Question 11 Mark
If $A=\left[\begin{array}{cc} {\alpha} & {\beta} \\ {\gamma} & {-\alpha} \end{array}\right]$ is such that $A^2 = I,$ then
AnswerIn the given question: $A=\left[\begin{array}{ll} {\alpha} & {\beta} \\ {\gamma} & {-\alpha} \end{array}\right]$
Calculating $A^2: A^2 = A.A$
= $\left[\begin{array}{cc} {\alpha} & {\beta} \\ {\gamma} & {-\alpha} \end{array}\right]\left[\begin{array}{cc} {\alpha} & {\beta} \\ {\gamma} & {-\alpha} \end{array}\right]$
= $\left[\begin{array}{cc} {\alpha \cdot \alpha+\beta \cdot \gamma} & {\alpha \cdot \beta+\beta \cdot(-\alpha)} \\ {\gamma \cdot \alpha+(-\alpha) \cdot \gamma} & {\gamma \cdot \beta+(-\alpha) \cdot(-\alpha)} \end{array}\right]$
= $\left[\begin{array}{cc} {\alpha^{2}+\beta \gamma} & {\alpha \cdot \beta-\alpha \cdot \beta} \\ {\gamma \cdot \alpha-\gamma \cdot \alpha} & {\gamma \cdot \beta+\alpha^{2}} \end{array}\right]$
= $\left[\begin{array}{cc} {\alpha^{2}+\beta \gamma} & {0} \\ {0} & {\beta \gamma+\alpha^{2}} \end{array}\right]$
And given that $A^2 = I$
Therefore, $\left[\begin{array}{cc} {\alpha^{2}+\beta \gamma} & {0} \\ {0} & {\beta \gamma+\alpha^{2}} \end{array}\right]=\left[\begin{array}{ll} {1} & {0} \\ {0} & {1} \end{array}\right]$
Comparing corresponding elements we obtain
$\Rightarrow \alpha^{2}+\beta \gamma=1$
$\Rightarrow 1-\alpha^{2}-\beta \gamma=0$
View full question & answer→Question 21 Mark
If A is square matrix such that $A^2 = A,$ then $(I + A)^3 – 7\ A$ is equal to
AnswerGiven that $A^2 = A$
Calculating value of $(I + A)^3 - 7A:$
$(I+A)^3 - 7A=I^{3}+A^{3}+3 I^{2} A+3 I A^{2}-7 A$
$= \mathrm{I}+\mathrm{A}^{2} \cdot \mathrm{A}+3 \mathrm{A}+3 \mathrm{A}^{2}-7 \mathrm{A}\left(\mathrm{I}^{\mathrm{n}}=\mathrm{I} \text { and } \mathrm{I} \cdot \mathrm{A}=\mathrm{A}\right)$
$= I+A \cdot A+3 A+3 A-7 A\left(A^{2}=A\right)$
$= I+A^{2}+3 A+3 A-7 A$
$= I +7A-7A$
Hence,$ (I + A)^3 - 7A = I$
View full question & answer→Question 31 Mark
If the matrix A is both symmetric and skew symmetric, then
AnswerOnly a null matrix can be symmetric as well as skew symmetric.
In Symmetric Matrix $A^T= A,$
Skew Symmetric Matrix $A^T= -A,$
Given that the matrix is satisfying both the properties.
Therefore, Equating the RHS we get $A = -A$ i.e, $2A = 0.$
Therefore $A = 0,$ which is a null matrix.
View full question & answer→Question 41 Mark
For the matrix $A=\left[\begin{array}{ll} {1} & {5} \\ {6} & {7} \end{array}\right]$, verify that (A – A′) is a skew-symmetric matrix
AnswerHere, $A=\left[\begin{array}{ll} {1} & {5} \\ {6} & {7} \end{array}\right] \text { and } A^{\prime}=\left[\begin{array}{ll} {1} & {6} \\ {5} & {7} \end{array}\right]$
subtracting A' from A, we get,
$A-A^{\prime}=\left[\begin{array}{ll} {1} & {5} \\ {6} & {7} \end{array}\right]-\left[\begin{array}{ll} {1} & {6} \\ {5} & {7} \end{array}\right]$
$\Rightarrow A-A^{\prime}=\left[\begin{array}{ll} {1-1} & {5-6} \\ {6-5} & {7-7} \end{array}\right]=\left[\begin{array}{ll} {0} & {-1} \\ {1} & ~~~{0} \end{array}\right]$
Clearly, $\left(\mathrm{A}-\mathrm{A}^{\prime}\right)^{\prime}=\left[\begin{array}{cc} {0} & {1} \\ {-1} & {0} \end{array}\right]~~~...(i)$
We can rewrite the above equation as
$\left(\mathrm{A}-\mathrm{A}^{\prime}\right)^\prime=(-1)\left[\begin{array}{cc} {0} & {-1} \\ {1} & {0} \end{array}\right]=(-1)(A-A^\prime) ~~~~~... using (i)$
$\therefore$ (A – A')' = -(A – A')
Hence we can say that matrix A is a Skew-symmetric matrix.
