Question 512 Marks
Examine the consistency of the system of equations:
x + 2y = 2
2x + 3y = 3
AnswerMatrix form of given equation is AX = B $\Rightarrow\ \begin{bmatrix}1&2\\2&3\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}2\\3\end{bmatrix}$
$\therefore \ \text{A}=\begin{bmatrix}1&2\\2&3\end{bmatrix}\text{and B}=\begin{bmatrix}2\\3\end{bmatrix}$
$\therefore\ \text{|A|}=\begin{vmatrix}1&2\\2&3\end{vmatrix}\text{and B}=3-4=-1\neq0$
Therefore, Unique solution and hence equations are consistent.
View full question & answer→Question 522 Marks
Find the inverse of the matrix $\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$.
Answer$\text{A}=\begin{bmatrix} 1 & -3 \\ 2 & 0 \end{bmatrix}$
Cofactors of A are:
C11 = 0, C21 = 3
C12 = -2 C22 = 1
$\text{Adj A}=\begin{bmatrix} 0 & 3 \\ -2 & 1 \end{bmatrix}$
View full question & answer→Question 532 Marks
If |A| = 2, where A is 2 × 2 matrix, find |adj A|.
AnswerFor any square matrix A of order n, |adj A| = |A|n-1
Given, |A| = 2
Here, order is 2
⇒ |adj A| = |2|2-1 = 2
View full question & answer→Question 542 Marks
Using the property of determinants and without expanding, prove that:
$\begin{vmatrix}0&a&-b\\-a&0&-c\\b&c&0\end{vmatrix}=0$
Answer$\text{Let}\ \triangle=\begin{vmatrix}0&a&-b\\-a&0&-c\\b&c&0\end{vmatrix}$
$=\triangle=(-1)^3\begin{vmatrix}0&-a&b\\a&0&c\\-b&-c&0\end{vmatrix}$ [Taking (-1) common from each row]
Interchanging rows and columns in the determinants on R.H.S.,
$\triangle=-\begin{vmatrix}0&a&-b\\-a&0&-c\\b&c&0\end{vmatrix}\ \Rightarrow\triangle=-\triangle\ \Rightarrow\triangle+\triangle=0$
$\Rightarrow 2\triangle=0\ \Rightarrow\triangle=0$ Proved.
View full question & answer→Question 552 Marks
Write the value of the determinant $\begin{vmatrix}2&3&4\\2\text{x}&3\text{x}&4\text{x}\\5&6&8\end{vmatrix}$
AnswerLet $\triangle=\begin{vmatrix}2&3&4\\2\text{x}&3\text{x}&4\text{x}\\5&6&8\end{vmatrix}$
$=\text{x}\begin{vmatrix}2&3&4\\2&3&4\\5&6&8\end{vmatrix}$ [Taking out x common from R2]
$=0$
View full question & answer→Question 562 Marks
write the value of the determinant $\begin{vmatrix}2&-3&5\\4&-6&10\\6&-9&15\end{vmatrix}$
Answer$\text{A}=\begin{vmatrix}2&-3&5\\4&-6&10\\6&-9&15\end{vmatrix}$
$=\begin{vmatrix}2&-3&5\\4-4&-6+6&10-10\\6&-9&15\end{vmatrix}$ [Applying R2 → R2 - 2R1]
$=\begin{vmatrix}2&-3&5\\0&0&0\\6&-9&15\end{vmatrix}$
$=0$
View full question & answer→Question 572 Marks
Find the value of x, if:
$\begin{vmatrix}2&4\\5&1\end{vmatrix}=\begin{vmatrix}2\text{x}&4\\6&\text{x}\end{vmatrix}$
AnswerGiven, $\begin{vmatrix}2&4\\5&1\end{vmatrix}=\begin{vmatrix}2\text{x}&4\\6&\text{x}\end{vmatrix}$
$\Rightarrow2-20=2\text{x}^2-24$
$\Rightarrow-18=2\text{x}^2-24$
$\Rightarrow2\text{x}=6$
$\Rightarrow\text{x}^2=3$
$\Rightarrow\text{x}=\pm\sqrt{3}$
View full question & answer→Question 582 Marks
Evaluate $\begin{vmatrix}1&\text{x}&\text{y}\\1&\text{x}+\text{y}&\text{y}\\1&\text{x}&\text{x+y}\end{vmatrix}$
Answer$\triangle=\begin{vmatrix}1&\text{x}&\text{y}\\1&\text{x}+\text{y}&\text{y}\\1&\text{x}&\text{x+y}\end{vmatrix}$
Applying R2 → R2 - R1 and R3 → R3 - R1, we have:
$\triangle=\begin{vmatrix}1&\text{x}&\text{y}\\0&\text{y}&0\\0&0&\text{x}\end{vmatrix}$
Expanding along C1, we have:
$\triangle$ = 1(xy - 0) = xy
View full question & answer→Question 592 Marks
If $\begin{vmatrix}\text{x}+1&\text{x}-1\\\text{x}-3&\text{x}+2\end{vmatrix}=\begin{vmatrix}4&-1\\1&3\end{vmatrix},$ then write the value of x.
