If $A$ is an invertible matrix of order $2,$ then det $(A–1)$ is equal to:
- Adet $(A)$
- ✓$\frac{1}{\text{det}\ (\text{A})}$
- C$1$
- D$0$
Answer: B.
View full solution →Question types
DETERMINANTS question types
697 questions across 9 question groups — pick any mix to generate a Maths paper with step-by-step answer keys.
697
Questions
9
Question groups
5
Question types
01
M.C.Q (1 Marks)
170 Q→02Assertion (A) & Reason (B) MCQ
17 Q→031 Marks Question
27 Q→042 Marks Questions
110 Q→053 Marks Question
73 Q→065 Marks Questions
269 Q→07Case study (4 Marks)
10 Q→08Fill In The Blanks[1 Marks ]
10 Q→09True False[1 Marks ]
11 Q→Sample Questions
DETERMINANTS questions
One sample from each question group in this chapter. Select any group above to see the full set with answer keys.
Choose the correct answer from given four options in each of the Exercise:
The value of $\begin{vmatrix}\text{a}-\text{b}&\text{b}+\text{c}&\text{a}\\\text{b}-\text{a}&\text{c}+\text{a}&\text{b}\\\text{c}-\text{a}&\text{a}+\text{b}&\text{c}\end{vmatrix}$ is:
The value of $\begin{vmatrix}\text{a}-\text{b}&\text{b}+\text{c}&\text{a}\\\text{b}-\text{a}&\text{c}+\text{a}&\text{b}\\\text{c}-\text{a}&\text{a}+\text{b}&\text{c}\end{vmatrix}$ is:
- A$a^3 + b^3 + c^3$
- B$3bc$
- C$a^3 + b^3 + c^3 - 3abc$
- ✓None of these.
Answer: D.
View full solution →Choose the correct answer from given four options in each of the Exercise: If $\text{A}=\begin{vmatrix}2&\lambda&-3\\0&2&5\\1&1&3\end{vmatrix},$ then $A^{-1}$ exists, if:
- A$\lambda=2$
- B$\lambda\neq2$
- C$\lambda\neq-2$
- ✓$\text{None of these}$
Answer: D.
View full solution →The value of $\begin{vmatrix}5^2&5^3&5^4\\5^3&5^4&5^5\\5^4&5^5&5^6\end{vmatrix}$ is:
- A$5^2$
- ✓$0$
- C$5^{13}$
- D$5^9$
Answer: B.
View full solution →The value of $\text{(adj } A)$ is equal to
- ✓$2A$
- B$4A$
- C$8A$
- D$16A$
Answer: A.
View full solution →Directions : In the following questions, a statement of assertion $(A)$ is followed by a statement of reason $(R).$ Mark the correct choice as:
Assertion : The value of $x$ for which $\begin{vmatrix}\text{x}&2\\18&\text{x}\end{vmatrix}=\begin{vmatrix}6&2\\18&6\end{vmatrix}$ is $\pm\ 6.$
Reason : The determinant of a matrix $A$ order $2\times 2,$ $\text{A}\begin{bmatrix}\text{a}&\text{b}\\\text{c}&\text{d}\end{bmatrix}$ is $= ab - dc.$
Assertion : The value of $x$ for which $\begin{vmatrix}\text{x}&2\\18&\text{x}\end{vmatrix}=\begin{vmatrix}6&2\\18&6\end{vmatrix}$ is $\pm\ 6.$
Reason : The determinant of a matrix $A$ order $2\times 2,$ $\text{A}\begin{bmatrix}\text{a}&\text{b}\\\text{c}&\text{d}\end{bmatrix}$ is $= ab - dc.$
- ABoth $A$ and $R$ are true and $R$ is the correct explanation of $A.$
- BBoth $A$ and $R$ are true but $R$ is not the correct explanation of $A.$
- ✓$A$ is true but $R$ is false.
- D$A$ is false but $R$ is true.
Answer: C.