View full question & answer→Question 51 Mark
Show that the matrix $A=\left[\begin{array}{rrr} {0} & {1} & {-1} \\ {-1} & {0} & {1} \\ {1} & {-1} & {0} \end{array}\right]$is a skew-symmetric matrix.
AnswerA matrix is said to be skew-symmetric if the transpose of the matrix is equal to the negative of the matrix. This means that A' = -A. Here,
$A=\left[\begin{array}{ccc} {0} & {1} & {-1} \\ {-1} & {0} & {1} \\ {1} & {-1} & {0} \end{array}\right]$ ...(1)
Therefore,we get
$A^{\prime}=\left[\begin{array}{ccc} {0} & {-1} & {1} \\ {1} & {0} & {-1} \\ {-1} & {1} & {0} \end{array}\right]$ ...(2)
Now if we carefully look at the equation 2 we can rewrite it as
$A^{\prime}=(-1)\left[\begin{array}{ccc} {0} & {1} & {-1} \\ {-1} & {0} & {1} \\ {1} & {-1} & {0} \end{array}\right]$(by taking - 1 common)
Therefore, A' = (-1) $\times$ A (from equation 1)
⇒ A' = -A,
Therefore, we can say that matrix A is a Skew-symmetric matrix.
View full question & answer→Question 61 Mark
Show that the matrix $A=\left[\begin{array}{rrr} {1} & {-1} & {5} \\ {-1} & {2} & {1} \\ {5} & {1} & {3} \end{array}\right]$is a symmetric matrix.
AnswerA matrix is said to be symmetric only if the transpose of a matrix and the matrix itself are equal or the same. This means that A = A'. Here,
$A=\left[\begin{array}{ccc} {1} & {-1} & {5} \\ {-1} & {2} & {1} \\ {5} & {1} & {3} \end{array}\right] $ ...(1)
Therefore,
$A^{\prime}=\left[\begin{array}{ccc} {1} & {-1} & {5} \\ {-1} & {2} & {1} \\ {5} & {1} & {3} \end{array}\right]$ ...(2)
Therefore,from equation (1) & (2) we obtain
A = A',
Therefore, we can say that matrix A is a Symmetric matrix.
View full question & answer→Question 71 Mark
Find the transpose of the matrix: $\left[\begin{array}{ccc} {-1} & {5} & {6} \\ {\sqrt{3}} & {5} & {6} \\ {2} & {3} & {-1} \end{array}\right]$
AnswerWe known that transpose of a matrix is obtained by interchanging the elements of the rows and columns. In other words, we can say, if ${A}=\left[{a}_{{i} j}\right]_{m\times n }$ then ${A}^{\prime}=\left[{a}_{\mathrm{ji}}\right]_{\mathrm{n} \times \mathrm{m}}$
So, let $\mathrm{C}=\left[\begin{array}{ccc} {-1} & {5} & {6} \\ {\sqrt{3}} & {5} & {6} \\ {2} & {3} & {-1} \end{array}\right]$
Therefore, the transpose of the given matrix C is denoted by C' and given by
$C^{\prime}=\left[\begin{array}{ccc} {-1} & {\sqrt{3}} & {2} \\ {5} & {5} & {3} \\ {6} & {6} & {-1} \end{array}\right]$
i.e, Transpose of the given matrix is $\left[\begin{array}{ccc} {-1} & {\sqrt{3}} & {2} \\ {5} & {5} & {3} \\ {6} & {6} & {-1} \end{array}\right]$
View full question & answer→Question 81 Mark
If $A = \left[\begin{array}{cc} {\cos \alpha} & {-\sin \alpha} \\ {\sin \alpha} & {\cos \alpha} \end{array}\right]$, and $A + A' = I,$ if the value of $\alpha$ is