Answer$\begin{vmatrix}\text{x}+1&\text{x}-1\\\text{x}-3&\text{x}+2\end{vmatrix}=\begin{vmatrix}4&-1\\1&3\end{vmatrix}$
⇒ (x + 1)(x + 2) - (x - 1)(x - 3) = 12 + 1
⇒ x2 + 3x + 2 - x2 + 4x - 3 = 13
⇒ 7x - 1 = 13
⇒ 7x = 14
⇒ x = 2
Hence, the value of x is 2
View full question & answer→Question 602 Marks
Write the value of the determinant $\begin{vmatrix}\text{a}&1&\text{b}+\text{c}\\\text{b}&1&\text{c}+\text{a}\\\text{c}&1&\text{a}+\text{b}\end{vmatrix}$
AnswerLet $\triangle=\begin{vmatrix}\text{a}&1&\text{b}+\text{c}\\\text{b}&1&\text{c}+\text{a}\\\text{c}&1&\text{a}+\text{b}\end{vmatrix}$
$=\begin{vmatrix}\text{a}+\text{b}+\text{c}&1&\text{b}+\text{c}\\\text{b}+\text{c}+\text{a}&1&\text{c}+\text{a}\\\text{c}+\text{a}+\text{b}&1&\text{a}+\text{b}\end{vmatrix}$ [Applying C1 → C1 + C3]
$=\text{a}+\text{b}+\text{c}\begin{vmatrix}1&1&\text{b}+\text{c}\\1&1&\text{c}+\text{a}\\1&1&\text{a}+\text{b}\end{vmatrix}$
$=(\text{a}+\text{b}+\text{c})\times0$
$=0$
View full question & answer→Question 612 Marks
If $\begin{bmatrix}1&0&0\\0&\text{y}&0\\0&0&1\end{bmatrix}\begin{bmatrix}\text{x}\\-1\\\text{z}\end{bmatrix}=\begin{bmatrix}1\\0\\1\end{bmatrix}$, find x, y and z.
AnswerHere,
$\begin{bmatrix}1&0&0\\0&\text{y}&0\\0&0&1\end{bmatrix}\begin{bmatrix}\text{x}\\-1\\\text{z}\end{bmatrix}=\begin{bmatrix}1\\0\\1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\-\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}1\\0\\1\end{bmatrix}$
$\therefore\ \text{x}=1,\text{y}=0\text{ and }\text{z}=1$
View full question & answer→Question 622 Marks
Find the adjoint of the following matrices: $\text{C}=\begin{bmatrix} \cos\alpha & \sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$ Verify that (adjoint A) A = |A|I = A (adjoint A) for the above matrices.
Answer$\text{adjoint C}=\begin{bmatrix} \cos\alpha & -\sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}$
$\text{(adjoint C)C}=\begin{bmatrix}\cos^2\alpha-\sin^2\alpha & 0 \\ 0 & \cos^2\alpha-\sin^2\alpha \end{bmatrix}$
$|\text{C}|=\cos^2\alpha-\sin^2\alpha$
$|\text{C}|\text{I}=\begin{bmatrix}\cos^2\alpha-\sin^2\alpha & 0 \\ 0 & \cos^2\alpha-\sin^2\alpha \end{bmatrix}$
$\text{C(adjoint C)}=\begin{bmatrix}\cos^2\alpha-\sin^2\alpha & 0 \\ 0 & \cos^2\alpha-\sin^2\alpha \end{bmatrix}$
$\therefore\ \text{(adjoint C)}=|\text{C}|\text{I}=\text{C(adjoint C)}$
Hence verified.
View full question & answer→Question 632 Marks
If $\begin{vmatrix}3\text{x}&7\\-2&4\end{vmatrix}=\begin{vmatrix}8&7\\6&4\end{vmatrix},$ find the value of x.
