\theta=n\pi,$ where $n$ is an integer

  2. always

  3. never

  4. if $a\cos { \theta  } \neq b\sin { \theta  } $

Show answer
Correct Option: A
Explanation:

$\begin{bmatrix} \cos { \theta  }  & -\sin { \theta  }  \ \sin { \theta  }  & \cos { \theta  }  \end{bmatrix}\begin{bmatrix} a & 0 \ 0 & b \end{bmatrix}=\begin{bmatrix} a\cos { \theta  }  & -b\sin { \theta  }  \ a\sin { \theta  }  & b\cos { \theta  }  \end{bmatrix}$   ...(1)

And $\begin{bmatrix} a & 0 \ 0 & b \end{bmatrix}\begin{bmatrix} \cos { \theta  }  & -\sin { \theta  }  \ \sin { \theta  }  & \cos { \theta  }  \end{bmatrix}=\begin{bmatrix} a\cos { \theta  }  & -a\sin { \theta  }  \ b\sin { \theta  }  & b\cos { \theta  }  \end{bmatrix}$   ...(2)
From (1) and (2), we get
$a\sin { \theta  } =b\sin { \theta  } \Rightarrow \left( a-b \right) \sin { \theta  } =0$
either $a=b$ or $\sin { \theta  } =0$
$\Rightarrow\theta=n\pi;n\in Z$

If $A$ and $B$ are two square matrices of order $3 \times  3$ which satisfy $AB = A$ and $BA = B$, then Which of the following is true?

business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

  1. If matrix $A$ is singular, then matrix $B$ is non singular.

  2. If matrix $A$ is nonsingular, then matrix $B$ is singular.

  3. If matrix $A$ is singular, then matrix $B$ is also singular.

  4. Cannot say anything.

Show answer
Correct Option: C
Explanation:
$A$ and $B$ are two square matrices of order $3\times3$ which satisfy $AB=A$ and $BA=B,$ then if matrix $A$ is singular, then matrix $B$ is also singular .
$A$ singular matrix is that matrix whose determinants is zero and which is non-irreversible.
$\therefore A$ and $B$ both are singular matrices, then only the conditions $AB=A$ and $BA=B$ holds true.
Hence, the answer is if matrix $A$ is singular, then matrix $B$ is also singular.

The multiplication of matrices is distributive with respect to the matrix addition.

State true or false.

business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

It is well known fact that, Multiplication of matrices is distributive with respect to the matrix addition. i.e $ A(B+C)=AB+AC$

The inverse of the matrix $\begin{bmatrix}3 & 5 & 7 \ 2 & -3 & 1 \ 1 & 1 & 2\end{bmatrix}$ is $\begin{bmatrix}7 & -3 & 26 \ 3 & 1 & 11 \ -5 & -2 & 0\end{bmatrix}$.
State true or false.

business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

  1. True

  2. False

Show answer
Correct Option: A

 In matrices $AB = O$ does not necessarily mean that 

business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

  1. $A=0$

  2. $B=0$

  3. Both $ A = 0$ and $B=0$

  4. all of the above

Show answer
Correct Option: D
Explanation:

Let $\quad A = \begin{bmatrix}1 & -1 & 1 \ -3 & 2 & -1

\ -2 & 1 & 0\end{bmatrix}$ and $B = \begin{bmatrix}1&2

& 3\2&4&6 \ 1&2 &3\end{bmatrix}$

$\therefore\quad

AB = \begin{bmatrix}1 & -1 & 1 \ -3 & 2 & -1 \ -2

& 1 & 0\end{bmatrix}\times\begin{bmatrix}1&2 &

3\2&4&6 \ 1&2 &3\end{bmatrix}$

$\quad                   = \begin{bmatrix}0&0&0 \ 0&0&0 \ 0&0&0\end{bmatrix} = O$

$\therefore \quad AB = O$
But neither $A = O$ nor $B = O$.