AnswerGiven $A=\left[\begin{array}{cc} {\cos \alpha} & {-\sin \alpha} \\ {\sin \alpha} & {\cos \alpha} \end{array}\right]$
Therefore, $A^{\prime}=\left[\begin{array}{cc} {\cos \alpha} & {\sin \alpha} \\ {-\sin \alpha} & {\cos \alpha} \end{array}\right]$
Also given that $A + A' = I ...(1)$
$($Putting the values in equation $(1))$
$\left[\begin{array}{cc} {\cos \alpha} & {-\sin \alpha} \\ {\sin \alpha} & {\cos \alpha} \end{array}\right]+\left[\begin{array}{cc} {\cos \alpha} & {\sin \alpha} \\ {-\sin \alpha} & {\cos \alpha} \end{array}\right]=\left[\begin{array}{ll} {1} & {0} \\ {0} & {1} \end{array}\right]$
$\Rightarrow$ $\left[\begin{array}{cc} {\cos \alpha+\cos \alpha} & {-\sin \alpha+\sin \alpha} \\ {\sin \alpha-\sin \alpha} & {\cos \alpha+\cos \alpha} \end{array}\right]=\left[\begin{array}{ll} {1} & {0} \\ {0} & {1} \end{array}\right]$
$\Rightarrow$ $\left[\begin{array}{cc} {2 \cos \alpha} & {0} \\ {0} & {2 \cos \alpha} \end{array}\right]=\left[\begin{array}{ll} {1} & {0} \\ {0} & {1} \end{array}\right]$
We know the two matrices are equal only when all their corresponding elements or entries are equal i.e. if $A = B,$ then $a_{ij}$ and $b_{ij}$ for all $i$ and $j.$
This implies,
$2 \cos \alpha=1$
$\Rightarrow$ $\cos \alpha=\frac{1}{2}$
$\Rightarrow$ $\cos \alpha=\cos \frac{\pi}{3}~~~ ...\left(\because \cos \frac{\pi}{3}=\frac{1}{2}\right)$
$\Rightarrow$ $\alpha=\frac{\pi}{3}$
View full question & answer→Question 91 Mark
Find the transpose of the matrix: $\left[\begin{array}{rr} {1} & {-1} \\ {2} & {3} \end{array}\right]$
AnswerWe know that transpose of a matrix is obtained by interchanging the elements of the rows and columns. In other words, we can say, if
$\mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right]_{m \times \mathrm{n}} \text { then } \mathrm{A}^{\prime}=\left[\mathrm{a}_{\mathrm{ji}}\right]_{\mathrm{n\times m}}$
So, let $B =$$\left[\begin{array}{cc} {1} & {-1} \\ {2} & {3} \end{array}\right]$
Therefore, transpose of the given matrix B denoted by B’ is given by
$ B' $= $\left[\begin{array}{cc} {1} & {2} \\ {-1} & {3} \end{array}\right]$
So,Transpose of the given matrix is $\left[\begin{array}{cc} {1} & {2} \\ {-1} & {3} \end{array}\right]$
View full question & answer→Question 101 Mark
If A, B are symmetric matrices of same order, then AB – BA is a
AnswerGiven A and B are symmetric matrix of same order
$\Rightarrow$ A = A' .......(i)
$\Rightarrow$ B = B' ......(ii)
So, AB - BA = A'B' - B'A' ...(from eqn (i) and (ii) )
$\Rightarrow$ AB - BA = (BA)' - (AB)' ... ($\because$ (AB)' = B'A')
$\Rightarrow$ AB - BA = (-1) ((AB)' - (BA)') ...(taking -1 common)
$\Rightarrow$ AB - BA = -(AB - BA)' ...($\because$ (A - B)' = A' - B')
Here we see that the relation between (AB - BA) and its transpose i.e. (AB - BA)' is (AB - BA) = - (AB -BA)', this implies that (AB - BA) is a skew symmetric matrix.