View full solution →Directions : In the following questions, a statement of assertion $(A)$ is followed by a statement of reason $(R).$ Mark the correct choice as :
Assertion : For two matrices $A$ and $B$ of order $3, \mid\text{A}\mid=2\mid\text{B}\mid=-3$ then if $\mid2\text{AB}\mid$ is $-48.$
Reason : For a square matrix $A, \text{A}(\text{adj}\ \text{A})=(\text{adj}\ \text{A})\text{A}=\mid\text{A}\mid.$
Assertion : For two matrices $A$ and $B$ of order $3, \mid\text{A}\mid=2\mid\text{B}\mid=-3$ then if $\mid2\text{AB}\mid$ is $-48.$
Reason : For a square matrix $A, \text{A}(\text{adj}\ \text{A})=(\text{adj}\ \text{A})\text{A}=\mid\text{A}\mid.$
- ABoth $A$ and $R$ are true and $R$ is the correct explanation of $A.$
- ✓Both $A$ and $R$ are true but $R$ is not the correct explanation of $A.$
- C$A$ is true but $R$ is false.
- D$A$ is false but $R$ is true.
Answer: B.
View full solution →Directions: In the following questions, a statement of assertion $(A)$ is followed by a statement of reason $(R).$ Mark the correct choice as:
Assertion: The value of $x$ for which $\begin{vmatrix}3&\text{x}\\\text{x}&1\end{vmatrix}=\begin{vmatrix}3&2\\4&1\end{vmatrix}$ is $\pm2\sqrt{2}.$
Reason: The determinant of a matrix $A$ order $2 \times 2, \text{A}\begin{bmatrix}\text{a}&\text{b}\\\text{c}&\text{d}\end{bmatrix}$ is $= ad - bc.$
Assertion: The value of $x$ for which $\begin{vmatrix}3&\text{x}\\\text{x}&1\end{vmatrix}=\begin{vmatrix}3&2\\4&1\end{vmatrix}$ is $\pm2\sqrt{2}.$
Reason: The determinant of a matrix $A$ order $2 \times 2, \text{A}\begin{bmatrix}\text{a}&\text{b}\\\text{c}&\text{d}\end{bmatrix}$ is $= ad - bc.$
- ✓Both $A$ and $R$ are true and $R$ is the correct explanation of $A.$
- BBoth $A$ and $R$ are true but $R$ is not the correct explanation of $A.$
- C$A$ is true but $R$ is false.
- D$A$ is false but $R$ is true.
Answer: A.
View full solution →Directions : In the following questions, a statement of assertion $(A) $ is followed by a statement of reason $(R).$ Mark the correct choice as :
Assertion : The points $A(a, b + c), B(b, c +a )$ and $C(c, a + b)$ are collinear.
Reason : Three points $A (x_1, y_1) , B(x_2, y_2)$ and $C(x_3, y_3)$ are collinear if area of a triangle $\text{ABC}$ is zero.
Assertion : The points $A(a, b + c), B(b, c +a )$ and $C(c, a + b)$ are collinear.
Reason : Three points $A (x_1, y_1) , B(x_2, y_2)$ and $C(x_3, y_3)$ are collinear if area of a triangle $\text{ABC}$ is zero.
- ✓Both $A$ and $R$ are true and $R$ is the correct explanation of $A$.
- BBoth $A$ and $R$ are true but $R$ is not the correct explanation of $A.$
- C$A$ is true but $R$ is false.
- D$A$ is false but $R$ is true.
Answer: A.
