Question 512 Marks
If $|A|=2$, where $A$ is $2 \times 2$ matrix, find $|\operatorname{adj} A|$.
AnswerFor any square matrix $A$ of order $n,|\operatorname{adj} A|=|A|^{n-1}$
Given, $|A|=2$
Here, order is 2
$\Rightarrow|\operatorname{adj} A|=|2|^{2-1}=2$
View full question & answer→Question 522 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 532 Marks
If $A$ is a non-singular square matrix such that $|A| = 10$, find $|A^{-1}|$.
Answer$\big|\text{A}^{-1}\big|=\Big|\frac{1}{\text{A}}\Big|$
$=\Big|\frac{1}{\text{A}}\Big|$
$=\frac{1}{10}\ \big[\because|\text{A}|=10\text{ (Given})\big]$
Hence, $\big|\text{A}^{-1}\big|=\frac{1}{10}$
View full question & answer→Question 542 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 $R_2$]
$=0$
View full question & answer→Question 552 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 $R_2 → R_2 - 2R_1$]
$=\begin{vmatrix}2&-3&5\\0&0&0\\6&-9&15\end{vmatrix}$
$=0$
View full question & answer→Question 562 Marks
Evaluate the determinants.
- $\begin{vmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{vmatrix}$
- $\begin{vmatrix}x^2-x+1&x-1\\x+1&x+1\end{vmatrix}$
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 $R_2 → R_2 - R_1$ and $R_3 → R_3 - R_1$, we have:
$\triangle=\begin{vmatrix}1&\text{x}&\text{y}\\0&\text{y}&0\\0&0&\text{x}\end{vmatrix}$
Expanding along $C_1$, 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$\begin{vmatrix}\text{x}+1&\text{x}-1\\\text{x}-3&\text{x}+2\end{vmatrix}=\begin{vmatrix}4&-1\\1&3\end{vmatrix}$
$\Rightarrow (x + 1)(x + 2) - (x - 1)(x - 3) = 12 + 1$
$\Rightarrow x^2 + 3x + 2 - x^2 + 4x - 3 = 13$
$\Rightarrow 7x - 1 = 13$
$\Rightarrow 7x = 14$
$\Rightarrow x = 2$
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 $C_1 → C_1 + C_3$]
$=\text{a}+\text{b}+\text{c}\begin{vmatrix}1&1&\text{b}+\text{c}\\1&1&\text{c}+\text{a}\\1&1&\text{a}+\text{b}\end{vmatrix}$
$=(\text{a}+\text{b}+\text{c})\times0$
$=0$
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
Evaluate the following determinant:
$\begin{vmatrix}67&19&21\\39&13&14\\81&24&26 \end{vmatrix}$
AnswerConsider the determinant
$\triangle=\begin{vmatrix}67&19&21\\39&13&14\\81&24&26 \end{vmatrix}$
Applying $C_1 → C_1- 4C_3$, We get,
$\triangle=\begin{vmatrix}4&19&21\\-3&13&14\\-3&24&26 \end{vmatrix}$
$\Rightarrow\triangle=\begin{vmatrix}4&19&21\\-3&13&14\\-3&24&26 \end{vmatrix}$
$\Rightarrow\triangle=\begin{vmatrix}1&32&35\\-3&13&14\\0&11&12\end{vmatrix}$ [Applying $R_3 → R_3 - R_2$ and $R_1 → R_1 + R_2$]
$\Rightarrow\triangle=\begin{vmatrix}1&32&35\\0&109&119\\0&11&12\end{vmatrix}$ [Applying $R_2 → 3R_1 + R_2$]
$\Rightarrow\triangle=1(109\times12-119\times11)$
$\Rightarrow\triangle=-1$
View full question & answer→Question 662 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, $D_1$ and $D_2$ are non zero. Thus the given system is inconsistent.
View full question & answer→Question 672 Marks
A matrix of order $3 \times 3$ has determinant $2.$ What is the value of $|A(3I)|,$ where I is the identity matrix of order $3 \times 3.$
AnswerLet $A$ be the given matrix. Then,
$|A| = 2 [$Order $= n = 3]$
$|I| = 1 [I$ is an identity matrix$]$
$3(I) = 3$
$|A^3(I)| = |3A| = 3^3|A| [A$ being of order $3]$
$= 27 \times 2 = 54$
$|A^3(I)| = 54$
View full question & answer→Question 682 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 $R_1$
$1(-2)+5(-4-\lambda)+1(-28-5\lambda)=0$
$-2-20-5\lambda-28-5\lambda=0$
$-50-10\lambda=0$
$\lambda=5$
Hence, $\lambda=5$
View full question & answer→Question 692 Marks
If A is a square matrix such that A (adj A) = 5I, where I denotes the identity matrix of the same order. Then, find the value of |A|.