Answer$\begin{vmatrix}3\text{x}&7\\-2&4\end{vmatrix}=\begin{vmatrix}8&7\\6&4\end{vmatrix}$
⇒ 12x + 14 = 32 - 42
⇒ 12x + 14 = -10
⇒ 12x = -24
⇒ x = -2
$\therefore$ x = -2
View full question & answer→Question 642 Marks
Find the value of x, if:
$\begin{vmatrix}\text{x}+1&\text{x}-1\\\text{x}-3&\text{x}+2\end{vmatrix}=\begin{vmatrix}4&-1\\1&3\end{vmatrix}$
AnswerGiven,
$\begin{vmatrix}\text{x}+1&\text{x}-1\\\text{x}-3&\text{x}+2\end{vmatrix}=\begin{vmatrix}4&-1\\1&3\end{vmatrix}$
$\Rightarrow(\text{x}+1)(\text{x}+2)-(\text{x}-3)(\text{x}-1)=12+1$
$\Rightarrow\text{x}^2+3\text{x}+2-\text{x}^2+4\text{x}-3=13$
$\Rightarrow7\text{x}-1=13$
$\Rightarrow7\text{x}=14$
$\Rightarrow\text{x}=2$
View full question & answer→Question 652 Marks
Show that the following systems of linear equations is inconsistent:
2x - y = 5,
4x - 2y = 7
AnswerConsider,
2x - y = 5
4x - 2y = 7
$\text{D}=\begin{vmatrix}2&-1\\4&-2\end{vmatrix}=-4+4=0$
$\text{D}_1=\begin{vmatrix}5&-1\\7&-2\end{vmatrix}=-10+7=-3$
$\text{D}_2=\begin{vmatrix}2&5\\4&7\end{vmatrix}=14-20=-6$
Hence, D1 and D2 are non zero. Thus the given system is inconsistent.
View full question & answer→Question 662 Marks
A matrix of order 3 × 3 has determinant 2. What is the value of |A(3I)|, where I is the identity matrix of order 3 × 3.
AnswerLet A be the given matrix. Then,
|A| = 2 [Order = n = 3]
|I| = 1 [I is an identity matrix]
3(I) = 3
|A3(I)| = |3A| = 33|A| [A being of order 3]
= 27 × 2 = 54
|A3(I)| = 54
View full question & answer→Question 672 Marks
Find the value of $\lambda$ so that the points (1, - 5), (-4, 5) and $(\lambda,7)$ are collinear.
AnswerIf the points are collinear, then the area of the triangle must be zero.
Hence,
$\begin{vmatrix}1&-5&1\\-4&5&1\\\lambda&7&1\end{vmatrix}=0$
Expanding along R1
$1(-2)+5(-4-\lambda)+1(-28-5\lambda)=0$
$-2-20-5\lambda-28-5\lambda=0$
$-50-10\lambda=0$
$\lambda=5$
Hence, $\lambda=5$
View full question & answer→Question 682 Marks
Evaluate the following determinant:
$\begin{vmatrix}102&18&36\\1&3&4\\17&3&6\end{vmatrix}$
AnswerLet $\triangle=\begin{vmatrix}102&18&36\\1&3&4\\17&3&6\end{vmatrix}$
Applying R3 → 17R2 - R3, we get
$\triangle=\begin{vmatrix}102&18&36\\1&3&4\\0&48&62\end{vmatrix}$
Applying R2 → 102R2 - R1, we get
$\triangle=\begin{vmatrix}102&18&36\\0&288&327\\0&48&62\end{vmatrix}$
Thus, $\triangle=102(288\times62-372\times48)$
$\triangle=0$
View full question & answer→Question 692 Marks
Evaluate $\begin{vmatrix}\cos\alpha\cos\beta&\cos\alpha\sin\beta&-\sin\alpha\\-\sin\beta&\cos\beta&0\\\sin\alpha\cos\beta&\sin\alpha\sin\beta&\cos\alpha\end{vmatrix}.$
Answer$\triangle=\begin{vmatrix}\cos\alpha\cos\beta&\cos\alpha\sin\beta&-\sin\alpha\\-\sin\beta&\cos\beta&0\\\sin\alpha\cos\beta&\sin\alpha\sin\beta&\cos\alpha\end{vmatrix}$
Expanding along C3, we have:
$\triangle= -\sin\alpha(-\sin\alpha\sin^2\beta-\cos^2\beta\sin\alpha)+\cos\alpha(\cos\alpha\cos^2\beta+\cos\alpha\sin^2\beta)$
$=\sin^2\alpha(\sin^2\beta+\cos^2\beta)+\cos^2\alpha(\cos^2\beta+\sin^2\beta)$
$=\sin^2\alpha(1)+\cos^2\alpha(1)$
= 1
View full question & answer→Question 702 Marks
For what value of x, the following matrix is singular?
