Tag: multiplication of matrices

Questions Related to multiplication of matrices

Let $A, : B : and : C$ be $2\times 2$ matrices with entries from the set of real numbers. Define $\ast $ as follows:
  $\displaystyle A \ast B=\frac{1}{2}(AB\,'+A'B)$. Which of the given is true?

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

  1. $A\ast B= B \ast A$

  2. $A\ast A=A^2$

  3. $A\ast (B+C)=A\ast B+A \ast C$

  4. $A\ast I =A+A'$

Show answer
Correct Option: A,C
Explanation:

$\displaystyle A \ast B=\frac{1}{2}(AB'+A'B)$

1) $\displaystyle B \ast A=\frac{1}{2}(BA'+B'A)=\frac{1}{2}(AB'+A'B)=A \ast B$

2)$\displaystyle A \ast A=\frac{1}{2}(AA'+A'A)$

3)$\displaystyle A \ast (B+C)=\frac{1}{2}(A(B+C)'+A'(B+C))$

                             $=\displaystyle\frac{1}{2}(AB'+A'B)+\frac{1}{2}(AC'+A'C)$

                             $=A\ast B+A \ast C$

4)$\displaystyle A \ast I=\frac{1}{2}(AI'+A'I)=\frac{1}{2}(A+A')$

Hence, options A and C.

Say true or false:

Let A, B be two matrices such that they commute, then $(AB)^n = A^nB^n$.

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

  1. True

  2. False

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Correct Option: A
Explanation:

$A$ and $B$ commute each other then $AB=BA$
${ \left( AB \right)  }^{ n }={ \left( BA \right)  }^{ n }\ \Rightarrow { \left( AB \right)  }^{ n }={ B }^{ n }{ A }^{ n }={ A }^{ n }{ B }^{ n }={ \left( BA \right)  }^{ n }$

Say true or false:
Let A, B be two matrices, such that $AB = A$ and $BA = B$, then $A^2 = A$ and $B^2 = B$.

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

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

$AB=A\Rightarrow B\left( AB \right) =B\left( A \right) \Rightarrow BAB=BA\Rightarrow BB=B\Rightarrow { B }^{ 2 }=B\ BA=B\Rightarrow A\left( BA \right) =A\left( B \right) \Rightarrow ABA=AB\Rightarrow AA=A\Rightarrow { A }^{ 2 }=A$

If $A$ is a non-singular matrix, then 

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

  1. ${ A }^{ -1 }$ is symmetric if $A$ is symmetric

  2. ${ A }^{ -1 }$ is skew-symmetric if $A$ is symmetric

  3. $\left| { A }^{ -1 } \right| =\left| A \right| $

  4. $\left| { A }^{ -1 } \right| ={ \left| A \right|  }^{ -1 }$

Show answer
Correct Option: A,D
Explanation:

Since $\left| A \right| \neq 0$, therefore ${ A }^{ -1 }$ exists.

Now, $A{ A }^{ -1 }=I={ A }^{ -1 }A$
$\Rightarrow \left( A{ A }^{ -1 } \right) '=I'=\left( { A }^{ -1 }A \right) '\Rightarrow \left( { A }^{ -1 } \right) 'A'=I=A'\left( { A }^{ -1 } \right) '\quad \quad \quad \left( \because A'=A \right) $
$\Rightarrow \left( { A }^{ -1 } \right) 'A=I=A\left( { A }^{ -1 } \right) '\Rightarrow { A }^{ -1 }=\left( { A }^{ -1 } \right) '\Rightarrow { A }^{ -1 }$ is symmetric
Also, since $\left| A \right| \neq 0,\therefore { A }^{ -1 }$ exists such that
$A{ A }^{ -1 }=I={ A }^{ -1 }A\Rightarrow \left| A{ A }^{ -1 } \right| =\left| I \right| $
$\Rightarrow \left| A \right| \left| { A }^{ -1 } \right| =1\quad \quad \left( \because \left| AB \right| =\left| A \right| \left| B \right|  \right) $
$\displaystyle \Rightarrow \left| { A }^{ -1 } \right| =\frac { 1 }{ \left| A \right|  } $

The inverse of a skew-symmetric matrix of an odd order is

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

  1. a symmetric matrix

  2. a skew-symmetric matrix

  3. diagonal matrix

  4. does not exists

Show answer
Correct Option: D
Explanation:

Let A be a skew-symmeteic matric of order $n.$

By definition $\displaystyle { A }^{ T }=-A$ 
$\displaystyle\Rightarrow \left| { A }^{ T } \right| =\left| -A \right| \Rightarrow \left| A \right| ={ \left( -1 \right)  }^{ n }\left| A \right| \$
$\displaystyle \Rightarrow \left| A \right| =-\left| A \right|\quad\quad[\because $ n is odd $]$
$\displaystyle \Rightarrow 2\left| A \right| =0\Rightarrow\left| A \right| =0$
$\therefore{ A }^{ -1 }$ does not exist. 

