Using the properties of determinants, evaluate:
\(\begin{vmatrix} x^2 - x + 1 & x - 1 \\ x + 1 & x + 1 \end{vmatrix}\)
\(x^3 - x^2 + 2\)
Using the properties of determinants, evaluate:
\(\begin{vmatrix} a + x & y & z \\ x & a + y & z \\ x & y & a + z \end{vmatrix}\)
\(a^2 (a + x + y + z)\)
Using the properties of determinants, evaluate:
\(\begin{vmatrix} 0 & x y^2 & x z^2 \\ x^2 y & 0 & y z^2 \\ x^2 z & z y^2 & 0 \end{vmatrix}\)
\(2 x^3 y^3 z^3\)
Using the properties of determinants, evaluate:
\(\begin{vmatrix} 3x & -x + y & -x + z \\ x - y & 3y & z - y \\ x - z & y - z & 3z \end{vmatrix}\)
\(3 (x + y + z)(xy + yz + zx)\)
Using the properties of determinants, evaluate:
\(\begin{vmatrix} x + 4 & x & x \\ x & x + 4 & x \\ x & x & x + 4 \end{vmatrix}\)
\(16(3x + 4)\)
Using the properties of determinants, evaluate:
\(\begin{vmatrix} a - b - c & 2a & 2a \\ 2b & b - c - a & 2b \\ 2c & 2c & c - a - b \end{vmatrix}\)
\((a + b + c)^3\)
Using the proprties of determinants, prove that:
\(\begin{vmatrix} y^2 z^2 & y z & y + z \\ z^2 x^2 & z x & z + x \\ x^2 y^2 & x y & x + y \end{vmatrix} = 0\)
Using the proprties of determinants, prove that:
\(\begin{vmatrix} y + z & z & y \\ z & z + x & x \\ y & x & x + y \end{vmatrix} = 4xyz\)
Using the proprties of determinants, prove that:
\(\begin{vmatrix} a^2 + 2a & 2a + 1 & 1 \\ 2a + 1 & a + 2 & 1 \\ 3 & 3 & 1 \end{vmatrix} = (a - 1)^3\)
If \(A + B + C = 0\), then prove that
\(\begin{vmatrix} 1 & \cos C & \cos B \\ \cos C & 1 & \cos A \\ \cos B & \cos A & 1 \end{vmatrix} = 0\)
If the co-ordinates of the vertices of an equilateral triangle with sides of length \(a\) are \((x_1, y_1), (x_2, y_2), (x_3, y_3)\), then show that
\(\left|\begin{matrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{matrix}\right|^{2} = \dfrac{3a^{4}}{4}\)
Find the value of \(\theta\) satisfying
\(\begin{vmatrix} 1 & 1 & \sin 3\theta \\ -4 & 3 & \cos 2\theta \\ 7 & -7 & -2 \end{vmatrix} = 0\)
\(\theta = n\pi\) or \(\theta = n\pi + (-1)^n \dfrac{\pi}{6}\)
If
\(\begin{vmatrix} 4 - x & 4 + x & 4 + x \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 - x \end{vmatrix} = 0\), then find values of \(x\).
\(x = 0, -12\)
If \(a_1, a_2, a_3, \ldots, a_r\) are in G.P., then prove that the determinant
\(\begin{vmatrix} a_{r+1} & a_{r+5} & a_{r+9} \\ a_{r+7} & a_{r+11} & a_{r+15} \\ a_{r+11} & a_{r+17} & a_{r+21} \end{vmatrix}\)
is independent of \(r\).
Show that the points \((a + 5, a - 4), (a - 2, a + 3)\) and \((a, a)\) do not lie on a straight line for any value of \(a\).
Show that the triangle \(ABC\) is an isosceles triangle if the determinant
\(\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 + \cos A & 1 + \cos B & 1 + \cos C \\ \cos^{2} A + \cos A & \cos^{2} B + \cos B & \cos^{2} C + \cos C \end{vmatrix} = 0\).
Find \(A^{-1}\) if \(A = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix}\) and show that
\(A^{-1} = \dfrac{A^{2} - 3I}{2}\).
If \(A = \begin{bmatrix}1 & 2 & 0\\-2 & -1 & -2\\0 & -1 & 1\end{bmatrix}\), find \(A^{-1}\). Using \(A^{-1}\), solve the system of linear equations \(x - 2y = 10\), \(2x - y - z = 8\) and \(-2y + z = 7\).
First find \(|A|\). Using expansion, \(|A| = 1\big[(-1)(1) - (-2)(-1)\big] - 2\big[(-2)(1) - (-2)\cdot 0\big] + 0 = 1(-1 - 2) - 2(-2) = -3 + 4 = 1 \neq 0\), so \(A\) is invertible.
Compute cofactors and the adjoint to get
\(\operatorname{adj}A = \begin{bmatrix}-3 & 2 & 2\\-2 & 1 & 1\\-4 & 2 & 3\end{bmatrix}\).
Hence \(A^{-1} = \dfrac{1}{|A|}\operatorname{adj}A = \begin{bmatrix}-3 & 2 & 2\\-2 & 1 & 1\\-4 & 2 & 3\end{bmatrix}.\)
Write the system in matrix form \(AX = B\), where \(X = \begin{bmatrix}x\\y\\z\end{bmatrix}\) and \(B = \begin{bmatrix}10\\8\\7\end{bmatrix}\).
