Question 12 Marks
If $\begin{vmatrix}2\text{x}+5&3\\5\text{x}+2&9\end{vmatrix}=0,$ find x.
Answer$\begin{vmatrix}2\text{x}+5&3\\5\text{x}+2&9\end{vmatrix}=0$
$\Rightarrow9(2\text{x}+5)-3(5\text{x}+2)=0$
$\Rightarrow18\text{x}+45-15\text{x}-6=0$
$\Rightarrow3\text{x}+39=0$
$\Rightarrow3\text{x}=-39$
$\Rightarrow\text{x}=\frac{-39}{3}$
$\Rightarrow\text{x}=-13$
View full question & answer→Question 22 Marks
Find the value of the determinant $\begin{vmatrix}4200&1201\\4205&4203\end{vmatrix}$
AnswerLet $\triangle=\begin{vmatrix}4200&1201\\4205&4203\end{vmatrix}$
$\triangle=\begin{vmatrix}4200&1\\4205&1\end{vmatrix}$ [Applying $C_2 → C_2 - C_1$]
$\triangle=4200 - 4202$
$\triangle=-2$
View full question & answer→Question 32 Marks
If $A$ is a square matrix of order $3$ with determinant $4$, then write the value of $|-A|.$
Answer$|A| = 4$
Here,
Order of the matrix $(n) = 3$
Using properties of matrices, we get
$|kA| = k^n|A| [$For a square matrix of order n and constant $k]$
$\Rightarrow |-A| = (-1)^3 |A| = (-1) \times 4 = -4$
View full question & answer→Question 42 Marks
Examine the consistency of the system of equations:
2x - y = 5
x + y = 4
AnswerMatrix form of given equations is AX = B $\Rightarrow \ \begin{bmatrix}2&-1\\1&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}5\\4\end{bmatrix}$
$\therefore\ \text{A}=\begin{bmatrix}2&-1\\1&1\end{bmatrix}\text{and B}=\begin{bmatrix}5\\4\end{bmatrix}$
$\therefore\ \text{|A|}=\begin{vmatrix}2&-1\\1&1\end{vmatrix}=2-(-1)=3\neq0$
Therefore, Unique solution and hence equations are consistent.
View full question & answer→Question 52 Marks
If $\text{A}=\begin{bmatrix}0&\text{i}\\\text{i}&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}0&1\\1&0\end{bmatrix},$ find the value of $|\text{A}|+|\text{B}|.$
Answer$\text{A}=\begin{bmatrix}0&\text{i}\\\text{i}&1\end{bmatrix}$$\Rightarrow|\text{A}|= 0-\text{i}^2$
$\Rightarrow|\text{A}|=-(-1)=1$
Also,
$\text{B}=\begin{bmatrix}0&1\\1&0\end{bmatrix}$
$\Rightarrow|\text{B}|=0-1=-1$
So,
$\Rightarrow|\text{A}|+|\text{B}|=1-1=0$
View full question & answer→Question 62 Marks
If $A$ is a square matrix of order $3$ such that $|adj\ A| = 64$, find $|A|.$
AnswerFor any square matrix of order $n$,
$|adj\ A| = |A|^{n-1}$
$\Rightarrow 64 = |A|^2 [\because |adj\ A| = 64]$
$\Rightarrow|\text{A}|=\pm8$
View full question & answer→Question 72 Marks
If $\text{x}\in\text{N}$ and $\begin{vmatrix}\text{x}+3&-2\\-3\text{x}&2\text{x} \end{vmatrix}=8,$ then find the value of x.