View full question & answer→Question 111 Mark
Find the transpose of the matrix: $\left[\begin{array}{c} {5} \\ {\frac{1}{2}} \\ {-1} \end{array}\right]$
AnswerWe know that transpose of a matrix is obtained by interchanging the elements of the rows and columns. In other words, we can say, if
$\mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right]_{m \times \mathrm{n}} \text { then } \mathrm{A}^{\prime}=\left[\mathrm{a}_{\mathrm{ji}}\right]_{\mathrm{n\times m}}$
So, let $\left[\begin{array}{c} {5} \\ {\frac{1}{2}} \\ {-1} \end{array}\right]=A$
Therefore, transpose of the given matrix A is denoted by A’
Hence, A' = $\left[\begin{array}{ccc} {5} & {\frac{1}{2}} & {-1} \end{array}\right]$
The transpose of the given matrix is $\left[\begin{array}{ccc} {5} & {\frac{1}{2}} & {-1} \end{array}\right]$
View full question & answer→Question 121 Mark
Compute $\left[\begin{array}{ccc} {-1} & {4} & {-6} \\ {8} & {5} & {16} \\ {2} & {8} & {5} \end{array}\right]+\left[\begin{array}{ccc} {12} & {7} & {6} \\ {8} & {0} & {5} \\ {3} & {2} & {4} \end{array}\right]$
Answer$\left[\begin{array}{ccc} {-1} & {4} & {-6} \\ {8} & {5} & {16} \\ {2} & {8} & {5} \end{array}\right]+\left[\begin{array}{ccc} {12} & {7} & {6} \\ {8} & {0} & {5} \\ {3} & {2} & {4} \end{array}\right]$
= $\left[\begin{array}{ccc} {-1+12} & {4+7} & {-6+6} \\ {8+8} & {5+0} & {16+5} \\ {2+3} & {8+2} & {5+4} \end{array}\right]$
= $\left[\begin{array}{ccc} {11} & {11} & {0} \\ {16} & {5} & {21} \\ {5} & {10} & {9} \end{array}\right]$
View full question & answer→Question 131 Mark
Assume X, Y, Z, W, and P are matrices of order 2 $\times$ n, 3 $\times$ k, 2 $\times$ p, n $\times$ 3 and p $\times$ k, respectively.
If n = p, then the order of the matrix 7X – 5Z is
AnswerMatrix X is of the order 2 $\times$ n.
Therefore, matrix 7X is also of the same order.
Matrix Z is of order 2 $\times$ p $= 2\times n ~~~~~...(\because p=n)$
$\Rightarrow$ Matrix 5Z is also of the same order.
Now, both the matrices 7X and 5Z are of the order 2 $\times$ n.
Thus, matrix 7X – 5Z is well- defined and is of the order 2 $\times$ n.
View full question & answer→Question 141 Mark
Assume X, Y, Z, W, and P are matrices of order 2 $\times$ n, 3 $\times$ k, 2 $\times$ p, n $\times$ 3 and p $\times$ k, respectively.
The restriction on n, k and p so that PY + WY will be defined are
AnswerMatrices P and Y are of the orders p $\times$ k and 3 $\times$ k respectively.
Therefore, matrix PY will be defined if k = 3.
Then, PY will be of the order p $\times$ k = p$\times$ 3.
Matrices W and Y are of the orders n $\times$ 3 and 3 $\times$ k = 3 $\times$ 3 respectively.
As, the number of columns in W is equal to the number of rows in Y, Matrix WY is well defined and is of the order n $\times$ 3.
Matrices PY and WY can be added only when their orders are the same.
Therefore, PY is of the order p $\times$ 3 and WY is of the order n $\times$ 3.
Thus, we must have p = n.
Therefore, k = 3 and p = n are the restrictions on n, k and p so that
PY + WY will be defined.
View full question & answer→Question 151 Mark
A trust fund has ₹ 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 ₹ 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of: ₹ 2000
AnswerLet ₹ x be invested in the first bond.
Then, the sum of money invested in the second bond will be ₹ (30000 – x).
It is given that the first bond pays 5% interest per year, and the second bond pays 7% interest per year.
Hence, the amount invested in each type of the bonds can be represented in matrix form with each column corresponding to a different type of bond as:
$X=\left[\begin{array}{ll} x & 30000-x \end{array}\right]$
Hence, the interest obtained after one year can be expressed in matrix representation as:
$\left[\begin{array}{ll} {x} & {(30000-x)} \end{array}\right]$$\left[\begin{array}{c} {\frac{5}{100}} \\ {\frac{7}{100}} \end{array}\right]$ = 2000
$\Rightarrow \frac{5 x}{100}+\frac{7(30000-x)}{100}$ = 2000
$\Rightarrow$ 5x + 210000 -7x = 200000
$\Rightarrow$ 210000 -2x = 200000
$\Rightarrow$ 2x = 210000 – 200000
$\Rightarrow$ 2x = 10000
$\Rightarrow$ x = 5000
Therefore, in order to obtain an annual total interest of ₹ 2000, the trust fund should invest ₹ 5000 in the first bond and the remaining ₹ (30000−5000) = ₹ 25000 in the second bond.