If inverse of $A=\left[ \begin{matrix} 1 & 1 & 1 \ 2 & -1 & -1 \ 1 & -1 & 1 \end{matrix} \right] $ is $\cfrac { -1 }{ 6 } \left[ \begin{matrix} -2 & -2 & 0 \ -3 & 0 & \alpha  \ -1 & 2 & -3 \end{matrix} \right] $ then $\alpha=$

business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

  1. $0$

  2. $-3$

  3. $3$

  4. $2$

Show answer
Correct Option: C
Explanation:
Given,

$A=\begin{bmatrix}1&1&1\\ 2&-1&-1\\ 1&-1&1\end{bmatrix}$

$A^{-1}=\begin{bmatrix}1&1&1\\ 2&-1&-1\\ 1&-1&1\end{bmatrix}^{-1}$

$=\begin{bmatrix}1&1&1&\mid \:&1&0&0\\ 2&-1&-1&\mid \:&0&1&0\\ 1&-1&1&\mid \:&0&0&1\end{bmatrix}$

$\:R _1\:\leftrightarrow \:R _2$

$=\begin{bmatrix}2&-1&-1&\mid \:&0&1&0\\ 1&1&1&\mid \:&1&0&0\\ 1&-1&1&\mid \:&0&0&1\end{bmatrix}$

$R _2\:\leftarrow \:R _2-\frac{1}{2}\cdot \:R _1$

$R _3\:\leftarrow \:R _3-\frac{1}{2}\cdot \:R _1$

$=\begin{bmatrix}2&-1&-1&\mid \:&0&1&0\\ 0&\frac{3}{2}&\frac{3}{2}&\mid \:&1&-\frac{1}{2}&0\\ 0&-\frac{1}{2}&\frac{3}{2}&\mid \:&0&-\frac{1}{2}&1\end{bmatrix}$

$R _3\:\leftarrow \:R _3+\frac{1}{3}\cdot \:R _2$

$=\begin{bmatrix}2&-1&-1&\mid \:&0&1&0\\ 0&\frac{3}{2}&\frac{3}{2}&\mid \:&1&-\frac{1}{2}&0\\ 0&0&2&\mid \:&\frac{1}{3}&-\frac{2}{3}&1\end{bmatrix}$

$R _3\:\leftarrow \frac{1}{2}\cdot \:R _3$

$R _2\:\leftarrow \:R _2-\frac{3}{2}\cdot \:R _3$

$=\begin{bmatrix}2&-1&-1&\mid \:&0&1&0\\ 0&\frac{3}{2}&0&\mid \:&\frac{3}{4}&0&-\frac{3}{4}\\ 0&0&1&\mid \:&\frac{1}{6}&-\frac{1}{3}&\frac{1}{2}\end{bmatrix}$

$R _1\:\leftarrow \:R _1+1\cdot \:R _3$

$R _2\:\leftarrow \frac{2}{3}\cdot \:R _2$

$=\begin{bmatrix}2&-1&0&\mid \:&\frac{1}{6}&\frac{2}{3}&\frac{1}{2}\\ 0&1&0&\mid \:&\frac{1}{2}&0&-\frac{1}{2}\\ 0&0&1&\mid \:&\frac{1}{6}&-\frac{1}{3}&\frac{1}{2}\end{bmatrix}$

$R _1\:\leftarrow \:R _1+1\cdot \:R _2$

$=\begin{bmatrix}2&0&0&\mid \:&\frac{2}{3}&\frac{2}{3}&0\\ 0&1&0&\mid \:&\frac{1}{2}&0&-\frac{1}{2}\\ 0&0&1&\mid \:&\frac{1}{6}&-\frac{1}{3}&\frac{1}{2}\end{bmatrix}$

$R _1\:\leftarrow \frac{1}{2}\cdot \:R _1$

$=\begin{bmatrix}1&0&0&\mid \:&\frac{1}{3}&\frac{1}{3}&0\\ 0&1&0&\mid \:&\frac{1}{2}&0&-\frac{1}{2}\\ 0&0&1&\mid \:&\frac{1}{6}&-\frac{1}{3}&\frac{1}{2}\end{bmatrix}$

$=\begin{bmatrix}\frac{1}{3}&\tfrac{1}{3}&0\\ \tfrac{1}{2}&0&-\tfrac{1}{2}\\ \tfrac{1}{6}&-\tfrac{1}{3}&\tfrac{1}{2}\end{bmatrix}$