Answer$\begin{vmatrix}\text{x}+3&-2\\-3\text{x}&2\text{x} \end{vmatrix}=8$
$\Rightarrow(\text{x}+3)2\text{x}-(-2)(-3\text{x})=8$
$\Rightarrow2\text{x}^2+6\text{x}-6\text{x}=8$
$\Rightarrow2\text{x}^2=8$
$\Rightarrow\text{x}^2-4=0$
$\Rightarrow\text{x}^2=4$
$\Rightarrow\text{x}=2$ $[\text{x}\neq-2\ \because\text{x}\in\text{N}]$
View full question & answer→Question 82 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}8&2&7\\12&3&5\\16&4&3 \end{vmatrix}$
Answer$\triangle=\begin{vmatrix}8&2&7\\12&3&5\\16&4&3 \end{vmatrix}$
$\Rightarrow\triangle=\begin{vmatrix}0&2&7\\12&3&5\\16&4&3 \end{vmatrix}$ [Applying $C_1 → C_1 - 4C_2$]
$\Rightarrow\triangle=0$
View full question & answer→Question 92 Marks
Let $A = [a_{ij}]$ be a square matrix of order $3 \times 3$ and $C_{ij}$ denote cofactor of $a_{ij}$ in $A.$ if $|A| = 5,$ write the value of $a_{13}C_{13} + a_{23}C_{23} + a_{33}C_{33.}$
AnswerIf $A = a_{ij}$ is a square matrix of order $n$ and $C_{ij}$ is a cofactor of $a_{ij},$ then
$\sum\limits_{\text{i}=1}^{\text{n}}\text{a}_{\text{ij}}\text{C}_\text{ij}=|\text{A}|$ and $\sum\limits_{\text{i}=1}^{\text{n}}\text{a}_{\text{ij}}\text{C}_\text{ij}=|\text{A}|$
Given, $|A| = 5$ and matrix $A$ is of order $3 × 3$
Since $a_{13}C_{13} + a_{23}C_{23} + a_{33}C_{33}$ represent expansion of $A$ along third column, we get
$\Rightarrow a_{13}C_{13} + a_{23}C_{23} + a_{33}C_{33} = |A| = 5$
$\Rightarrow a_{13}C_{13} + a_{23}C_{23} + a_{33}C_{33} = 5$
View full question & answer→Question 102 Marks
Show that the following systems of linear equations has infinite number of solutions and solve:
x + 2y = 5,
3x + 6y = 15
AnswerUsing the equations, we get
$\text{D}=\begin{vmatrix}1&2\\3&6\end{vmatrix}=6-6=0$
$\text{D}_1=\begin{vmatrix}5&2\\15&6\end{vmatrix}=30-30=0$
$\text{D}_2=\begin{vmatrix}1&5\\3&15\end{vmatrix}=15-15=0$
$\therefore\text{D}=\text{D}_1=\text{D}_2$
Hence, the system of linear equation has infinitely many solutions.
View full question & answer→Question 112 Marks
If $\text{A}=\begin{bmatrix}1&2\\4&2\end{bmatrix}$, then show that $\left|2\text{A}\right|=4\left|\text{A}\right|$
AnswerThe given matrix is $\text{A}=\begin{bmatrix}1&2\\4&2\end{bmatrix}$
$\therefore2\text{A}=2\begin{bmatrix}1&2\\4&2\end{bmatrix}=\begin{bmatrix}2&4\\8&4\end{bmatrix}$
$\therefore\text{L.H.S.}=|2\text{A}|=\begin{vmatrix}2&4\\8&4\end{vmatrix}=2\times4-4\times8=8-32=-24$
$\text{Now},|\text{A}|=\begin{vmatrix}1&2\\4&2\end{vmatrix}=2-8=-6$
$\therefore\text{R.H.S.}=4|\text{A}|=4\times\left(-6\right)=-24$
$\therefore\text{L.H.S.}=\text{R.H.S.}$
View full question & answer→Question 122 Marks
Write the value of the determinant $\begin{vmatrix}2&3&4\\5&6&8\\6\text{x}&9\text{x}&12\text{x}\end{vmatrix}$
Answer$\begin{vmatrix}2&3&4\\5&6&8\\6\text{x}&9\text{x}&12\text{x}\end{vmatrix}$
$=\begin{vmatrix}2&3&4\\5&6&8\\2&3&4\end{vmatrix}$ [Taking 2x common from $R_3$]
$=0$
View full question & answer→Question 132 Marks
Write the adjoint of the matrix $\text{A}=\begin{bmatrix} -3 & 4 \\ 7 & -2 \end{bmatrix}$.
AnswerLet $C_{ij}$ be a cofactor of $a_{ij}$ in A.