View full solution →Directions : In the following questions, a statement of assertion $(A)$ is followed by a statement of reason $(R).$ Mark the correct choice as :
Assertion : For a matrix $\begin{bmatrix}2&-1\\-3&4\end{bmatrix}, A. \text{adj}$ $\text{A}=\begin{bmatrix}4&0\\0&4\end{bmatrix}.$
Reason : For a square matrix $A, \text{A}(\text{adj}\text{A})=(\text{adj}\text{A})\text{A}=\mid\text{A}\mid.$
Assertion : For a matrix $\begin{bmatrix}2&-1\\-3&4\end{bmatrix}, A. \text{adj}$ $\text{A}=\begin{bmatrix}4&0\\0&4\end{bmatrix}.$
Reason : For a square matrix $A, \text{A}(\text{adj}\text{A})=(\text{adj}\text{A})\text{A}=\mid\text{A}\mid.$
- ABoth $A$ and $R$ are true and $R$ is the correct explanation of $A.$
- BBoth $A$ and $R$ are true but $R$ is not the correct explanation of $A.$
- C$A$ is true but $R$ is false.
- ✓$A$ is false but $R$ is true.
Answer: D.
View full solution →If $\begin{vmatrix} 3 \text{x}& 7 \$0.3em] -2 & 4\$0.3em] \end{vmatrix} = \begin{vmatrix} 8 & 7 \$0.3em] 6 & 4\$0.3em] \end{vmatrix}, $find the value of x.
View full solution →If $\begin{bmatrix} 3 \text{x}& 7 \$0.3em] -2 & 4\$0.3em] \end{bmatrix} = \begin{bmatrix} 8 & 7 \$0.3em] 6 & 4\$0.3em] \end{bmatrix}, $find the value of x.
View full solution →If $\begin{bmatrix} \text{x - y }& \text{z} \$0.3em] 2\text{x - y }& \text{w} \$0.3em] \end{bmatrix} = \begin{bmatrix} -1& 4 \$0.3em] 0 & 5\$0.3em] \end{bmatrix},$find the value of x + y.
View full solution →$\text{If }x \in \text{N and} \begin{bmatrix} \text{x + 3} & -2 \\ \text{-3x} & \text{2x} \\ \end{bmatrix} = 8, $ then find the value of $x.$
View full solution →What positive value of x makes the following pair of determinants equal? .
$\begin{vmatrix}\text{2x}&3\\5&\text{x} \end{vmatrix}, \begin{vmatrix}\text{16}&3\\5&\text{2} \end{vmatrix}$
View full solution →$\begin{vmatrix}\text{2x}&3\\5&\text{x} \end{vmatrix}, \begin{vmatrix}\text{16}&3\\5&\text{2} \end{vmatrix}$
If $\begin{vmatrix}2\text{x}+5&3\\5\text{x}+2&9\end{vmatrix}=0,$ find x.
View full solution →Find the value of the determinant $\begin{vmatrix}4200&1201\\4205&4203\end{vmatrix}$
View full solution →If $A$ is a square matrix of order $3$ with determinant $4$, then write the value of $|-A|.$
View full solution →Examine the consistency of the system of equations:
2x - y = 5
x + y = 4
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 $|\text{A}|+|\text{B}|.$
View full solution →Using the properties of determinants, prove that
$ \begin{vmatrix} \text{a + b} & \text{b + c} & \text{c + a} \\ \text{b + c} & \text{c + a} & \text{a + b} \\ \text{c + a} & \text{a + b} & \text{b + c} \end{vmatrix}=2 \begin{vmatrix} \text{a} & \text{b} & \text{c} \\ \text{b} & \text{c} & \text{a} \\ \text{c} & \text{a} & \text{b} \end{vmatrix}$.
View full solution →$ \begin{vmatrix} \text{a + b} & \text{b + c} & \text{c + a} \\ \text{b + c} & \text{c + a} & \text{a + b} \\ \text{c + a} & \text{a + b} & \text{b + c} \end{vmatrix}=2 \begin{vmatrix} \text{a} & \text{b} & \text{c} \\ \text{b} & \text{c} & \text{a} \\ \text{c} & \text{a} & \text{b} \end{vmatrix}$.
If $\text{A}= \begin{bmatrix} 3 & 1 \\ -1 & 2 \\ \end{bmatrix},$ show the $\text{A}^{2}-\text{5A}+\text{7I}=0$. Hence find $A^{-1}$.