$\begin{vmatrix}5-\text{x}&\text{x}+1\\2&4\end{vmatrix}$
AnswerIf a matrix A is singular, then |A| = 0
$\therefore\begin{vmatrix}5-\text{x}&\text{x}+1\\2&4\end{vmatrix}=0$
⇒ 4(5 - x) - 2(x + 1) = 0
⇒ 20 - 4x - 2x - 2
⇒ 18 - 6x = 0
⇒ 18 = 6x
⇒ x = 3
View full question & answer→Question 712 Marks
A matrix A of order 3 × 3 is such that |A| = 4. Find the value of |2A|.
Answer|KA| = kn |A|
Here, n is the order of A.
Given, |A| = 4
⇒ |2A| = 23 × 4 = 32
View full question & answer→Question 722 Marks
If A is a square matrix satisfying AT A = l, write the value of |A|.
AnswerLet $|\text{A}|=|\text{A}|^{\text{T}} $ [By property of determinants]
Given,
$\text{A}^{\text{T}}\text{A}=\text{I}$
$\Rightarrow|\text{A}^{\text{T}}\text{A}|=1$
Then,
$|\text{A}^{\text{T}}\text{A}|=|\text{A}^{\text{T}}||\text{A}|$ [Since the determinants are of the same order]
$\Rightarrow|\text{A}^{\text{T}}||\text{A}|=1$
$\Rightarrow|\text{A}|=\frac{1}{|\text{A}^{\text{T}}|}$
$\Rightarrow|\text{A}|=\frac{1}{|\text{A}|}$ $\big[\therefore|\text{A}|=|\text{A}^{\text{T}}|\big]$
$\Rightarrow|\text{A}|^2=1$
$\Rightarrow|\text{A}|=\pm1$
View full question & answer→Question 732 Marks
Write the cofactor of a12 in the following matrix $\begin{bmatrix}2&-3&5\\6&0&4\\1&5&-7\end{bmatrix}$
AnswerGiven,
$\begin{bmatrix}2&-3&5\\6&0&4\\1&5&-7\end{bmatrix}$
Here, $\text{a}_{12}=-3$
Cofactor of $\text{a}_{12}=(-1)^{1+2}\begin{vmatrix}6&4\\1&-7\end{vmatrix}$
$\text{a}_{12}=-(-42-4)=46$
View full question & answer→Question 742 Marks
Find equation of line joining (1, 2) and (3, 6) using determinants.
AnswerLet P (x, y) be any points on the line joining the points (1, 2) and (3, 6).
Then, Area of triangle that could be formed by these points is zero.
$\therefore\ \ \text{Area of triangle}=\text{Modulus of}\ \frac{1}{2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}=0$
$\Rightarrow\ \text{Modulus of}\ \frac{1}{2}\begin{vmatrix}x&y&1\\1&2&1\\3&6&1\end{vmatrix}=0$
$\Rightarrow\ \frac{1}{2}\left[x(2-6)-y(1-3)+1(6-6)\right]=0$
$\Rightarrow\ \ -4x+2y=0\ \Rightarrow\ \ -2x+y=0$
$\Rightarrow\ \ y=2x$ Which is required line.
View full question & answer→Question 752 Marks
If $\text{A}=\begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix},$ write A-1 in terms of A.
Answer$|\text{A}|=\begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}=-19\neq0$
A is a non-singular matrix. Therefore, it is invertible.
Let Cij be a cofactor of aij in A.
The cofactors of element A are given by
C11 = -2
C12 = -5
C21 = -3
C22 = 2
$\text{adj A}=\begin{bmatrix} -2 & -5 \\ -3 & 2 \end{bmatrix}^\text{T}=\begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix},$
$\therefore\text{A}^{-1}=\frac{1}{|\text{A}|}\text{ adj A}=\begin{bmatrix} \frac{2}{19} & \frac{3}{19} \\ \frac{5}{19} & \frac{-2}{19} \end{bmatrix}$
$\Rightarrow\text{A}^{-1}=\frac{1}{19}\text{A}$
View full question & answer→Question 762 Marks
If A is a non-singular square matrix such that $\text{A}^{-1}=\begin{bmatrix} 5 & 3 \\ -2 & -1 \end{bmatrix},$ then find (AT)-1.
AnswerFor any invertible matrix A.