Now,
$C_{11} = -2$
$C_{12} = -7$
$C_{21} = -4$
$C_{22} = -3$
$\therefore\ \text{adj A}=\begin{bmatrix} -2 & -7 \\ -4 & -3 \end{bmatrix}^\text{T}=\begin{bmatrix} -2 & -4 \\ -7 & -3 \end{bmatrix}$
View full question & answer→Question 142 Marks
Evaluate the following determinant:
$\begin{vmatrix}1&3&5\\2&6&10\\31&11&38\end{vmatrix}$
Answer$\triangle=\begin{vmatrix}1&3&5\\2&6&10\\31&11&38\end{vmatrix}$
$=1\begin{vmatrix}6&10\\11&38\end{vmatrix}-3\begin{vmatrix}2&10\\31&38\end{vmatrix}+5\begin{vmatrix}2&6\\31&11\end{vmatrix}$
$=(228-110)-3(76-310)+5(22-186)$
$=1(118)-3(-234)+5(-164)$
$=118+702-820$
$=0$
View full question & answer→Question 152 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, $A = A^T$. Since they are of same order, $AA^T = AA^T$.
Given, A = 2
$\Rightarrow AA^T= 2^2 = 4$
View full question & answer→Question 162 Marks
A is a skew-symmetric of order 3, write the value of |A|.
AnswerWe know that if a skew symmetric matrix A is of odd order, then |A| = 0
Since the order of the given matrix is 3, |A| = 0.
View full question & answer→Question 172 Marks
Using the property of determinants and without expanding, prove that:
$\begin{vmatrix}1&bc&a(b+c)\\1&ca&b(c+a)\\1&ab&c(a+b)\end{vmatrix}=0$
Answer$\text{Given}:\ \begin{vmatrix}1&bc&a(b+c)\\1&ca&b(c+a)\\1&ab&c(a+b)\end{vmatrix}=\begin{vmatrix}1&bc&ab+ac\\1&ca&bc+ba\\1&ab&ca+cb\end{vmatrix}$
$\text{Operating}\ \text{C}_3\rightarrow\text{C}_3+\text{C}_2\ \begin{vmatrix}1&bc&ab+bc+ac\\1&ca&ab+bc+ca\\1&ab&ab+bc+ca\end{vmatrix}$
$=(ab+bc+ca)\begin{vmatrix}1&bc&1\\1&ca&1\\1&ab&1\end{vmatrix}$
$=(ab+bc+ca)(0)=0$ $\left[\because\text{two columns are identical Proved.}\right]$
View full question & answer→Question 182 Marks
If A and B are square matrices of the same order such that |A| = 3 and AB = I, then write the value of |B|.
AnswerSince A & B are square matrix of the same order, by the property of determinants we get
|AB| = |A| × |B|
|A| = 3, AB = I
⇒ |AB| = 1
⇒ |A| × |B| = 1
⇒ 3 × |B| = 1
$\Rightarrow|\text{B}|=\frac{1}{3}$
View full question & answer→Question 192 Marks
Write the value of $a_{11}C_{21} + a_{12}C_{22} + a_{13}C_{23}$.
AnswerWe know that in a square matrix of order n, the sum of the products of elements of a row (or a column) with the cofactors of the corresponding elements of some other row (or column) is zero. Therefore,
$A = [a_{ij}]$ is a square matrix of order n.
$\Rightarrow\sum\limits_{\text{n}}^{\text{i}=1}\text{a}_{\text{ij}}\text{C}_\text{kj}=0$ and $\sum\limits_{\text{i}=1}^{\text{n}}\text{a}_{\text{ij}}\text{C}_\text{ik}=0$
$\Rightarrow a_{11}C_{21} + a_{12}C_{22} + a_{13}C_{23} = 0$
[Since the elements are of first row and the cofactors are of elements of second row]
$\Rightarrow a_{11}C_{21} + a_{12}C_{22} + a_{13}C_{23} = 0$
View full question & answer→Question 202 Marks
If A is a non-singular symmetric matrix, write whether $A^{-1}$ is symmetric or skew-symmetric.
AnswerLet A be an invertible symmetric matrix. Then,
$|\text{A}|\neq0\text{ and }\text{A}^\text{T}=\text{A}$
Now, $(A^{-1})^T = (A^T)^{-1}$
$\Rightarrow (A^{-1})^T = A^{-1} [\because A^T = A]$
Thus, $A^{-1}$ is symmetric matrix.
View full question & answer→Question 212 Marks
If $A = [A_{ij}]$ is a $3 \times 3$ scalar matrix such that $a_{11} = 2$, then write the value of $|A|$.