View full question & answer→Question 161 Mark
A trust fund has ₹ 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 ₹ 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of: ₹ 1800
AnswerLet Rs x be invested in the first bond.
Then, the sum of money invested in the second bond will be ₹ (30000 – x).
It is given that the first bond pays 5% interest per year, and the second bond pays 7% interest per year.
Thus, in order to obtain an annual total interest of ₹ 1800, we get:
$\left[\begin{array}{ll} {x} & {(30000-x)} \end{array}\right]$$\left[\begin{array}{c} {\frac{5}{100}} \\ {\frac{7}{100}} \end{array}\right]$ = 1800
$\Rightarrow \frac{5 \mathrm{x}}{100}+\frac{7(30000-\mathrm{x})}{100}$ = 1800
$\Rightarrow$ 5x + 210000 -7x = 180000
$\Rightarrow$ 210000 -2x = 180000
$\Rightarrow$ 2x = 210000 – 180000
$\Rightarrow$ 2x = 30000
$\Rightarrow$ x = 15000
Therefore, in order to obtain an annual total interest of ₹ 1800, the trust fund should invest ₹ 15000 in the first bond and the remaining ₹ 15000 in the second bond.
View full question & answer→Question 171 Mark
Let A = $\left[\begin{array}{ll} {2} & {4} \\ {3} & {2} \end{array}\right]$, B = $\left[\begin{array}{ll} {1} & {3} \\ {-2} & {5} \end{array}\right]$, C = $\left[\begin{array}{cc} {-2} & {5} \\ {3} & {4} \end{array}\right]$. Find each of the following:
- A + B
- A - B
- 3A - C
- AB
- BA
Answer - A + B = $\left[\begin{array}{ll} {2} & {4} \\ {3} & {2} \end{array}\right]+\left[\begin{array}{cc} {1} & {3} \\ {-2} & {5} \end{array}\right]$
= $\left[\begin{array}{cc} {2+1} & {4+3} \\ {3-2} & {2+5} \end{array}\right]$
= $\left[\begin{array}{ll} {3} & {7} \\ {1} & {7} \end{array}\right]$ - A – B = $\left[\begin{array}{ll} {2} & {4} \\ {3} & {2} \end{array}\right]-\left[\begin{array}{cc} {1} & {3} \\ {-2} & {5} \end{array}\right]$
= $\left[\begin{array}{cc} {2-1} & {4-3} \\ {3-(-2)} & {2-5} \end{array}\right]$
= $\left[\begin{array}{cc} {1} & {1} \\ {5} & {-3} \end{array}\right]$ - We have,
3A - C = $3\left[\begin{array}{cc} {2} & {4} \\ {3} & {2} \end{array}\right]-\left[\begin{array}{cc} {-2} & {5} \\ {3} & {4} \end{array}\right]$
= $\left[\begin{array}{ll} {3 \times 2} & {3 \times 4} \\ {3 \times 3} & {3 \times 2} \end{array}\right]-\left[\begin{array}{cc} {-2} & {5} \\ {3} & {4} \end{array}\right]$
= $\left[\begin{array}{cc} {6} & {12} \\ {9} & {6} \end{array}\right]-\left[\begin{array}{cc} {-2} & {5} \\ {3} & {4} \end{array}\right]$
= $\left[\begin{array}{cc} {6+2} & {12-5} \\ {9-3} & {6-4} \end{array}\right]$
= $\left[\begin{array}{ll} {8} & {7} \\ {6} & {2} \end{array}\right]$ - We have,
AB = $\left[\begin{array}{cc} {2} & {4} \\ {3} & {2} \end{array}\right]\left[\begin{array}{cc} {1} & {3} \\ {-2} & {5} \end{array}\right]$