View full solution →Using properties of determinants, prove the following:
$\begin{vmatrix} 3a & -a + b & -a + c \\ a - b & 3b & c - b \\ a - c & b - c & 3c \end{vmatrix} = 3(a + b + c) (ab + bc + ca) $
View full solution →$\begin{vmatrix} 3a & -a + b & -a + c \\ a - b & 3b & c - b \\ a - c & b - c & 3c \end{vmatrix} = 3(a + b + c) (ab + bc + ca) $
If $A = \begin{bmatrix} 2 & -3 & \\ 3 & 4 & \\ \end{bmatrix} $ show that $\text{A^{2} - 6 A + 17 I = 0.}$ Hense find $A^{-1}.$
View full solution →Using properties of determinants, prove the following:$ \begin{vmatrix} a - b -c & 2a & 2a \\ 2b & b- c - a & 2b \\ 2c & 2c & c- a -b \end{vmatrix} = (a + b + c)^{3}$
View full solution →Using properties of determinants, prove that
$\begin{vmatrix} \text{a}^{2} + \text{2a} & \text{2a + 1} & 1 \\ \text{2a + 1} & \text{a + 2} & 1 \\ 3 & 3 & 1 \end{vmatrix} = \text{(a - 1)}^{3}$
View full solution →$\begin{vmatrix} \text{a}^{2} + \text{2a} & \text{2a + 1} & 1 \\ \text{2a + 1} & \text{a + 2} & 1 \\ 3 & 3 & 1 \end{vmatrix} = \text{(a - 1)}^{3}$
Using properties of determinants, prove that:
$\begin{vmatrix} \text{1 + a } & \text{1} & \text{1} 0.3em] \text{1} & \text{1 + b} & \text{1} 0.3em]\text{1} & 1 &\text{1 + c} \end{vmatrix}= \text{ abc + bc + ca + ab}$
View full solution →$\begin{vmatrix} \text{1 + a } & \text{1} & \text{1} 0.3em] \text{1} & \text{1 + b} & \text{1} 0.3em]\text{1} & 1 &\text{1 + c} \end{vmatrix}= \text{ abc + bc + ca + ab}$
Using properties of determinants, prove that
$\begin{vmatrix} \text{b + c } & \text{c + a} & \text{a + b} 0.3em] \text{q } + \text{r} & \text{r + p} & \text{p + q} 0.3em] \text{y + z} & \text{z + x} &\text{x + y} \end{vmatrix}= \text{2}$
$\begin{vmatrix} \text{a } & \text{b} & \text{c} 0.3em] \text{p} & \text{q} & \text{r} 0.3em] \text{x} & \text{y} &\text{z} \end{vmatrix}$
View full solution →$\begin{vmatrix} \text{b + c } & \text{c + a} & \text{a + b} 0.3em] \text{q } + \text{r} & \text{r + p} & \text{p + q} 0.3em] \text{y + z} & \text{z + x} &\text{x + y} \end{vmatrix}= \text{2}$
$\begin{vmatrix} \text{a } & \text{b} & \text{c} 0.3em] \text{p} & \text{q} & \text{r} 0.3em] \text{x} & \text{y} &\text{z} \end{vmatrix}$
Using properties of determinants, show that $\triangle\text{ABC}$ is isosceles if:
$\begin{vmatrix} 1 & 1 & 1 \\ 1 + \cos\text{A} & 1 + \cos\text{B} & 1 + \cos\text{C} \\ \cos^{2}\text{A} + \cos\text{A} & \cos^{2}\text{B}+\cos\text{B} & \cos^{2}\text{C} + \cos\text{C} \end{vmatrix} = 0 $
View full solution →$\begin{vmatrix} 1 & 1 & 1 \\ 1 + \cos\text{A} & 1 + \cos\text{B} & 1 + \cos\text{C} \\ \cos^{2}\text{A} + \cos\text{A} & \cos^{2}\text{B}+\cos\text{B} & \cos^{2}\text{C} + \cos\text{C} \end{vmatrix} = 0 $
Using properties of determinants, prove the following: $ \begin{bmatrix} \text{ x}&\text{x + y }&\text{x} + 2\text{y}\\ \text{x} + 2\text{y} & \text{x}& \text{x + y }\\\text{x + y}&\text{x} + 2\text{y}& \text{x} \end{bmatrix} = 9\text{y}^{2}(\text{x} + \text{y}). $
View full solution →If there is a statement involving the natural number n such that:
Also, if $A$ is a square matrix of order n, then $A^2$ is defined as $AA$. In general, $A^m = AA .... A (m$ times$)$. where m is any positive integer.
