Tag: properties of multiplication of matrix

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Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

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

  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 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $\sin { \theta  } =0$
$\Rightarrow\theta=n\pi;n\in Z$

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

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?

  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.
Reveal answer Fill a bubble to check yourself
C Correct answer
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.
Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

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

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

  4. all of the above
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $B = O$.

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

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 $\alpha=$

  1. $0$
  2. $-3$
  3. $3$
  4. $2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given,

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\therefore \alpha =3$
Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

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

  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. $\begin{bmatrix}-2&5 \\1 &-3\end{bmatrix}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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$\left(A\right)={\begin{bmatrix}2& -5\\-1 & 3\end{bmatrix}}^{T}$
$=\begin{bmatrix}2& -1\\-5 & 3\end{bmatrix}$
${A}^{-1}=\dfrac{adj\left(A\right)}{\left|A\right|}=\dfrac{1}{1}\begin{bmatrix}2& -1\\-5 & 3\end{bmatrix}$
$\therefore {A}^{-1}=\begin{bmatrix}2& -1\\-5 & 3\end{bmatrix}$

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

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

  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 $\displaystyle AB = BA.$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $AB=BA$.

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

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

  1. $I$
  2. $2A$
  3. $O$
  4. $-I$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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


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

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

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

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

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