AnswerA scalar matrix is a digonal matrix, in which all the diagonal elements are equal to a given scalar number.
Given, $A = [a_{ij}]$ is $3 \times 3$ matrix, where $a_{11} = 2$
$\Rightarrow\text{A}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}$
$\Rightarrow\text{A}=\begin{vmatrix}2&0&0\\0&2&0\\0&0&2\end{vmatrix}$
$\Rightarrow|\text{A}|=2\times\begin{vmatrix}2&0\\0&2\end{vmatrix}$ [Expanding along $C_1$]
$\Rightarrow|\text{A}|=2\times2\times2$
$\Rightarrow|\text{A}|=8$
View full question & answer→Question 222 Marks
Write rthe value of the determinant $\begin{vmatrix}\text{p}&\text{p}+1\\\text{p}-1&\text{p}\ \end{vmatrix}$
Answer$\begin{vmatrix}\text{p}&\text{p}+1\\\text{p}-1&\text{p}\ \end{vmatrix}=\text{p}^2-(\text{p}+1)(\text{p}-1)$
$=\text{p}^2-(\text{p}^2-1)$
$=\text{p}^2-\text{p}^2+1$
$=1$
View full question & answer→Question 232 Marks
Find the maximum value of $\begin{vmatrix}1&1&1\\1&1+\sin\theta&1\\1&1&1+\cos\theta \end{vmatrix}$
AnswerLet $\triangle=\begin{vmatrix}1&1&1\\1&1+\sin\theta&1\\1&1&1+\cos\theta \end{vmatrix}$
Applying $R_2 → R_2 - R_1$ and $R_3 → R_3 - R_1$ we get
$\triangle=\begin{vmatrix}1&1&1\\0&\sin\theta&0\\0&0&\cos\theta \end{vmatrix}$
$=\sin\theta\cos\theta$
$=\frac{\sin2\theta}{2}$
We know that $-1\leq\sin2\theta\leq1$
$\therefore$ Maximum value of $\triangle=\frac{1}{2}\times1=\frac{1}{2}$
View full question & answer→Question 242 Marks
Evaluate the following determinant:
$\begin{vmatrix}\text{a}&\text{h}&\text{g}\\\text{h}&\text{b}&\text{f}\\\text{g}&\text{f}&\text{c}\end{vmatrix}$
Answer$\triangle=\begin{vmatrix}\text{a}&\text{h}&\text{g}\\\text{h}&\text{b}&\text{f}\\\text{g}&\text{f}&\text{c}\end{vmatrix}$
$=\text{a}\begin{vmatrix}\text{b}&\text{f}\\\text{f}&\text{c} \end{vmatrix}-\text{h}\begin{vmatrix}\text{h}&\text{f}\\\text{g}&\text{c} \end{vmatrix}+\text{g}\begin{vmatrix}\text{h}&\text{b}\\\text{g}&\text{f} \end{vmatrix}$
$=\big(\text{bc}-\text{f}^2\big)-\text{h}\big(\text{hc}-\text{fg}\big)+\text{g}\big(\text{hf}-\text{gb}\big)$
$=\text{abc}-\text{af}^2-\text{h}^2\text{c}+\text{fgh}+\text{fgh}-\text{g}^2\text{b}$
$=\text{abc}+2\text{fgh}-\text{af}^2-\text{ch}^2-\text{bg}^2$
View full question & answer→Question 252 Marks
If A is a square matrix of order $3$ such that $|A| = 5$, write the value of $|adj\ A|$.
AnswerFor any square matrix of order n,
$|adj\ A| = |A|^{n-1}$
$\Rightarrow |adj\ A| = |A|^2= 5^2= 25$
View full question & answer→Question 262 Marks
If $\text{A}=\begin{bmatrix}1&2\\3&-1 \end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&-4\\3&-2\end{bmatrix},$ find |AB|.
Answer$\Rightarrow\text{A}=\begin{bmatrix}1&2\\3&-1 \end{bmatrix}$
⇒ |A| = -1 - 6 = -7
$\Rightarrow\text{B}=\begin{bmatrix}1&-4\\3&-2\end{bmatrix}$
⇒ |B| = -2 + 12 = 10
If A and B are square matrix of the same order, then |AB| = |A| |B|.
