# Q&A - Ask Doubts and Get Answers

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Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k,
respectively. Choose the correct answer in Exercises 21 and 22.

Q22.    If n = p, then the order of the matrix $7X - 5Z$ is:
(A) p × 2
(B) 2 × n
(C) n × 3
(D) p × n

X has of order   .   7X also  has of order  .       Z has of order   .   5Z also  has of order   . Mtarices 7X and 5Z can only be subtracted  if they both have same order  i.e  =    and it is given that  p=n. We can say that both matrices have order of . Thus, order of   is . Option (B) is correct.

Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k, respectively. Choose the correct answer in Exercises 21 and 22.

Q21.    The restriction on n, k and p so that PY + WY will be defined are:
(A) $k = 3, p = n$

(B) k is arbitrary,$p = 2$

(C) p is arbitrary, $k = 3$

(D) $k = 2, p = 3$

P and Y are of order   and  respectivly.   PY will be defined only if k=3, i.e. order of PY is . W and Y are of order   and  respectivly.   WY is  defined because the number of columns of W is equal to the number of rows of Y which is 3, i.e. the order of WY is  Matrices PY and WY can only be added if they both have same order i.e =    implies p=n. Thus, are restrictions on n, k and p so that...

Q20.     The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are Rs 80, Rs 60 and Rs 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

The bookshop has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are Rs 80, Rs 60 and Rs 40 each respectively. The total amount the bookshop will receive from selling all the books:   The total amount the bookshop will receive from selling all the books is 20160.

Q19.    A trust fund has Rs. 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of:

(ii) Rs. 2000

Let Rs. x be invested in the first bond. Money invested in second bond = Rs (3000-x) The first bond pays 5% interest per year and the second bond pays 7% interest per year. To obtain an annual total interest of  Rs. 1800,  we have                           (simple interest for 1 year  )   Thus, to obtain an annual total interest of  Rs. 2000, the trust fund should invest Rs 5000 in the...

Q19.    A trust fund has Rs. 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of:

(a) Rs. 1800

Let Rs. x be invested in the first bond. Money invested in second bond = Rs (3000-x) The first bond pays 5% interest per year and the second bond pays 7% interest per year. To obtain an annual total interest of  Rs. 1800,  we have                           (simple interest for 1 year  )   Thus, to obtain an annual total interest of  Rs. 1800, the trust fund should invest Rs 15000 in the...

Q18.     If $A = \begin{bmatrix} 0&-\tan\frac{\alpha}{2}\\\tan\frac{\alpha}{2}& 0\end{bmatrix}$ and I is the identity matrix of order 2, show that$I + A = (I- A)\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$

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

Q17.    If

$A = \begin{bmatrix}3 &-2\\4&-2 \end{bmatrix}$ and

$I = \begin{bmatrix}1 &0\\0&1 \end{bmatrix}$,

find k so that $A^{2} = kA - 2I$.

We have,                Hence, the value of k is 1.

Q16.    If

$A = \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix}$

prove that $A^3 - 6A^2 + 7A + 2I = 0$.

First, find the square of matrix A and then multiply it with A to get the cube of matrix A                        L.H.S :                                            Hence, L.H.S = R.H.S    i.e..

Q15.    Find$A^2 -5A + 6I$, if

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

First, we will find ou the value of the square of matrix A

Q14.    Show that

(ii)    $\begin{bmatrix}1& 2&3\\0&1&0\\1&1&0 \end{bmatrix} \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix} \neq \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix}1& 2&3\\0&1&0\\1&1&0 \end{bmatrix}$

To prove the following multiplication of three by three matrices are not equal                                                                                  Hence,    i.e. .

Q14.    Show that

(i)    $\begin{bmatrix}5&-1\\6&7 \end{bmatrix}\begin{bmatrix} 2 & 1\\ 3 & 4 \end{bmatrix}\neq \begin{bmatrix} 2 & 1\\ 3 & 4 \end{bmatrix}\begin{bmatrix}5&-1\\6&7 \end{bmatrix}$

To prove:                               Hence, the right-hand side not equal to the left-hand side, that is  .

Q13.    If $F(x) = \begin{bmatrix} \cos x & -\sin x& 0\\\sin x &\cos x & 0 \\ 0 &0&1\end{bmatrix}$, show that $F(x) F(y) = F(x + y)$.

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

Q12.    Given $3\begin{bmatrix}x & y \\ z & w \end{bmatrix} = \begin{bmatrix} x & 6 \\ -1 & 2w \end{bmatrix} + \begin{bmatrix} 4 &x + y \\ z + w & 3 \end{bmatrix}$, find the values of x, y, z and w.

If two matrices are equal than corresponding elements are also equal. Thus, we have                                            Put the value of x Hence, we have

Q11.    If $x\begin{bmatrix}2\\3 \end{bmatrix} + y\begin{bmatrix} -1\\1 \end{bmatrix} = \begin{bmatrix} 10\\5 \end{bmatrix}$, find the values of x and y.

Adding both the matrix in LHS and rewriting                                Adding equation 1 and 2, we get                      Put the value of x in equation 2, we have

Q10.    Solve the equation for x, y, z and t, if $2\begin{bmatrix}x & z \\ y &t \end{bmatrix} + 3\begin{bmatrix} 1 & -1\\ 0 & 2 \end{bmatrix} = 3\begin{bmatrix} 3 & 5\\ 4 & 6 \end{bmatrix}$

Multiplying with constant terms and rearranging we can rewrite the matrix as Dividing by 2 on both sides

Q9.    Find x and y, if $2\begin{bmatrix} 1 & 3\\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0\\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6\\ 1 & 8\end{bmatrix}$

Now equating LHS and RHS we can write the following equations

Q8.    Find X, if $Y = \begin{bmatrix} 3 &2 \\ 1 & 4 \end{bmatrix}$ and $2X+ Y = \begin{bmatrix} 1 &0 \\ -3 & 2 \end{bmatrix}$

Substituting the value of Y in the above equation

Q7.    Find X and Y, if

(ii) $2X + 3Y = \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix}$ and $3X + 2Y = \begin{bmatrix} 2 &-2 \\ -1 & 5 \end{bmatrix}$

(ii)  and                                    Multiply equation 1 by 3 and equation 2 by 2 and subtract them,                                                          Putting value of Y in equation 1 , we get

Q7.    Find X and Y, if

(i) $X + Y = \begin{bmatrix} 7 &0 \\ 2 &5 \end{bmatrix}$ and $X - Y = \begin{bmatrix} 3 &0 \\ 0 &3 \end{bmatrix}$

(i) The given matrices are       and                                  Adding equation 1 and 2, we get                                                Putting the value of X in equation 1, we get

Q6.    Simplify $\cos\theta\begin{bmatrix} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta \end{bmatrix} + \sin\theta\begin{bmatrix} \sin\theta & -\cos\theta\\ \cos\theta & \sin\theta \end{bmatrix}$.

The simplification is explained in the following step     the final answer is an identity matrix of order 2
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