= $\left[\begin{array}{ll} {2(1)+4(-2)} & {2(3)+4(5)} \\ {3(1)+2(-2)} & {3(3)+2(5)} \end{array}\right]$
= $\left[\begin{array}{ll} {2-8} & {6+20} \\ {3-4} & {9+10} \end{array}\right]$
= $\left[\begin{array}{cc} {-6} & {26} \\ {-1} & {19} \end{array}\right]$ - We have
BA = $\left[\begin{array}{cc} {1} & {3} \\ {-2} & {5} \end{array}\right]\left[\begin{array}{ll} {2} & {4} \\ {3} & {2} \end{array}\right]$
= $\left[\begin{array}{cc} {1(2)+3(3)} & {1(4)+3(2)} \\ {-2(2)+5(3)} & {-2(4)+5(2)} \end{array}\right]$
= $\left[\begin{array}{rr} {2+9} & {4+6} \\ {-4+15} & {-8+10} \end{array}\right]$
= $\left[\begin{array}{ll} {11} & {10} \\ {11} & {2} \end{array}\right]$
View full question & answer→Question 181 Mark
Which of the given values of x and y make the pair of matrices equal
$\left[\begin{array}{cc} {3 x+7} & {5} \\ {y+1} & {2-3 x} \end{array}\right],\left[\begin{array}{cc} {0} & {y-2} \\ {8} & {4} \end{array}\right]$
AnswerIn the given question:
$\left[\begin{array}{cc} {3 x+7} & {5} \\ {y+1} & {2-3 x} \end{array}\right]=\left[\begin{array}{cc} {0} & {y-2} \\ {8} & {4} \end{array}\right]$
Since, the given matrices are equal, their corresponding elements are also equal.
Comparing the corresponding element, we have:
3x + 7 = 0
$\Rightarrow x=-\frac{7}{3}$
Also, 5 = y – 2
$\Rightarrow$ y = 7
Again, y + 1 = 8
$\Rightarrow$ y = 7
And 2 – 3x = 4
$\Rightarrow ~x=-\frac{2}{3}$
Therefore, on comparing the corresponding elements of the two matrices, we get different values of x, which is not possible.
Thus, it is not possible to find the values of x and y for which the given matrices are equal.
View full question & answer→Question 191 Mark
$A = [a_{ij}]_{ m \times n}$ is a square matrix, if
AnswerWe know that if a given matrix is said to be a square matrix if the number of rows is equal to the number of columns.
Therefore, $A = [a_{ij}]_{ m \times n}$ is a square matrix if $m = n.$
View full question & answer→Question 201 Mark
Find the values of x, y, and z from the following equation:
$\left[\begin{array}{ll} {4} & {3} \\ {x} & {5} \end{array}\right]=\left[\begin{array}{ll} {y} & {z} \\ {1} & {5} \end{array}\right]$
Answer$\left[\begin{array}{ll} {4} & {3} \\ {x} & {5} \end{array}\right]=\left[\begin{array}{ll} {y} & {z} \\ {1} & {5} \end{array}\right]$
Since, the given matrices are equal, their corresponding elements are also equal.
Comparing the corresponding element, we have:
x = 1, y = 4 and z = 3
View full question & answer→Question 211 Mark
Construct a $2 \times 2$ matrix $A = [a_{ij}]$, whose element $a_{ij} = \frac{(i+2j)^{2}}{2}$.
AnswerSince it is a $ 2 \times 2$ matrix
it has $2$ rows and $2$ column.