$=-\dfrac{1}{6}\begin{bmatrix}-2 &-2  &0 \\  -3& 0 &3 \\  -1& 2 &-3 \end{bmatrix}$

$\therefore \alpha =3$

Inverse of $\begin{bmatrix}3& 1\5&2\end{bmatrix}$ is:

business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

  1. $\begin{bmatrix}3&-1 \-5 &-3\end{bmatrix}$

  2. $\begin{bmatrix}2&-1 \-5 &3\end{bmatrix}$

  3. $\begin{bmatrix}-3&5 \1 &-2\end{bmatrix}$

  4. $\begin{bmatrix}-2&5 \1 &-3\end{bmatrix}$

Show answer
Correct Option: B
Explanation:
Let $A=\begin{bmatrix}3& 1\\5&2\end{bmatrix}$
$\left|A\right|=6-5=1\neq 0$
$\therefore {A}^{-1}$ exists.
${C} _{11}={\left(-1\right)}^{1+1}{M} _{11}={\left(-1\right)}^{2}2=2$
${C} _{12}={\left(-1\right)}^{1+2}{M} _{12}={\left(-1\right)}^{3}5=-5$
${C} _{13}={\left(-1\right)}^{1+3}{M} _{13}={\left(-1\right)}^{4}1=1$
${C} _{14}={\left(-1\right)}^{1+4}{M} _{14}={\left(-1\right)}^{5}3=-3$
${C} _{ij}=\begin{bmatrix}2& -5\\-1 & 3\end{bmatrix}$
Adj$\left(A\right)={\begin{bmatrix}2& -5\\-1 & 3\end{bmatrix}}^{T}$
$=\begin{bmatrix}2& -1\\-5 & 3\end{bmatrix}$
${A}^{-1}=\dfrac{adj\left(A\right)}{\left|A\right|}=\dfrac{1}{1}\begin{bmatrix}2& -1\\-5 & 3\end{bmatrix}$
$\therefore {A}^{-1}=\begin{bmatrix}2& -1\\-5 & 3\end{bmatrix}$

Let $\displaystyle A=\begin{pmatrix}1 &2 \3  &4
\end{pmatrix}$ and $\displaystyle B=\begin{pmatrix}a &0 \0  &b \end{pmatrix} a,b \epsilon N.$Then

business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

  1. there cannot exist any B such that $\displaystyle AB = BA $

  2. there exist more than one but finite number of B's such that $\displaystyle AB = BA$

  3. there exists exactly One B such that $\displaystyle AB = BA$

  4. there exist infinitely many B's such that $\displaystyle AB = BA.$

Show answer
Correct Option: D
Explanation:

$A=\begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}$ and $B=\begin{bmatrix} a & 0 \ 0 & b \end{bmatrix}$

$AB = \begin{bmatrix} a & 2b \ 3a & 4b \end{bmatrix}$

$BA = \begin{bmatrix} a & 2a \ 3b & 4b \end{bmatrix}$

$AB\quad =\quad BA \Rightarrow a=b$

$\therefore$ there exist infinitely many  $B's$  such that $AB=BA$.

If matrix $A = [a _{ij}] _{2\times 2}$, where $a _{ij} = \left{\begin{matrix} 1,& \ \text{if}\ &i\neq j \ 0, & \ \text{if}\ & i + j\end{matrix}\right.$, then $A^{2}$ is equal to

business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

  1. $I$

  2. $2A$

  3. $O$

  4. $-I$

Show answer
Correct Option: A
Explanation:

$A=\begin{bmatrix}  a _{11}& a _{12} \  a _{21}&  a _{22}\end{bmatrix}$


$a _{11}:1\,\,\,a _{12}:0$

$a _{21}:0\,\,\,a _{22}:1$

$A=\begin{bmatrix}  1&  0\ 0 &  1\end{bmatrix}$

$A^2=\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \ 0 &  0\end{bmatrix}=\begin{bmatrix}  1& 0 \  0&  1\end{bmatrix}=I$

$\boxed{Hence\,A^2=I}$