(AT)-1 = (A-1)T
We have $\text{A}^{-1}=\begin{bmatrix} 5 & 3 \\ -2 & -1 \end{bmatrix}$
$\Rightarrow(\text{A}^\text{T})^{-1}=\begin{bmatrix} 5 & -2 \\ 3 & -1 \end{bmatrix}$
View full question & answer→Question 772 Marks
If A is a square matrix such that $\text{A (adj A)}=\begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix},$ then write the value of |adj A|.
AnswerGiven,
$\text{A (adj A)}=\begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix}$
$\Rightarrow|\text{A}|\text{I}_\text{n}=5\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$\Rightarrow|\text{A}|=5$
Now, |adj A| = |A|n-1 = 53-1 = 25.
View full question & answer→Question 782 Marks
Evaluate the following integrals:
$\int\limits^{-\frac{\pi}{2}}_{-\frac{3\pi}{2}}\Big\{\sin^2(3\pi+\text{x})+(\pi+\text{x})^3\Big\}\text{dx}$
AnswerLet $\text{I}=\int\limits^{-\frac{\pi}{2}}_{-\frac{3\pi}{2}}\Big\{\sin^2(3\pi+\text{x})+(\pi+\text{x})^3\Big\}\text{dx}$
Put $\pi+\text{x}=\text{z}$
$\Rightarrow\text{dx}=\text{dz}$
When $\text{x}\rightarrow-\frac{3\pi}{2},\text{ z}\rightarrow-\frac{\pi}{2}$
When $\text{x}\rightarrow-\frac{\pi}{2},\text{ z}\rightarrow\frac{\pi}{2}$
$\therefore\ \text{I}=\int\limits^{-\frac{\pi}{2}}_{-\frac{\pi}{2}}\Big[\sin^2(2\pi+\text{z})+\text{z}^3\Big]\text{dx}$
$=\int\limits^{-\frac{\pi}{2}}_{-\frac{\pi}{2}}\big(\sin^2\text{z}+\text{z}^3\big)\text{dz}$ $\big[\sin(2\pi+\theta)=\sin\theta\big]$
$=\frac{1}{2}\int\limits^{-\frac{\pi}{2}}_{-\frac{\pi}{2}}\frac{1-\cos2\text{z}}{2}\text{ dz}+\int\limits^{-\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{z}^3\text{ dz}$
$=\frac{1}{2}\int\limits^{-\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{dz}-\frac{1}{2}\int\limits^{-\frac{\pi}{2}}_{-\frac{\pi}{2}}\cos2\text{z dz}+\int\limits^{-\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{z}^3\text{ dz}$
$=\frac{1}{2}\Big[\text{z}\Big]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}-\frac{1}{2}\Big[\frac{\sin2\text{z}}{2}\Big]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}+\Big[\frac{\text{z}^4}{4}\Big]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}$
$=\frac{1}{2}\bigg[\frac{\pi}{2}-\Big(-\frac{\pi}{2}\Big)\bigg]-\frac{1}{4}\big[\sin\pi-\sin(-\pi)\big]+\frac{1}{4}\Big(\frac{\pi^4}{16}-\frac{\pi^4}{16}\big)$
$=\frac{1}{2}\times\pi-\frac{1}{4}(0+0)+\frac{1}{4}\times0$
$=\frac{\pi}{2}$
View full question & answer→Question 792 Marks
Evaluate the following determinant:
$\begin{vmatrix}\text{x}&-7\\\text{x}&5\text{x}+1 \end{vmatrix}$
AnswerLet $\text{A}=\begin{vmatrix}\text{x}&-7\\\text{x}&5\text{x}+1 \end{vmatrix}$
$|\text{A}|=\text{x}(5\text{x}+1)+7\times\text{x}$
$=5\text{x}^2+\text{x}+7\text{x}$
$=5\text{x}^2+8\text{x}$
Hence, $|\text{A}|=5\text{x}^2+8\text{x}$
View full question & answer→Question 802 Marks
Write A-1 for $\text{A}=\begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}$
Answer$|\text{A}|=\begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}=1\neq0$
Let Cij be the cofactor of aij in A.
