## Filters

Sort by :
Clear All
Q

Choose the correct answer in the Exercises 11 and 12.

Q12.    If $A = \begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha& \cos\alpha \end{bmatrix}$ and $A+A' =I$, then the value of $\alpha$ is

(A)    $\frac{\pi}{6}$

(B)    $\frac{\pi}{3}$

(C)    $\pi$

(D)    $\frac{3\pi}{2}$

Option B is correct.

Choose the correct answer in the Exercises 11 and 12.

Q11.    If A, B are symmetric matrices of same order, then AB – BA is a

(A) Skew symmetric matrix
(B) Symmetric matrix
(C) Zero matrix
(D) Identity matrix

If A, B are symmetric matrices then                                                          and       we have,                                                                                                                                                                                                                                 Hence, we have  Thus,( AB-BA)' is skew symmetric. Option A...

Q10.    Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

(iv)    $\begin{bmatrix} 1 & 5\\ -1 & 2 \end{bmatrix}$

Let                       Thus,    is a symmetric matrix.                Let  Thus,  is a skew-symmetric matrix. Represent   A  as the sum of B and C.

Q10.    Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

(iii)    $\begin{bmatrix} 3 & 3 & -1\\ -2 & -2 & 1\\ -4 & -5 & 2 \end{bmatrix}$

Let                       Thus,    is a symmetric matrix.                Let  Thus,  is a skew-symmetric matrix. Represent   A  as the sum of B and C.

Q10.    Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

(ii)    $\begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}$

Let                       Thus,    is a symmetric matrix.                Let  Thus,  is a skew-symmetric matrix. Represent   A  as the sum of B and C.

10.    Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

(i)    $\begin{bmatrix} 3 & 5\\ 1 & -1 \end{bmatrix}$

Let                       Thus,    is a symmetric matrix.                Let  Thus,  is a skew symmetric  matrix. Represent   A  as sum of B and C.

Q9.    Find $\frac{1}{2}(A+A')$ and $\frac{1}{2}(A-A')$, when $A = \begin{bmatrix} 0 & a & b\\ -a & 0 & c\\ -b & -c & 0 \end{bmatrix}$

the transpose of the matrix is obtained by interchanging rows and columns

Q8.    For the matrix $A = \begin{bmatrix} 1 & 5\\ 6 & 7 \end{bmatrix}$, verify that

(ii) $(A - A')$ is a skew symmetric matrix.

We have  Hence ,   is a skew-symmetric matrix.

Q8.    For the matrix $A = \begin{bmatrix} 1 & 5\\ 6 & 7 \end{bmatrix}$, verify that

(i) $(A + A')$ is a symmetric matrix.

We have  Hence ,   is a symmetric matrix.

Q7.    (ii) Show that the matrix $A = \begin{bmatrix} 0 & 1 & -1\\ -1 & 0 &1 \\ 1 & -1 &0 \end{bmatrix}$ is a skew-symmetric matrix.

The transpose of A is Since, so given matrix is a skew-symmetric matrix.

Q7.    (i) Show that the matrix $A = \begin{bmatrix} 1 &- 1& 5\\ -1 & 2 & 1\\ 5 & 1 & 3 \end{bmatrix}$ is a symmetric matrix.

the transpose of A is Since, so given matrix is a symmetric matrix.

Q6.    (ii) If $A = \begin{bmatrix} \sin\alpha & \cos\alpha \\ -\cos\alpha & \sin\alpha \end{bmatrix}$, then verify that $A'A = I$

By interchanging columns and rows of the matrix A we get the transpose of A To prove:  L.H.S :

Q6.    (i) If $A = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}$, then verify that $A'A =I$

By interchanging rows and columns we get transpose of A To prove:  L.H.S :

Q5.    For the matrices A and B, verify that $(AB)' = B'A'$, where

(ii)    $A = \begin{bmatrix} 0\\ 1\\ 2 \end{bmatrix}$$B = \begin{bmatrix} 1 & 5&7 \end{bmatrix}$

,  To prove :  Heence, L.H.S =R.H.S  i.e..