AnswerA (adj A) = 5I (Given)
|A| I = 5I [$\because$ A(adj A) = |A|I]
$\therefore |\text{A}|=5$
View full question & answer→Question 702 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 $R_3 → 17R_2 - R_3$, we get
$\triangle=\begin{vmatrix}102&18&36\\1&3&4\\0&48&62\end{vmatrix}$
Applying $R_2 → 102R_2 - R_1$, 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 712 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 $C_3$, 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 722 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 732 Marks
A matrix A of order $3 \times 3$ is such that $|A| = 4$. Find the value of $|2A|$.
Answer$|KA| = k^n |A|$
Here, $n$ is the order of $A$.
Given, $|A| = 4$
$\Rightarrow |2A| = 2^3 \times 4 = 32$
View full question & answer→Question 742 Marks
If $A$ is a square matrix satisfying $A^T A = l$, 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 752 Marks
Write the cofactor of $a_{12}$ 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 762 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 772 Marks
If A is a non-singular square matrix such that $\text{A}^{-1}=\begin{bmatrix} 5 & 3 \\ -2 & -1 \end{bmatrix},$ then find $(A^T)^{-1}$.
AnswerFor any invertible matrix A.
$(A^T)^{-1} = (A^{-1})^T$
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 782 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, $|adj A| = |A|^{n-1} = 5^{3-1} = 25$.
View full question & answer→Question 792 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 802 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 812 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 $C_{ij}$ be the cofactor of $a_{ij}$ in A.
The cofactors of element A are given by
$C_{11} = 3$
$C_{12} = -1$
$C_{21} = -5$
$C_{22} = 2$
$\text{adj A}=\begin{bmatrix} 3 & -1 \\ -5 & 2 \end{bmatrix}^\text{T}=\begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix}$
$|\text{A}|=6-5=1$
$\therefore\text{A}^{-1}=\frac{1}{|\text{A}|}\text{ adj A}=\begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix}$
View full question & answer→Question 822 Marks
If $A = [A_{ij}]$ is a $3 \times 3$ diaginal matrix such that $a_{11} = 1, a_{22} = 2, a_{33} = 3,$ then find $|A|.$
AnswerIf $A = [A_{ij}]$ is a diagonal matrix of order n, then $|A| = a_{11} \times a_{22} \times a_{33} \times ...... \times a_{mn}.$
Given, $a_{11} - 1, a_{22} - 2$ and $a_{33} - 3$
$\Rightarrow |A| = 1 \times 2 \times 3 = 6 [$Applying the above property$]$
View full question & answer→Question 832 Marks
Show that points:
A (a, b + c), B (b, c + a), C (c, a + b) are collinear.
AnswerArea of triangle ABC = Modulus of $\frac{1}{2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}=\begin{vmatrix}\frac{1}{2}\begin{bmatrix}a&b+c&1\\b&c+a&1\\c&a+b&1\end{bmatrix}\end{vmatrix}$$=\bigg|\frac{1}{2}\left[a(c+a-a-b)-(b+c)(b-c)+1\left\{b(a+b)-c(c+a)\right\}\right]\bigg|$
$=\bigg|\frac{1}{2}\left[a(c-b)-(b^2-c^2)+(ab+b^2-c^2-ac)\right]\bigg|$
$=\bigg|\frac{1}{2}(ac-ab-b^2+c^2+ab+b^2-c^2-ac)\bigg|$
$=\begin{vmatrix}\frac{1}{2}\times0\end{vmatrix}=0$
Therefore, points A, B and C are collinear.
View full question & answer→Question 842 Marks
If $A$ is symmetric matrix, write whether $A^T$ is symmetric or skew-symmetric.
AnswerFor any symmetric matrix, $A^T = A$.
Hence, $A^T$ is also symmetric.
View full question & answer→Question 852 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 862 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 872 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 $C_2 → C_2 - C_1$]
$=2\times\begin{vmatrix}4785&1\\4789&1\end{vmatrix}$
$=2\times(4785-4789)$
$=2\times(-4)=-8$
$\Rightarrow\begin{vmatrix}4785&4787\\4789&4791\end{vmatrix}=-8$
View full question & answer→Question 882 Marks
If $A$ is an invertible matrix such that $|A^{-1}| = 2$, find the value of $|A|$.