The cofactors of element A are given by
C11 = 3
C12 = -1
C21 = -5
C22 = 2
$\text{adj A}=\begin{bmatrix} 3 & -1 \\ -5 & 2 \end{bmatrix}^\text{T}=\begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix}$
$|\text{A}|=6-5=1$
$\therefore\text{A}^{-1}=\frac{1}{|\text{A}|}\text{ adj A}=\begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix}$
View full question & answer→Question 812 Marks
Evaluate the following determinant:
$\begin{vmatrix}1&-3&2\\4&-1&2\\3&5&2\end{vmatrix}$
Answer$\triangle=\begin{vmatrix}1&-3&2\\4&-1&2\\3&5&2\end{vmatrix}$
$=1\begin{vmatrix} -1&2\\5&2\end{vmatrix}-(-3)\begin{vmatrix}4&2\\3&2 \end{vmatrix}+2\begin{vmatrix}4&-1\\3&5 \end{vmatrix}$
$=1(-2-10)+3(8-6)+2(20+3)$
$=(-12)+6+46$
$=40$
View full question & answer→Question 822 Marks
Find value of k if area of triangle is 4 sq. units and vertices are:
(k, 0), (4, 0), (0, 2)
AnswerGiven: Area of triangle = Modulus of $\frac{1}{2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}=4$
$\Rightarrow\ \text{Modulus of}\ \frac{1}{2}\begin{vmatrix}k&0&1\\4&0&1\\0&2&1\end{vmatrix}=4$
$\Rightarrow\ \bigg|\frac{1}{2}\left[k(0-2)-0+1(8-0)\right]\bigg|=4$
$\Rightarrow\ \bigg|\frac{1}{2}(-2k+8)\bigg|=4$
$\Rightarrow\ \bigg|-k+4\bigg|=4\ \ \Rightarrow\ \ -k+4=\pm4$
Taking positive sign, -k + 4 = 4 $\ \Rightarrow\ \ \ k=0$
Taking negative sign, -k + 4 = -4 $\Rightarrow\ \ \ k=8$
View full question & answer→Question 832 Marks
Evaluate $\begin{vmatrix}4785&4787\\4789&4791\end{vmatrix}$
AnswerLet $\triangle=\begin{vmatrix}4785&4787\\4789&4791\end{vmatrix}$
$\Rightarrow\triangle=\begin{vmatrix}4785&2\\4789&2\end{vmatrix}$ [Applying C2 → C2 - C1]
$=2\times\begin{vmatrix}4785&1\\4789&1\end{vmatrix}$
$=2\times(4785-4789)$
$=2\times(-4)=-8$
$\Rightarrow\begin{vmatrix}4785&4787\\4789&4791\end{vmatrix}=-8$
View full question & answer→Question 842 Marks
Write the value of $\begin{vmatrix}\sin20^{\circ}&-\cos20^{\circ}\\\sin70^{\circ}&\cos70^{\circ}\end{vmatrix}$
AnswerLet $\triangle=\begin{vmatrix}\sin20^{\circ}&-\cos20^{\circ}\\\sin70^{\circ}&\cos70^{\circ}\end{vmatrix}$
$=\sin20^{\circ}\cos70^{\circ}+\cos20^{\circ}\sin70^{\circ}$
$=\sin(20^{\circ}+70^{\circ})$ [trignometric identity]
$=\sin90^{\circ}$
$=1$
View full question & answer→Question 852 Marks
Using determinants prove that the points (a, b), (a', b) and (a - a', b - b') are collinear if ab' = a'b.
Answer$\begin{vmatrix}\text{a}&\text{b}&1\\\text{a}'&\text{b}'&1\\\text{a}-\text{a}'&\text{b}-\text{b}'&1\end{vmatrix}=\begin{vmatrix}\text{a}&\text{b}&1\\\text{a}'-\text{a}&\text{b}'-\text{b}&0\\\text{a}-\text{a}'&\text{b}-\text{b}'&1\end{vmatrix}$ [Applying R2 → R2 - R1]
$=\begin{vmatrix}\text{a}&\text{b}&1\\\text{a}'-\text{a}&\text{b}'-\text{b}&0\\-\text{a}'&-\text{b}'&0\end{vmatrix}$ [Applying R3 → R3 - R1]
$=\begin{vmatrix}\text{a}'-\text{a}&\text{b}'-\text{b}\\-\text{a}'&-\text{b}'\end{vmatrix}$
$=-\text{b}'(\text{a}'-\text{a})+\text{a}'(\text{b}'-\text{b})$
$=-\text{b}'\text{a}'+\text{b}'\text{a}+\text{a}'\text{b}'-\text{a}'\text{b}$
$=\text{b}'\text{a}-\text{a}'\text{b}$
If the points are collinear then $\triangle=0$
$\text{a}\text{b}'-\text{a}'\text{b}=0$
Thus, $\text{a}\text{b}'=\text{a}'\text{b}$
View full question & answer→Question 862 Marks
Use elementary column operation C2 → C2 + 2C1 in the following matrix equation:
$\begin{pmatrix} 2 & 1 \\ 2 & 0 \end{pmatrix}=\begin{pmatrix} 3 & 1 \\ 2 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix}$