⇒ |AB| = |A| |B|
⇒ |AB| = -7 × 10 = -70
View full question & answer→Question 272 Marks
Prove that the determinant $\begin{vmatrix}x&\sin\theta&\cos\theta\\-\sin\theta&-x&1\\\cos\theta&1&x\end{vmatrix}$ is independent of θ.
Answer$\triangle=\begin{vmatrix}x&\sin\theta&\cos\theta\\-\sin\theta&-x&1\\\cos\theta&1&x\end{vmatrix}$
$= x(-x^2-1)-\sin\theta(-x\sin\theta-\cos\theta)+\cos\theta(-\sin\theta+x\cos\theta)$
$=-x^3-x+x\sin^2\theta+\sin\theta\cos\theta-\sin\theta\cos\theta+x\cos^2\theta$
$=-x^3-x+x(\sin^2\theta+\cos^2\theta)$
$=-x^3-x+x$
$=-x^3$ (Independent of $\theta$)
Hence, $\triangle$ is independent of $\theta.$
View full question & answer→Question 282 Marks
If $\text{A}=\begin{bmatrix} \text{a} & \text{b} \\ \text{c} & \text{d} \end{bmatrix},\text{B}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},$ find adj (AB).
Answer$\text{A}\times\text{B}=\begin{bmatrix} \text{a} & \text{b} \\ \text{c} & \text{d} \end{bmatrix}$
A × B is a non-singular matrix. Therefore, it is invertible.
Let $C_{ij}$ be a cofactor of $a_{ij}$ in A.
The cofactors of element A are given by
$C_{11} = d$
$C_{12} = -c$
$C_{21} = -b$
$C_{22} = a$
$\therefore\ \text{adj A}=\begin{bmatrix} \text{d} & -\text{c} \\ -\text{b} & \text{a} \end{bmatrix}^\text{T}=\begin{bmatrix} \text{d} & -\text{b} \\ -\text{c} & \text{a} \end{bmatrix}$
View full question & answer→Question 292 Marks
Write the value of $\begin{vmatrix}\text{a}+\text{ib}&\text{c}+\text{id}\\-\text{c}+\text{id}&\text{a}-\text{ib}\end{vmatrix}$
Answer$\begin{vmatrix}\text{a}+\text{ib}&\text{c}+\text{id}\\-\text{c}+\text{id}&\text{a}-\text{ib}\end{vmatrix}$
$=\text{a}^2-\text{iab}+\text{iab}-\text{i}^2\text{b}^2-(-\text{c}^2-\text{icd}+\text{icd}+\text{i}^2\text{d}^2)$
$=\text{a}^2-\text{i}^2\text{b}^2+\text{c}^2-\text{i}^2\text{d}^2$
Here, $\text{i}^2=-1$
$\begin{vmatrix}\text{a}+\text{ib}&\text{c}+\text{id}\\-\text{c}+\text{id}&\text{a}-\text{ib}\end{vmatrix}=\text{a}^2+\text{b}^2+\text{c}^2+\text{d}^2$
View full question & answer→Question 302 Marks
If w is an imaginary cube root of unity, find the value of $\begin{vmatrix}1&\text{w}&\text{w}^2\\\text{w}&\text{w}^2&1\\\text{w}^2&1&\text{w}\end{vmatrix}$
Answer$\begin{vmatrix}1&\text{w}&\text{w}^2\\\text{w}&\text{w}^2&1\\\text{w}^2&1&\text{w}\end{vmatrix}$
$=\begin{vmatrix}1+\text{w}+\text{w}^2&\text{w}&\text{w}^2\\\text{w}+\text{w}^2+1&\text{w}^2&1\\\text{w}^2+1+\text{w}&1&\text{w}\end{vmatrix}$ [Applying $C_1 → C_1 + C_2 + C_3$]
$=\begin{vmatrix}0&\text{w}&\text{w}^2\\0&\text{w}^2&1\\0&1&\text{w}\end{vmatrix}$ $[\because 1 + w + w^2 = 0, w$ is the imaginary cube root of unity$]$
View full question & answer→Question 312 Marks
Find area of the triangle with vertices at the point given: (2, 7), (1, 1), (10, 8)
AnswerArea of triangle = Modulus of $\frac{1}{2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}=\begin{vmatrix}\frac{1}{2}\begin{bmatrix}2&7&1\\1&1&1\\10&8&1\end{bmatrix}\end{vmatrix}$
$=\bigg|\frac{1}{2}\left[2(1-8)-7(1-10)+1(8-10)\right]\bigg|$
$=\bigg|\frac{1}{2}\left[2(-7)-7(-9)-2\right]\bigg|$
$=\bigg|\frac{1}{2}(-14+63-2)\bigg|=\bigg|\frac{1}{2}(63-16)\bigg|$
$=\begin{vmatrix}\frac{47}{2}\end{vmatrix}=\frac{47}{2}\text{sq.units}$
View full question & answer→Question 322 Marks
Find the value of x from the following: $\begin{vmatrix}\text{x}&4\\2&2\text{x}\end{vmatrix}=0$
Answer$\begin{vmatrix}\text{x}&4\\2&2\text{x}\end{vmatrix}=0$
$\Rightarrow2\text{x}^2-8=0$
$\Rightarrow2\text{x}^2=8$
$\Rightarrow\text{x}^2=\frac{8}{2}=4$
$\Rightarrow\text{x}=\sqrt{4}=\pm2$
View full question & answer→Question 332 Marks
If A is a square matrix, then write the matrix adj $(A^T) − (adj A)^T$.