Let matrix be $A$
Where $A=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right]$
Now it is given that
$\mathrm{a}_{\mathrm{ij}}=\frac{(\mathrm{i}+2 \mathrm{j})^2}{2}$
$a_{11}=\frac{(1+2(1))^2}{2}=\frac{(1+2)^2}{2}=\frac{(3)^2}{2}=\frac{9}{2}$
$a_{12}=\frac{(1+2(2))^2}{2}=\frac{(1+4)^2}{2}=\frac{(5)^2}{2}=\frac{25}{2}$
$\mathrm{a}_{21}=\frac{(2+2(1))^2}{2}=\frac{(2+2)^2}{2}=\frac{(4)^2}{2}=\frac{16}{2} = 8$
$\mathrm{a}_{22}=\frac{(2+2(2))^2}{2}=\frac{(2+4)^2}{2}=\frac{(6)^2}{2}=\frac{36}{2} = 18$
Hence,
the required matrix A is
$A\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right]=\left[\begin{array}{cc} \frac{9}{2} & \frac{25}{2} \\ 8 & 18 \end{array}\right]$
View full question & answer→Question 221 Mark
Construct a 2 $\times$ 2 matrix, A = [aij], whose element $a_{i j}=\frac{i}{j}$
AnswerIn general, a 2 $\times$ 2 matrix is given by A = $\left[\begin{array}{ll} {a_{11}} & {a_{12}} \\ {a_{21}} & {a_{22}} \end{array}\right]$
$a_{i j}=\frac{i}{j}; ~~~ i, j=1,2$
Therefore, $a_{11}=\frac{1}{1}=1$
$a_{12}=\frac{1}{2}$
$a_{21}=\frac{2}{1}=2$
$a_{22}=\frac{2}{2}=1$
Therefore, the required matrix is A = $\left[\begin{array}{ll} {1} & {\frac{1}{2}} \\ {2} & {1} \end{array}\right]$
View full question & answer→Question 231 Mark
Construct a $2\times 2$ matrix $A = [a_{ij}]$ whose element $a_{ij}$ is given by: $\frac{(i+j)^{2}}{2}$
AnswerLet $A = [a_{ij}]_{2 \times2}$
So, the elements in a $2\times2$ matrix are
$a_{11}, a_{12}, a_{21}, a_{22},$
$A = \begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix} ...(i)$
$a_{11} =\frac{\left(1+1\right)^2}2$ = $\frac42 = 2$
$a_{12} = \frac{(1+2)^{2}}{2}=\frac{3^{2}}{2}=\frac{9}{2}$
$a_{21} = \frac{(2+1)^{2}}{2}=\frac{3^{2}}{2}=\frac{9}{2}$
$a_{22} = \frac{(2+2)^{2}}{2}=\frac{4^{2}}{2}=\frac{16}{2} = 8$
So, (i) becomes
A = $\left(\begin{array}{cc} {2} & {\frac 92} \\ {\frac92} & {8} \end{array}\right)$
View full question & answer→Question 241 Mark
If a matrix has 18 elements, what are the possible orders it can have? What if it has 5 elements?
AnswerSince, a matrix having mn element is of order m × n.
- Therefore, there are 6 possible matrices having 18 elements of orders 1×18, 2×9, 3×6, 18×1, 9×2, 6×3.
- Prime number 5 = 1×5 and 5×1. Therefore, there are 2 possible matrices of order 1×5 (Row matrix) and 5×1 (Column matrix).
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If a matrix has 24 elements, what are possible orders it can have? What, if it has 13 elements?
AnswerSince, a matrix having mn element is of order m × n.
- Therefore, there are 8 possible matrices having 24 elements of orders 1 × 24, 2 × 12, 3 × 8, 4 × 6, 24 × 1, 12 × 2, 8 × 3, 6 × 4.
- Prime number 13 = 1 × 13 and 13 × 1. Therefore, there are 2 possible matrices of order 1 × 13 (Row matrix) and 13 × 1 (Column matrix).
View full question & answer→Question 261 Mark
The number of all possible matrices of order $3 \times 3$ with each entry $0$ or $1$ is
Answer$2^{3\times 3} = 2^9 = 512$.The number of elements in a 3 $\times$ 3 matrix is the product $3 \times 3 = 9.$
Each element can either be a $0$ or a $1$.
Given this, the total possible matrices that can be selected is $2^9= 512$
View full question & answer→Question 271 Mark
If $A=\left[\begin{array}{cc}8 & 0 \\ 4 & -2 \\ 3 & 6\end{array}\right]$ and $B=\left[\begin{array}{cc}2 & -2 \\ 4 & 2 \\ -5 & 1\end{array}\right]$, then find the matrix $X$ such that $2 A+3 X=5 B$.
AnswerGiven:3X = 5B - 2A
$=5\left[ {\begin{array}{*{20}{c}} 2 \\ 4 \\ { - 5} \end{array}\;\;\begin{array}{*{20}{c}} { - 2} \\ 2 \\ 1 \end{array}} \right] - 2\left[ {\begin{array}{*{20}{c}} 8 \\ 4 \\ 3 \end{array}\;\;\begin{array}{*{20}{c}} 0 \\ { - 2} \\ 6 \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} {10} \\ {20} \\ { - 25} \end{array}\;\;\begin{array}{*{20}{c}} { - 10} \\ {10} \\ 5 \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} { - 16} \\ { - 8} \\ { - 6} \end{array}\;\;\begin{array}{*{20}{c}} 0 \\ 4 \\ { - 12} \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} { - 6} \\ {12} \\ { - 31} \end{array}\;\;\begin{array}{*{20}{c}} { - 10} \\ {14} \\ { - 7} \end{array}} \right]$
$X = \frac{1}{3}\left[ {\begin{array}{*{20}{c}} { - 6} \\ {12} \\ { - 31} \end{array}\;\;\begin{array}{*{20}{c}} { - 10} \\ {14} \\ { - 7} \end{array}} \right]$
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If A = $\left[\begin{array}{lll} {1} & {2} & {3} \\ {2} & {3} & {1} \end{array}\right]$and B = $\left[\begin{array}{ccc} {3} & {-1} & {3} \\ {-1} & {0} & {2} \end{array}\right]$ then find 2A – B.