Answer$\begin{pmatrix} 2 & 1 \\ 2 & 0 \end{pmatrix}=\begin{pmatrix} 3 & 1 \\ 2 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix}$
Applying C2 → C2 + 2C1
$\begin{pmatrix} 2 & 5 \\ 2 & 4 \end{pmatrix}=\begin{pmatrix} 3 & 1 \\ 2 & 0 \end{pmatrix}\begin{pmatrix} 1 & 2 \\ -1 & -1 \end{pmatrix}$
View full question & answer→Question 872 Marks
In the following matrix equation use elementary operation R2 → R2 + R1 and the equation thus obtained:
$\begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 2 & -1 \end{bmatrix}\begin{bmatrix} 8 & -3 \\ 9 & -4 \end{bmatrix}$
Answer$\begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 2 & -1 \end{bmatrix}\begin{bmatrix} 8 & -3 \\ 9 & -4 \end{bmatrix}$
By applying elementary operation R2 → R2 + R1, we get
$\begin{bmatrix} 2 & 3 \\ 3 & 7 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 2 & -1 \end{bmatrix}\begin{bmatrix} 8 & -3 \\ 17 & -7 \end{bmatrix}$
(Every row operation is equlvalent to left-multiplication be an elementary matrix.)
View full question & answer→Question 882 Marks
If $\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}1\\-1\\0\end{bmatrix}$, find x, y and z.
AnswerHere,
$\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}1\\-1\\0\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}1\\-1\\0\end{bmatrix}$
$\therefore\ \text{x}=1,\ \text{y}=-1\text{ and }\text{z}=0$
View full question & answer→Question 892 Marks
If A and B are non-singular matrices of the same order, write whether AB is singular or non-singular.
AnswerLet A & B be non-singular matrices of order n.
A ≠ 0 and B ≠ 0 By definition.
Since they are of same order, AB = AB, AB = 0 if either A = 0 or B = 0 But it is not the case here. Thus, AB is non-zero and AB is non-singular matrix.
View full question & answer→Question 902 Marks
Write Minors and Cofactors of the elements of following determinant:
$\begin{vmatrix}a&c\\b&d\end{vmatrix}$
AnswerThe given determinant is $\begin{vmatrix}a&c\\b&d\end{vmatrix}.$
Minor of element aij is Mij.
$\therefore$ M11 = minor of element a11 = d
M12 = minor of element a12 = b
M21 = minor of element a21 = c
M22 = minor of element a22 = a
cofactor of aij is Aij = (-1)i+j Mij.
$\therefore$ A11 = (-1)1+1 M11 = (-1)2 (d) = d
A12 = (-1)1+2 M12 = (-1)3 (b) = -b
A21 = (-1)2+1 M21 = (-1)3 (c) = -c
A22 = (-1)2+2 M22 = (-1)4 (a) = a
View full question & answer→Question 912 Marks
Let A be a 3 × 3 square matrix, such that A (adj A) = 2I, where I is the identity matrix. Write the value of |adj A|.
Answer$\because$ A(adj (A)) = |A|I
2I = |A|I (Given A(adj A) = 2I)
|A| = 2
Also, |adj A| = |A|n-1
= (2)3-1
= (2)2
= 4
|adj A| = 4
View full question & answer→Question 922 Marks
If I3 denotes identity matrix of order 3 × 3, write the value of its determinant.
AnswerIn an identity matrix, all the diagonal elements are 1 and rest of the elements are 0.
Here,
$\text{I}_3=\begin{vmatrix}1&0&0\\0&1&0\\0&0&1\end{vmatrix}$
$\text{I}_3=1\times\begin{vmatrix}1&0\\0&1\end{vmatrix}$ [Expanding along C1]
$\text{I}_3=1$
$\text{I}_3=1$
View full question & answer→Question 932 Marks
If $\text{A}=\begin{bmatrix} 1 & -3 \\ 2 & 0 \end{bmatrix},$ write adj A.
Answer$|\text{A}|=\begin{bmatrix} 1 & -3 \\ 2 & 0 \end{bmatrix}=6\neq0$
A is a non-singular matrix. Therefore, it is invertible.
Let Cij be a cofactor of aij in A.