Based on the above information, answer the following questions.
View full solution →- The statement is true for $n = 1$
- When the statement is true for $n = k ($where $k$ is some positive integer$),$ then the statement is also true for $n = k + 1.$
Also, if $A$ is a square matrix of order n, then $A^2$ is defined as $AA$. In general, $A^m = AA .... A (m$ times$)$. where m is any positive integer.
Based on the above information, answer the following questions.
- If $\text{A}=\begin{bmatrix}3&-4\\1&-1\end{bmatrix},$ then for any positive integer n,
- $\text{A}^\text{n}=\begin{bmatrix}3\text{n}&-4\text{n}\\\text{n}&-\text{n}\end{bmatrix}$
- $\text{A}^\text{n}=\begin{bmatrix}1+2\text{n}&-4\text{n}\\\text{n}&1-2\text{n}\end{bmatrix}$
- $\text{A}^\text{n}=\begin{bmatrix}3\text{n}&-8\text{n}\\1&-\text{n}\end{bmatrix}$
- $\text{A}^\text{n}=\begin{bmatrix}1+3\text{n}&-4\text{n}\\\text{n}&1-3\text{n}\end{bmatrix}$
- If $\text{A}=\begin{bmatrix}1&2\\0&1\end{bmatrix},$ then $|A^n|$, where $\text{n}\in\text{ N},$ is equal to:
- $2^n$
- $3^n$
- $n$
- $1$
- If $\text{A}=\begin{bmatrix}1&0\\1&1\end{bmatrix}$ and $\text{I}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$ then which of the following holds for all natural numbers $\text{n}\geq1?$
- $A^n= nA - (n - 1)I$
- $A^n = 2^{n-1} A - (n - 1)I$
- $A^n= nA + (n - 1)I$
- $A^n = 2^{n-1} A + (n - 1)I$
- Let $\text{A}=\begin{bmatrix}\text{a}&0&0\\0&\text{a}&0\\0&0&\text{a}\end{bmatrix}$ and $\text{A}^\text{n}=[\text{a}_{\text{ij}}]_{3\times3}$ for some positive integer n, then the cofactor of $a_{13}$ is:
- $a^n$
- $-a^n$
- $2a^n$
- $0$
- If $A$ is a square matrix such that $|A| = 2,$ then for any positive integer n, $|A^n|$ is equal to:
- $0$
- $2n$
- $2^n$
- $n^2$
Area of a triangle whose vertices are $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$ is given by the determinant:
$\Delta=\frac{1}{2}\begin{vmatrix}\text{x}_1&\text{y}_1&1\\\text{x}_2&\text{y}_2&1\\\text{x}_3&\text{y}_3&1\end{vmatrix}$
Since, area is a positive quantity, so we always take the absolute value of the determinant $\Delta.$ Also, the area of the triangle formed by three collinear points is zero.
Based on the above information, answer the following questions.
View full solution →$\Delta=\frac{1}{2}\begin{vmatrix}\text{x}_1&\text{y}_1&1\\\text{x}_2&\text{y}_2&1\\\text{x}_3&\text{y}_3&1\end{vmatrix}$
Since, area is a positive quantity, so we always take the absolute value of the determinant $\Delta.$ Also, the area of the triangle formed by three collinear points is zero.
Based on the above information, answer the following questions.