If $AB=A$ and $BA=B$, where $A$ and $B$ are square matrices, then 

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

  1. ${ B }^{ 2 }=B$ and ${ A }^{ 2 }=A$

  2. ${ B }^{ 2 }=A$ and ${ A }^{ 2 }=B$

  3. $AB=BA$

  4. none of these

Show answer
Correct Option: A
Explanation:

We have 

$AB=A\Rightarrow A\left( BA \right) =A\quad$, subsitute $BA=B$ 
${$ $A(BA) =(AB)A$ $}$
$\Rightarrow \left( AB \right) A=A$
Subsitute $AB = A$
$\Rightarrow AA=A\quad \quad \left[ \therefore AB=A \right] \ \Rightarrow { A }^{ 2 }=A$
Again $BA=B$

$\Rightarrow B\left( AB \right) =B\quad \quad \left[ \because AB=A \right] $
$\Rightarrow \left( BA \right) B=B$
$\Rightarrow BB=B$
$\Rightarrow { B }^{ 2 }=B$

If $A=\begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix}$, $B=\begin{bmatrix} 0 & -i \ i & 0 \end{bmatrix}$ then ${(A+B)}^{2}$ equals

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

  1. ${A}^{2}+{B}^{2}$

  2. ${A}^{2}+{B}^{2}+2AB$

  3. ${A}^{2}+{B}^{2}+AB-BA$

  4. none of these

Show answer
Correct Option: A
Explanation:

Given, $A=\begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix},B=\begin{bmatrix} 0 & -i \ i & 0 \end{bmatrix}$


$ A+B=\begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix}+\begin{bmatrix} 0 & -i \ i & 0 \end{bmatrix}=\begin{bmatrix} 0 & 1-i \ i+1 & 0 \end{bmatrix}$

$ { \left( A+B \right)  }^{ 2 }=\begin{bmatrix} 0 & 1-i \ i+1 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1-i \ i+1 & 0 \end{bmatrix}=\begin{bmatrix} 2 & 0 \ 0 & 2 \end{bmatrix}$

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

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

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

If $D=diag({d} _{1}, {d} _{2}, {d} _{3}........{d} _{n})$, where ${d} _{1}\ne 0$ for all $i=1, 2,.....n$, then ${D}^{-1}$ is equal to

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

  1. $D$

  2. ${I} _{n}$

  3. diag $({d} _{1}^{-1}, {d} _{2}^{-1}, ........{d} _{n}^{-1})$

  4. None of these

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Correct Option: C

If for suitable matrices $A, B$; $AB=A$ and $BA=B$; then ${A}^{2}$ equals-

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

  1. $I$

  2. $A$

  3. $B$

  4. $0$

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Correct Option: B,C
Explanation:

Taking $AB=A$


Multiplying by a matrix $A$

$ABA=A^2$

$\Rightarrow A(BA)=A^2$      (Associative property)

$AB=A^2$      ($\because BA=B$)

$\Rightarrow A=A^2$    ($\because AB=A$)

lf $\mathrm{A}$ is $\left{\begin{array}{lll}
8 & -6 & 2\
-6 & 7 & -4\
2 & -4 & \lambda
\end{array}\right}$  is a singular matrix then  $\lambda =$ 

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

  1. 3

  2. 4

  3. 2

  4. 5

Show answer
Correct Option: A
Explanation:

Given, $A=\begin{pmatrix}
8 & -6 & 2\
-6 & 7 & -4\
2 & -4 & \lambda
\end{pmatrix}$ is a singular matrix
So, det A=0
$\therefore $ BY operation of matrix (s),
$det A=8(7 \lambda-16)+6[-6 \lambda + 8]+2[24-14]$
$=56 \lambda - 128 -36 \lambda +48 +20$
$=20 \lambda - 60$
So, $det A = 0= 20 \lambda -60$
$\lambda =3$