The cofactors of element A are given by
C11 = 0
C12 = -2
C21 = 3
C22 = 1
$\therefore\ \text{adj A}=\text{A}=\begin{bmatrix} 0 & -2 \\ 3 & 1 \end{bmatrix}^\text{T}=\begin{bmatrix} 0 & 3 \\ -2 & 1 \end{bmatrix}$
View full question & answer→Question 942 Marks
If Cij is the cofactor of the element aij of the matrix $\text{A}=\begin{bmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{bmatrix},$ then write the value of a32C32.
AnswerIn the given matrix $\text{A}=\begin{bmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{bmatrix}$
C32 = (-1)3+2 (8 - 30) = 22
Therefore, a32C32 = 5 × 22 = 110.
Hence, the value of a32C32 is 110.
View full question & answer→Question 952 Marks
A, B, C are three non-null square matrices of the same order, write the condition on A such that AB = AC ⇒ B = C.
AnswerConsider AB = AC. On multiplying both sides by A-1, we get AA-1
B = AA-1 ⇒ IB = IC [Because AA-1 = I where I is the identity matrix] ⇒ B = C Therefore, the required condition is A must be invertible or $|\text{A}|\neq0$. View full question & answer→Question 962 Marks
Show that the following systems of linear equations is inconsistent:
3x + y = 5,
-6x - 2y = 9
Answer$\text{D}=\begin{vmatrix}3&1\\-6&-2\end{vmatrix}=-6+6=0$
$\text{D}_1=\begin{vmatrix}5&1\\9&-2\end{vmatrix}=-10-9=-19\neq0$
Since D = 0 but $\text{D}_1\neq0$
Hence the given system of equations is inconsistent.
View full question & answer→Question 972 Marks
If $\text{A}=\begin{bmatrix}5&3&8\\2&0&1\\1&2&3\end{bmatrix}.$ Write the cofactor of element a32.
AnswerMinor of $\text{a}_{32}=\text{M}_{32}=\begin{vmatrix}5&8\\2&1\end{vmatrix}=5-16=-11$
Cofactor of $\text{a}_{\text{n}}=\text{A}_{32}=(-1)^{3+2}\text{M}_{32}=11$
Hence, the cofactor of the elements an is 11.
View full question & answer→Question 982 Marks
If the points (a, 0), (0, b) and (1, 1) are collinear, prove that a + b = ab.
AnswerIf the points (a, 0), (0, b) and (1, 1) are collinear, then
$\begin{vmatrix}\text{a}&0&1\\0&\text{b}&1\\1&1&1\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{a}&0&1\\-\text{a}&\text{b}&0\\1&1&1\end{vmatrix}=0$ [Applying R2 → R2 - R1]
$\Rightarrow\begin{vmatrix}\text{a}&0&1\\-\text{a}&\text{b}&0\\1-\text{a}&1&0\end{vmatrix}=0$ [Applying R3 → R3 - R1]
$\Rightarrow\triangle=\begin{vmatrix}-\text{a}&\text{b}\\1-\text{a}&1\end{vmatrix}=0$
$\Rightarrow-\text{a}-\text{b}(1-\text{a})=0$
$\Rightarrow\text{a}+\text{b}=\text{ab}$
View full question & answer→Question 992 Marks
Find the adjoint of the following matrices: $\begin{bmatrix}1 & \frac{\tan\alpha}{2} \\ -\frac{\tan\alpha}{2} & 1 \end{bmatrix}$ Verify that (adjoint A) A = |A|I = A (adjoint A) for the above matrices.
Answer$\text{D}=\begin{bmatrix}1 & \frac{\tan\alpha}{2} \\ -\frac{\tan\alpha}{2} & 1 \end{bmatrix}$
$\text{adjoint D}=\begin{bmatrix}1 & -\frac{\tan\alpha}{2} \\ \frac{\tan\alpha}{2} & 1 \end{bmatrix}$
$(\text{adjoint D)D}=\begin{bmatrix} 1+\tan^2\frac{\alpha}{2} & 0 \\ 0 &1+\tan^2\frac{\alpha}{2} \end{bmatrix}$
$|\text{D}|=1+\tan^2\frac{\alpha}{2}$
$|\text{D}|\text{I}=\begin{bmatrix} 1+\tan^2\frac{\alpha}{2} & 0 \\ 0 &1+\tan^2\frac{\alpha}{2} \end{bmatrix}$
View full question & answer→Question 1002 Marks
If A is a square matrix such that |A| = 2, write the value of $\big|\text{A}\text{A}^{\text{T}}\big|.$
AnswerIn a square matrix, A = AT. Since they are of same order, AAT = AAT.
Given, A = 2
⇒ AAT= 22 = 4
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