- Find the area of the triangle whose vertices are $(-2, 6), (3, -6) $ and $(1, 5).$
- $30$ sq. units
- $35 $ sq. units
- $40$ sq. units
- $15.5$ sq. units
- If the points $(2, -3), (k, -1)$ and $(0, 4)$ are collinear, then find the value of $4k.$
- $4$
- $\frac{7}{140}$
- $47$
- $\frac{40}{7}$
- If the area of a triangle $ABC,$ with vertices $A(1, 3), B(0, 0)$ and $C(k, 0)$ is 3 sq. units, then a value of $k$ is:
- $2$
- $3$
- $4$
- $5$
- Using determinants, find the equation of the tine joining the points $A(1, 2)$ and $B(3, 6).$
- $y = 2x$
- $x = 3y$
- $y = x$
- $4x - y = 5$
- If $\text{A}\equiv(11,7),\text{B}\equiv(5,5)$ and $\text{C}\equiv(-1,3),$ then:
- $\Delta\text{ABC}$ is scalene triangle.
- $\Delta\text{ABC}$ is equilateral triangle.
- $A, B$ and $C$ are collinear.
- None of these.
A company produces three products every day. Their production on certain day is $45$ tons. It is found that the production of third product exceeds the production of first product by $8$ tons while the total production of first and third product is twice the production of second product.
Using the concepts of matrices and determinants, answer the following questions.
View full solution →Using the concepts of matrices and determinants, answer the following questions.
- If $x, y$ and $z$ respectively denotes the quantity (in tons) of first, second and third product produced, then which of the following is true?
- $x + y + z = 45$
- $x + 8 = z$
- $x - 2y + z = 0$
- All of these.
- If $\begin{pmatrix}1&1&1\\1&0&-2\\1&-1&1\end{pmatrix}^{-1}=\frac{1}{6}\begin{pmatrix}2&2&2\\3&0&-3\\1&-2&1\end{pmatrix}$ then the inverse of $\begin{pmatrix}1&1&1\\1&0&-1\\1&-2&1\end{pmatrix}$ is:
- $\begin{pmatrix}\frac{1}{3}&\frac{1}{3}&\frac{1}{3}\\\frac{1}{2}&0&\frac{-1}{2}\\\frac{1}{6}&\frac{-1}{3}&\frac{1}{6}\end{pmatrix}$
- $\begin{pmatrix}\frac{1}{2}&0&-\frac{1}{2}\\\frac{1}{3}&\frac{1}{3}&\frac{1}{3}\\\frac{1}{6}&\frac{-1}{3}&\frac{1}{6}\end{pmatrix}$
- $\begin{pmatrix}\frac{1}{3}&\frac{1}{2}&\frac{1}{6}\\\frac{1}{3}&0&\frac{-1}{3}\\\frac{1}{3}&\frac{-1}{2}&\frac{1}{6}\end{pmatrix}$
- None of these.
- $x : y : z$ is equal to:
- $12 : 13 : 20$
- $11 : 15 : 19$
- $15 : 19 : 11$
- $13 : 12 : 20$
- Which of the following is not true?
- $|A| = |A'|$
- $(A')^{-1} = (A^{-1})'$
- $A$ is skew synunetric matrix of odd order, then $|A| = 0$
- $|AB| = |A| + |B|$
- Which of the following is not true in the given determinant of $A,$ where A $=[\text{a}_\text{ij}]_{3\times3}?$
- Order of minor is less than order of the det $(A).$
- Minor of an element can never be equal to cofactor of the same element.
- Value of a determinant is obtained by multiplying elements of a row or column by corresponding cofactors.
- Order of minors and cofactors of same elements of $A$ is same.
Let $\text{A}=\begin{bmatrix}1&0\\2&1\end{bmatrix},$ and $U_1, U_2$ are e first and second columns respectively of a $2 \times 2$ matrix $U$. Also, let the column matrices $U_1$ and $U_2$ satisfying $\text{AU}_1=\begin{bmatrix}1\\0\end{bmatrix}$ and $\text{AU}_2=\begin{bmatrix}2\\3\end{bmatrix}.$
Based on the above information, answer the following questions.