AnswerWe have
2A – B = 2$\left[\begin{array}{lll} {1} & {2} & {3} \\ {2} & {3} & {1} \end{array}\right]-\left[\begin{array}{ccc} {3} & {-1} & {3} \\ {-1} & {0} & {2} \end{array}\right]$
= $\left[\begin{array}{lll} {2} & {4} & {6} \\ {4} & {6} & {2} \end{array}\right]+\left[\begin{array}{rrr} {-3} & {1} & {-3} \\ {1} & {0} & {-2} \end{array}\right]$
= $\left[\begin{array}{ccc} {2-3} & {4+1} & {6-3} \\ {4+1} & {6+0} & {2-2} \end{array}\right]$
= $\left[\begin{array}{ccc} {-1} & {5} & {3} \\ {5} & {6} & {0} \end{array}\right]$
View full question & answer→Question 291 Mark
Given A = $\left[\begin{array}{rrr} {\sqrt{3}} & {1} & {-1} \\ {2} & {3} & {0} \end{array}\right]$ and B = $\left[\begin{array}{ccc} {2} & {\sqrt{5}} & {1} \\ {-2} & {3} & {\frac{1}{2}} \end{array}\right]$ find A + B
AnswerSince A, B are of the same order 2 $\times$ 3. Therefore, the addition of A and B is defined and is given by
A + B = $\left[\begin{array}{ccc} {2+\sqrt{3}} & {1+\sqrt{5}} & {1-1} \\ {2-2} & {3+3} & {0+\frac{1}{2}} \end{array}\right]=\left[\begin{array}{ccc} {2+\sqrt{3}} & {1+\sqrt{5}} & {0} \\ {0} & {6} & {\frac{1}{2}} \end{array}\right]$
View full question & answer→Question 301 Mark
If a matrix has 8 elements, what are the possible orders it can have?
Answer$1 \times 8,8 \times 1,4 \times 2,2 \times 4$
View full question & answer→Question 311 Mark
Find the values of x and y from the following equation: 2$\left[\begin{array}{cc}{x} & {5} \\ {7} & {y-3}\end{array}\right]$ + $\left[\begin{array}{ll}{3} & {-4} \\ {1} & {2}\end{array}\right]$ = $\left[\begin{array}{cc}{7} & {6} \\ {15} & {14}\end{array}\right]$
AnswerWe have,
2$\left[\begin{array}{cc}{x} & {5} \\ {7} & {y-3}\end{array}\right]$ + $\left[\begin{array}{ll}{3} & {-4} \\ {1} & {2}\end{array}\right]$ $=\left[\begin{array}{cc}{7} & {6} \\ {15} & {14}\end{array}\right]$
$\Rightarrow$ $\left[\begin{array}{cc}{2 x} & {10} \\ {14} & {2 y-6}\end{array}\right]$ + $\left[\begin{array}{ll}{3} & {-4} \\ {1} & {2}\end{array}\right]$ = $\left[\begin{array}{cc}{7} & {6} \\ {15} & {14}\end{array}\right]$
$\Rightarrow$ $\left[\begin{array}{cc}{2 x+3} & {10 - 4} \\ {14 + 1} & {2 y-6 + 2}\end{array}\right]$ = $\left[\begin{array}{cc}{7} & {6} \\ {15} & {14}\end{array}\right]$
$\Rightarrow$ $\left[\begin{array}{cc}{2 x+3} & {6} \\ {15} & {2 y-4}\end{array}\right]$ = $\left[\begin{array}{cc}{7} & {6} \\ {15} & {14}\end{array}\right]$
or 2x + 3 = 7 and 2y - 4 = 14 (why?)
or 2x = 7 - 3 and 2y = 18
or x = $\frac{4}{2}$ and y = $\frac{18}{2}$
i.e. x = 2 and y = 9
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