AnswerIn a non-singular matrix, adj $A^T = (adj\ A)^T$.
$\Rightarrow (adj\ A^T) - (adj\ A)^T$ = Null matrix
View full question & answer→Question 342 Marks
Evaluate: $\begin{vmatrix}\cos15^\circ&\sin15^\circ\\\sin75^\circ&\cos75^\circ\end{vmatrix}$
Answer$\begin{vmatrix}\cos15^\circ&\sin15^\circ\\\sin75^\circ&\cos75^\circ\end{vmatrix}$
$=\cos15^\circ\cos75^\circ-\sin15^\circ\sin75^\circ$
$=\cos(15^\circ+75^\circ)$ $\big[\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}=\cos(\text{A}+\text{B})\big]$
$=\cos90^\circ$
$=0$
$\begin{vmatrix}\cos15^\circ&\sin15^\circ\\\sin75^\circ&\cos75^\circ\end{vmatrix}=0$
View full question & answer→Question 352 Marks
Find the value of x, if:
if $\begin{vmatrix}3\text{x}&7\\2&4\end{vmatrix}=10,$ find the value of x.
AnswerGiven, $\begin{vmatrix}3\text{x}&7\\2&4\end{vmatrix}=10$
$\Rightarrow12\text{x}-14=10$
$\Rightarrow12\text{x}=24$
$\Rightarrow\text{x}=2$
View full question & answer→Question 362 Marks
If A is a singular matrix, then write the value of |A|.
AnswerSince A is a singular matrix
Thus, |A| = 0
View full question & answer→Question 372 Marks
If the matrix $\begin{bmatrix}5\text{x}&2\\-10&1\end{bmatrix}$ is a singular, find the value of x.
AnswerA matrix is said to be singular if its determinant is zero. since the given matrix is singular, we get
$\text{A}=\begin{bmatrix}5\text{x}&2\\-10&1\end{bmatrix}$
$\Rightarrow|\text{A}|=\begin{bmatrix}5\text{x}&2\\-10&1\end{bmatrix}$
$\Rightarrow|\text{A}|=0$
$\Rightarrow5\text{x}+20=0$ [Expanding]
$\Rightarrow\text{x}=-\frac{20}{5}$
$\Rightarrow\text{x}=-4$
View full question & answer→Question 382 Marks
Examine the consistency of the system of equations:
x + 3y = 5
2x + 6y = 8
AnswerMatrix from of given equations is AX = B $\Rightarrow\ \begin{bmatrix}1&3\\2&6\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}5\\8\end{bmatrix}$
$\therefore\ \text{A}=\begin{bmatrix}1&3\\2&6\end{bmatrix}\text{and B}=\begin{bmatrix}5\\8\end{bmatrix}$
$\therefore\ \text{|A|}=\begin{vmatrix}1&3\\2&6\end{vmatrix}=6-6=0$
$\text{Now},\ \text{(adj. A)B}=\begin{bmatrix}6&-3\\-2&1\end{bmatrix}\begin{bmatrix}5\\8\end{bmatrix}=\begin{bmatrix}33-24\\-10+8\end{bmatrix}=\begin{bmatrix}6\\-2\end{bmatrix}\neq0$
Therefore, given equations are inconsistent, i.e., have no common solution.