Answer$\big|\text{A}^{-1}\big|=2$
$\Big|\frac{1}{\text{A}}\Big|=2$
$\frac{1}{|\text{A}|}=2$
$\therefore|\text{A}|=\frac{1}{2}$
Hence, $|\text{A}|=\frac{1}{2}$
View full question & answer→Question 892 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 902 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 $R_2 → R_2 - R_1$]
$=\begin{vmatrix}\text{a}&\text{b}&1\\\text{a}'-\text{a}&\text{b}'-\text{b}&0\\-\text{a}'&-\text{b}'&0\end{vmatrix}$ [Applying $R_3 → R_3 - R_1$]
$=\begin{vmatrix}\text{a}'-\text{a}&\text{b}'-\text{b}\\-\text{a}'&-\text{b}'\end{vmatrix}$
$=-\text{b}'(\text{a}'-\text{a})+\text{a}'(\text{b}'-\text{b})$
$=-\text{b}'\text{a}'+\text{b}'\text{a}+\text{a}'\text{b}'-\text{a}'\text{b}$
$=\text{b}'\text{a}-\text{a}'\text{b}$
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 912 Marks
Use elementary column operation $C_2 → C_2 + 2C_1$ 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 $C_2 → C_2 + 2C_1$
$\begin{pmatrix} 2 & 5 \\ 2 & 4 \end{pmatrix}=\begin{pmatrix} 3 & 1 \\ 2 & 0 \end{pmatrix}\begin{pmatrix} 1 & 2 \\ -1 & -1 \end{pmatrix}$
View full question & answer→Question 922 Marks
Find adjoint of each of the matrix.
$\begin{bmatrix}1&2\\3&4\end{bmatrix}$
AnswerHere $\text{A}=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}=\begin{bmatrix}1&2\\3&4\end{bmatrix}\ \Rightarrow\ |\text{A}|=\begin{vmatrix}1&2\\3&4\end{vmatrix}$
$\therefore A_{11}$ = Confactor of $a_{11} = (-1)^2 (4) = 4$
$A_{12}$ = Confactor of $a_{12} = (-1)^3 (3) = -3$
$A_{21}$ = Confactor of $a_{21} = (-1)^3 (2) = -2$
$A_{22}$ = Confactor of $a_{22} = (-1)^4 (1) = 1$
$\therefore\ \text{adj.}\ \text{A}=\begin{bmatrix}\text{A}_{11}&\text{A}_{12}\\\text{A}_{21}&\text{A}_{22}\end{bmatrix}=\begin{bmatrix}4&-3\\-2&1\end{bmatrix}=\begin{bmatrix}4&-2\\-3&1\end{bmatrix}$
View full question & answer→Question 932 Marks
In the following matrix equation use elementary operation $R_2 → R_2 + R_1$ 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 $R_2 → R_2 + R_1$, 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 942 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 952 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 962 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 $a_{ij}$ is $M_{ij}$.
$\therefore M_{11}$ = minor of element $a_{11} = d$
$M_{12}$ = minor of element $a_{12} = b$
$M_{21}$ = minor of element $a_{21}= c$
$M_{22}$ = minor of element $a_{22} = a$
cofactor of $a_{ij}$ is $A_{ij} = (-1)^{i+j} M_{ij}$.
$\therefore A_{11}= (-1)^{1+1} M_{11} = (-1)^2(d) = d$
$A_{12}= (-1)^{1+2}M_{12} = (-1)^3 (b) = -b$
$A_{21} = (-1)^{2+1} M_{21} = (-1)^3(c) = -c$
$A_{22} = (-1)^{2+2} M_{22} = (-1)^4 (a) = a$
View full question & answer→Question 972 Marks
Let A be a $3 \times 3$ 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 982 Marks
Find the value of the determinant $\begin{vmatrix}243&156&300\\81&52&100\\-3&0&4\end{vmatrix}$
Answer[Applying $R_1 → R_1 - 3R_2$]
$=\begin{vmatrix}0&0&0\\81&52&100\\-3&0&4\end{vmatrix}$
$=0$
View full question & answer→Question 992 Marks
If $I_3$ denotes identity matrix of order $3 \times 3$, 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 $C_1$]
$\text{I}_3=1$
$\text{I}_3=1$
View full question & answer→Question 1002 Marks
If A is a square matrix of order n × n such that |A| = λ, then write the value of |-A|.
Answer$|\text{A}|=\lambda$ [Order of A is n]
$\Rightarrow|-\text{A}|=(-1)^{\text{n}}|\text{A}|=(-1)^{\text{n}}\lambda$
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