View full solution →Based on the above information, answer the following questions.
- The matrix $U_1 + U_2$ is equal to:
- $\begin{bmatrix}1\\-1\end{bmatrix}$
- $\begin{bmatrix}2\\-2\end{bmatrix}$
- $\begin{bmatrix}3\\-3\end{bmatrix}$
- $\begin{bmatrix}4\\-4\end{bmatrix}$
- The value of $|U|$ is:
- $2$
- $-2$
- $3$
- $-3$
- If $\text{X}=\begin{bmatrix}3&2\end{bmatrix}\text{U}\begin{bmatrix}3\\2\end{bmatrix},$ then the value of |X| =
- $3$
- $-3$
- $-5$
- $5$
- The minor of element at the position $a_{22}$ in $U$ is:
- $1$
- $2$
- $-2$
- $-1$
- If $\text{U}=[\text{a}_\text{ij}]_{2\times2},$ then the value of $a_{11}A_{11}+ a_{12}A_{12},$ where $A_{ij} $ denotes the cofactor of $a_{ij},$ is:
- $1$
- $2$
- $-3$
- $3$
Gaurav purchased $5$ pens, $3$ bags and $1$ instrument box and pays $₹ 16$. From the same shop, Dheeraj purchased $2$ pens, $1$ bag and $3$ instrument boxes and pays $₹ 19,$ while Ankur purchased $1$ pen, $2$ bags and $4$ instrument boxes and pays $₹ 25.$
Using the concept of matrices and determinants, answer the following questions.
View full solution →Using the concept of matrices and determinants, answer the following questions.
- The cost of one pen is:
- $₹ 2$
- $₹ 5$
- $₹ 1$
- $₹ 3$
- What is the cost of one pen and one bag?
- $₹ 3$
- $₹ 5$
- $₹ 7$
- $₹ 8$
- What is the cost of one pen and one instrument box?
- $₹ 7$
- $₹ 6$
- $₹ 8$
- $₹ 9$
- Which of the following is correct?
- Determinant is a square matrix.
- Determinant is a number associated to a matrix.
- Determinant is a number associated to a square matrix.
- All of the above.
- From the matrix equation $AB = AC,$ it can be concluded that $B = C$ provided:
- $A$ is singular.
- $A$ is non-singular.
- $A$ is symmetric.
- $A$ is square.
Fill in the blanks:
If A is a matrix of order $3 × 3,$ then $(A^2)^{-1}= \_\_\_\_\_\_\_\_.$
View full solution →If A is a matrix of order $3 × 3,$ then $(A^2)^{-1}= \_\_\_\_\_\_\_\_.$
Fill in the blanks:
If A is a matrix of order 3 × 3, then number of minors in determinant of A are ________.
If A is a matrix of order 3 × 3, then number of minors in determinant of A are ________.
Fill in the blanks:
If $\cos2\theta=0,$ then $\begin{vmatrix}0&\cos\theta&\sin\theta\\\cos\theta&\sin\theta&0\\\sin\theta&0&\cos\theta\end{vmatrix}^2=$ _________.
View full solution →If $\cos2\theta=0,$ then $\begin{vmatrix}0&\cos\theta&\sin\theta\\\cos\theta&\sin\theta&0\\\sin\theta&0&\cos\theta\end{vmatrix}^2=$ _________.
Fill in the blanks:
$\begin{vmatrix}0&\text{xyz}&\text{x}-\text{z}\\\text{y}-\text{x}&0&\text{y}-\text{z}\\\text{z}-\text{x}&\text{z}-\text{y}&0\end{vmatrix}=$ ________.
View full solution →$\begin{vmatrix}0&\text{xyz}&\text{x}-\text{z}\\\text{y}-\text{x}&0&\text{y}-\text{z}\\\text{z}-\text{x}&\text{z}-\text{y}&0\end{vmatrix}=$ ________.
Fill in the blanks:
If A is a matrix of order 3 × 3, then |3A| = _______.