View full question & answer→Question 392 Marks
Find area of the triangle with vertices at the point given:
(1, 0), (6, 0), (4, 3)
AnswerArea of triangle = Modulus of $\frac{1}{2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}=\begin{vmatrix}\frac{1}{2}\begin{bmatrix}1&0&1\\6&0&1\\4&3&1\end{bmatrix}\end{vmatrix}$
$=\bigg|\frac{1}{2}\left[1(0-3)-0(6-4)+1(18-0)\right]\bigg|$
$=\bigg|\frac{1}{2}(-3+18)\bigg|=\bigg|\frac{15}{2}\bigg|=\frac{15}{2}\text{sq.units}$
View full question & answer→Question 402 Marks
On expanding by first row, the value of the determinant of $3 \times 3$ square matrix $A = [a_{ij}] is a_{11} C_{11} + a_{12} C_{12} + a_{13} C_{13},$ where $[C_{ij}]$ is the cofactor of $a_{ij}$ in $A.$ Write the expression for its value on expanding by second column.
AnswerIf $A = a_{ij}$ is a square matrix of order $n,$ then the sum of the products of elements of a row $($or a column$)$ with their cofactors is always equal to det $(A).$ Therefore,
$\sum\limits_{\text{i}=1}^{\text{n}}\text{a}_{\text{ij}}\text{C}_\text{ij}=|\text{A}|$ and $\sum\limits_{\text{i}=1}^{\text{n}}\text{a}_{\text{ij}}\text{C}_\text{ij}=|\text{A}|$
Given, $|A| = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13} [$Expanding along $R_1]$
Now,
$|A| = a_{12}C_{12} + a_{22}C_{22} + a_{32}C_{32} [$Expanding along $R_2] [a_{12}, a_{22} $ and $a_{32} $ are elements of $C_3]$
View full question & answer→Question 412 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 $C_{ij}$ be a cofactor of $a_{ij}$ in A.
The cofactors of element A are given by
$C_{11}= -2$
$C_{12} = -5$
$C_{21} = -3$
$C_{22} = 2$
$\text{adj A}=\begin{bmatrix} -2 & -5 \\ -3 & 2 \end{bmatrix}^\text{T}=\begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix},$
$\therefore\text{A}^{-1}=\frac{1}{|\text{A}|}\text{ adj A}=\begin{bmatrix} \frac{2}{19} & \frac{3}{19} \\ \frac{5}{19} & \frac{-2}{19} \end{bmatrix}$
$\Rightarrow\text{A}^{-1}=\frac{1}{19}\text{A}$
View full question & answer→Question 422 Marks
State whether the matrix $\begin{vmatrix}2&3\\6&4\end{vmatrix}$ is singular or non-singular.
AnswerLet $\triangle=\begin{vmatrix}2&3\\6&4\end{vmatrix}$
= 2 × 4 - 6 × 3
= 18 - 18 = -10
A matrix is said to be singular if its determinant is equal to zero. Since $\triangle=-10\neq0,$ the given matrix is non-singular.
View full question & answer→Question 432 Marks
If $\text{A}=\begin{bmatrix}1&1&-2\\2&1&-3\\5&4&-9\end{bmatrix},$ find $\left|\text{A}\right|.$
Answer$\text{Let A}=\begin{bmatrix}1&1&-2\\2&1&-3\\5&4&-9\end{bmatrix}.$
By expanding along the first row, we have:
$\left|\text{A}\right|=1\begin{vmatrix}1&-3\\4&-9\end{vmatrix}-1\begin{vmatrix}2&-3\\5&-9\end{vmatrix}-2\begin{vmatrix}2&1\\5&4\end{vmatrix}$
$=1(-9+12)-1(-18+15)-2(8-5)$
$ =1(3)-1(-3)-2(3)$
$ =3+3-6$
$=6-6$
$=0$
View full question & answer→Question 442 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 452 Marks
Evaluate the following determinant:
$\begin{vmatrix}1&4&9\\4&9&16\\9&16&25 \end{vmatrix}$
AnswerLet $\triangle$ be the determinant.