Q5.    For the matrices A and B, verify that $(AB)' = B'A'$, where

(i)    $A = \begin{bmatrix} 1\\-4 \\3 \end{bmatrix}$$B = \begin{bmatrix} -1& 2 &1 \end{bmatrix}$

,     To prove :  Hence, L.H.S =R.H.S  so it is verified that .

Q4.    If $A' = \begin{bmatrix} -2 & 3\\ 1 & 2 \end{bmatrix}$ and $B= \begin{bmatrix} -1 & 0\\ 1 & 2 \end{bmatrix}$, then find $(A + 2B)'$

: Transpose is obtained by interchanging rows and columns and the transpose of A+2B is

Q3.    If $A = \begin{bmatrix} 3 & 4\\ -1 &2 \\ 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} -1 & 2 & 1\\ 1 &2 &3 \end{bmatrix}$, then verify

(ii)    $(A - B)' = A' - B'$

To prove:        R.H.S:      Hence, L.H.S = R.H.S i.e. .

Q3.    If $A' = \begin{bmatrix} 3 & 4\\ -1 &2 \\ 0 & 1 \end{bmatrix}$  and $B = \begin{bmatrix} -1 & 2 & 1\\ 1 &2 &3 \end{bmatrix}$, then verify

(i)    $(A + B)' = A' + B'$

$A' = \begin{bmatrix} 3 & 4\\ -1 &2 \\ 0 & 1 \end{bmatrix}$     $B = \begin{bmatrix} -1 & 2 & 1\\ 1 &2 &3 \end{bmatrix}$

$A=(A')' = \begin{bmatrix} 3 & -1&0\\ 4 &2 & 1 \end{bmatrix}$

To prove: $(A + B)' = A' + B'$

$L.H.S : (A + B)' =$

$A+B = \begin{bmatrix} 3 & -1&0\\ 4 &2 & 1 \end{bmatrix}$  $+ \begin{bmatrix} -1 & 2 & 1\\ 1 &2 &3 \end{bmatrix}$

$A+B = \begin{bmatrix} 3+(-1) & -1+(-1)&0+1\\ 4+1 &2+2 & 1+3 \end{bmatrix}$

$A+B = \begin{bmatrix} 2 & -2&1\\ 5 &4 & 4 \end{bmatrix}$

$\therefore \, \, \, (A+B)' = \begin{bmatrix} 2 & 5\\ 1 &4\\1 & 4 \end{bmatrix}$

R.H.S:  $A' + B'$

$A'+B' = \begin{bmatrix} 3 & 4\\ -1 &2 \\ 0 & 1 \end{bmatrix}$  $+ \begin{bmatrix} -1 & 1\\ 2 &2 \\ 1 & 3 \end{bmatrix}$

$A'+B' = \begin{bmatrix} 2 & 5\\ 1 &4 \\ 1 & 4 \end{bmatrix}$

Hence, L.H.S = R.H.S, hence verified that $(A + B)' = A' + B'$.

View More

Q2.    If $A = \begin{bmatrix} -1 & 2 & 3\\ 5 &7 &9 \\ -2 & 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} -4 & 1 & -5\\ 1 &2 &0 \\ 1 & 3 & 1 \end{bmatrix}$, then verify

(ii)    $(A - B)' = A' - B'$

and        L.H.S :        R.H.S :      Hence, L.H.S = R.H.S. so verified that   .

Q2.    If $A = \begin{bmatrix} -1 & 2 & 3\\ 5 &7 &9 \\ -2 & 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} -4 & 1 & -5\\ 1 &2 &0 \\ 1 & 3 & 1 \end{bmatrix}$, then verify

(i)    $(A + B)' = A' + B'$

and        L.H.S :        R.H.S :      Thus we find that the LHS is equal to RHS and hence verified.
Exams
Articles
Questions