If A is a matrix of order 3 × 3, then |3A| = _______.
State True or False for the statements of the following Exercise:
Let $ \begin{vmatrix}\text{a}&\text{p}&\text{x}\\\text{b}&\text{q}&\text{y}\\\text{c}&\text{r}&\text{z}\end{vmatrix}=16,$ then $\Delta_1=\begin{vmatrix}\text{p}+\text{x}&\text{a}+\text{x}&\text{a}+\text{p}\\\text{q}+\text{y}&\text{b} +\text{y}&\text{b}+\text{q}\\\text{r}+\text{z}&\text{c}+ \text{z}&\text{c}+\text{r}\end{vmatrix}=32.$
View full solution →Let $ \begin{vmatrix}\text{a}&\text{p}&\text{x}\\\text{b}&\text{q}&\text{y}\\\text{c}&\text{r}&\text{z}\end{vmatrix}=16,$ then $\Delta_1=\begin{vmatrix}\text{p}+\text{x}&\text{a}+\text{x}&\text{a}+\text{p}\\\text{q}+\text{y}&\text{b} +\text{y}&\text{b}+\text{q}\\\text{r}+\text{z}&\text{c}+ \text{z}&\text{c}+\text{r}\end{vmatrix}=32.$
State True or False for the statements of the following Exercise:
If the determinant $\begin{vmatrix}\text{x}+\text{a}&\text{p}+\text{u}&\text{l}+\text{f}\\\text{y}+\text{b}&\text{q}+\text{v}&\text{m}+\text{g}\\\text{z}+\text{c}&\text{r}+\text{w}&\text{n}+\text{h}\end{vmatrix}$ splits into exactly K determinants of order 3, each element of which contains only one term, then the value of K is 8.
View full solution →If the determinant $\begin{vmatrix}\text{x}+\text{a}&\text{p}+\text{u}&\text{l}+\text{f}\\\text{y}+\text{b}&\text{q}+\text{v}&\text{m}+\text{g}\\\text{z}+\text{c}&\text{r}+\text{w}&\text{n}+\text{h}\end{vmatrix}$ splits into exactly K determinants of order 3, each element of which contains only one term, then the value of K is 8.
State True or False for the statements of the following Exercise:
$\begin{vmatrix}\text{x}+1&\text{x}+2&\text{x}+\text{a}\\\text{x}+2&\text{x}+3&\text{x}+\text{b}\\\text{x}+3&\text{x}+4&\text{x}+\text{c}\end{vmatrix}=0,$ where a, b, c are in A.P.
View full solution →$\begin{vmatrix}\text{x}+1&\text{x}+2&\text{x}+\text{a}\\\text{x}+2&\text{x}+3&\text{x}+\text{b}\\\text{x}+3&\text{x}+4&\text{x}+\text{c}\end{vmatrix}=0,$ where a, b, c are in A.P.
State True or False for the statements of the following Exercise:
The determinant $\begin{vmatrix}\sin\text{A}&\cos\text{A}&\sin\text{A}+\cos\text{B}\\\sin\text{B}&\cos\text{A}&\sin\text{B}+\cos\text{B}\\\sin\text{C}&\cos\text{A}&\sin\text{C}+\cos\text{B}\end{vmatrix}$ is equal to zero.
View full solution →The determinant $\begin{vmatrix}\sin\text{A}&\cos\text{A}&\sin\text{A}+\cos\text{B}\\\sin\text{B}&\cos\text{A}&\sin\text{B}+\cos\text{B}\\\sin\text{C}&\cos\text{A}&\sin\text{C}+\cos\text{B}\end{vmatrix}$ is equal to zero.
State True or False for the statements of the following Exercise:
$\big(\text{A}^3\big)^{-1}=(\text{A}^{-1})^3,$ where A is square matrx and |A| ≠ 0.
View full solution →$\big(\text{A}^3\big)^{-1}=(\text{A}^{-1})^3,$ where A is square matrx and |A| ≠ 0.
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