$\triangle=\begin{vmatrix}1&4&9\\4&9&16\\9&16&25 \end{vmatrix}$
Applying $R_3 → R_3 - R_2$, we get
$\Rightarrow\triangle=\begin{vmatrix}1&4&9-4\\4&9&16-9\\9&16&25-16 \end{vmatrix}$
$\Rightarrow\triangle=\begin{vmatrix}1&4&5\\4&9&7\\9&16&9\end{vmatrix}$
$\Rightarrow\triangle=\begin{vmatrix}1&5&5\\4&13&7\\9&25&9 \end{vmatrix}$ [Applying $C_2 → C_1 + C_2$]
$\Rightarrow\triangle=\begin{vmatrix}1&0&0\\4&-7&-13\\9&-20&-36 \end{vmatrix}$ [Applying $C_2 → 5C_1 - C_2$ and $C_3 → 5C_1 - C_3$]
$\Rightarrow\triangle=1(7\times36-13\times20)$
$\Rightarrow\triangle=252-260=-8$
View full question & answer→Question 462 Marks
Evaluate the determinant:$ \begin{vmatrix}3&-4&5\\1&1&-2\\2&3&1\end{vmatrix}$
AnswerLet $\text{A}=\begin{vmatrix}3&-4&5\\1&1&-2\\2&3&1\end{vmatrix}.$
By expanding along the first row, we have:
$|\text{A}|=3\begin{vmatrix}1&-2\\3&1\end{vmatrix}+4\begin{vmatrix}1&-2\\2&1\end{vmatrix}+5\begin{vmatrix}1&1\\2&3\end{vmatrix}$
$=3(1+6)+4(1+4)+5(3-2)$
$=3(7)+4(5)+5(1)$
$=21+20+5=46$
View full question & answer→Question 472 Marks
A matrix A of order $3 \times 3$ has determinant $5.$ What is the value of $|3A|?$
AnswerIf $A = [a_{ij}]$ is a square matrix of order n and k is a constant, then
$|kA| = k^n|A|$
Here,
Number of rows $= n$
k is a common factor from each row of k
$|3A| = 3^3|A| = 27 \times 5 = 135 [$Given matrix is $3 \times 3$ such that $|A| = 5]$
Thus, $|3A| = 135$
View full question & answer→Question 482 Marks
Examine the consistency of the system of equations:
x + y + z = 1
2x + 3y + 2z = 2
ax + ay + 2az = 4
AnswerMatrix form of given equations is AX = B $\Rightarrow\ \begin{bmatrix}1&1&1\\2&3&2\\a&a&2a\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\2\\4\end{bmatrix}$
$\text{Here}\ \text{A}=\begin{bmatrix}1&1&1\\2&3&2\\a&a&2a\end{bmatrix}$ $\therefore\ \text{|A|}=\begin{vmatrix}1&1&1\\2&3&2\\a&a&2a\end{vmatrix}$
$\Rightarrow\ \text{|A|}=1(6a-2a)-1(4a-2a)+1(2a-3a)=4a-2a-a=a\neq0$
Therefore, Unique solution and hence equations are consistent.
View full question & answer→Question 492 Marks
If $A$ is a square matrix of order $3$ such that adj $(2A) = k\ adj\ (A),$ then write the value of $k.$
AnswerFor any matrix $A$ of order $n,$ adj $(\lambda\text{A})=\lambda^{\text{n}-1}=\lambda^{\text{n}-1} (adj\ A) $where $\lambda$ is a constant.
Thus, for matrix $A$ of order $3$, we have
$ \operatorname{adj}(2 A)=2^{3-1}(\operatorname{adj} A) $
$ \Rightarrow \operatorname{adj}(2 A)=2^2(\operatorname{adj} A) $
$ \Rightarrow \operatorname{adj}(2 A)=4(\operatorname{adj})(A) $
$ \Rightarrow k \operatorname{adj}(A)=4 \operatorname{adj}(A)[\because \operatorname{adj}(2 A)=k \operatorname{adj}(A)] $
$ \Rightarrow k=4$
View full question & answer→Question 502 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:
$C_{11} = 0, C_{21} = 3$
$C_{12} = -2 C_{22} = 1$
$\text{Adj A}=\begin{bmatrix} 0 & 3 \\ -2 & 1 \end{bmatrix}$
View full question & answer→