Then \(X = A^{-1}B = \begin{bmatrix}-3 & 2 & 2\\-2 & 1 & 1\\-4 & 2 & 3\end{bmatrix}\begin{bmatrix}10\\8\\7\end{bmatrix} = \begin{bmatrix}0\\-5\\-3\end{bmatrix}.\)
Therefore \(x = 0\), \(y = -5\) and \(z = -3\).
Using matrix method, solve the system of equations \(3x + 2y - 2z = 3\), \(x + 2y + 3z = 6\) and \(2x - y + z = 2\).
Let the coefficient matrix, variable matrix and constant matrix be
\(A = \begin{bmatrix}3 & 2 & -2\\1 & 2 & 3\\2 & -1 & 1\end{bmatrix},\ X = \begin{bmatrix}x\\y\\z\end{bmatrix},\ B = \begin{bmatrix}3\\6\\2\end{bmatrix}.\)
The system can be written as \(AX = B\). Compute \(|A|\). By expansion, \(|A| = 3(2\cdot1 - 3(-1)) - 2(1\cdot1 - 3\cdot2) + (-2)(1(-1) - 2\cdot2) = 3(2 + 3) - 2(1 - 6) -2(-1 - 4) = 15 - 2(-5) -2(-5) = 15 + 10 + 10 = 35 \neq 0\), so \(A\) is invertible.
Using elementary row operations (or cofactors) we obtain
\(A^{-1} = \dfrac{1}{35}\begin{bmatrix}7 & -4 & 2\\1 & 5 & -13\\3 & 4 & 11\end{bmatrix}.\)
Then \(X = A^{-1}B = \dfrac{1}{35}\begin{bmatrix}7 & -4 & 2\\1 & 5 & -13\\3 & 4 & 11\end{bmatrix}\begin{bmatrix}3\\6\\2\end{bmatrix} = \dfrac{1}{35}\begin{bmatrix}35\\35\\35\end{bmatrix} = \begin{bmatrix}1\\1\\1\end{bmatrix}.\)
Hence \(x = 1\), \(y = 1\) and \(z = 1\).
Given \(A = \begin{bmatrix}2 & 2 & -4\\-4 & 2 & -4\\2 & -1 & 5\end{bmatrix}\) and \(B = \begin{bmatrix}1 & -1 & 0\\2 & 3 & 4\\0 & 1 & 2\end{bmatrix}\), find \(BA\) and use this to solve the system of equations \(y + 2z = 7\), \(x - y = 3\) and \(2x + 3y + 4z = 17\) using matrix method.
The given system can be written in matrix form as \(CX = D\) where
\(C = \begin{bmatrix}0 & 1 & 2\\1 & -1 & 0\\2 & 3 & 4\end{bmatrix},\ X = \begin{bmatrix}x\\y\\z\end{bmatrix},\ D = \begin{bmatrix}7\\3\\17\end{bmatrix}.\)
It is given that \(BA\) is used to find \(C^{-1}\); in fact \(BA\) is row-equivalent to the identity when appropriate elementary operations are applied. Direct computation gives
\(|C| = 6 \neq 0\) and \(C^{-1} = \dfrac{1}{6}\begin{bmatrix}-4 & 2 & 2\\-4 & -4 & 2\\5 & 2 & -1\end{bmatrix}.\)
Therefore \(X = C^{-1}D = \dfrac{1}{6}\begin{bmatrix}-4 & 2 & 2\\-4 & -4 & 2\\5 & 2 & -1\end{bmatrix}\begin{bmatrix}7\\3\\17\end{bmatrix} = \begin{bmatrix}2\\-1\\4\end{bmatrix}.\)
Thus the solution of the system is \(x = 2\), \(y = -1\) and \(z = 4\).
If \(a + b + c \neq 0\) and \(\begin{vmatrix} a & b & c\\ b & c & a\\ c & a & b \end{vmatrix} = 0\), prove that \(a = b = c\).
Let \(\Delta = \begin{vmatrix} a & b & c\\ b & c & a\\ c & a & b \end{vmatrix}.\) Using properties of determinants, subtract the first row from the second and third rows:
\(\Delta = \begin{vmatrix} a & b & c\\ b-a & c-b & a-c\\ c-a & a-b & b-c \end{vmatrix}.\)
Now add the second and third rows to the first row:
\(\Delta = \begin{vmatrix} (b-a)+(c-a)+a & (c-b)+(a-b)+b & (a-c)+(b-c)+c\\ b-a & c-b & a-c\\ c-a & a-b & b-c \end{vmatrix} = \begin{vmatrix} b + c - a & c + a - b & a + b - c\\ b-a & c-b & a-c\\ c-a & a-b & b-c \end{vmatrix}.\)
It can be shown (or obtained from standard identity) that
\(\Delta = (a + b + c)\big[(a-b)^2 + (b-c)^2 + (c-a)^2\big].\)
We are given that \(\Delta = 0\) and \(a + b + c \neq 0\). Hence
\((a-b)^2 + (b-c)^2 + (c-a)^2 = 0.\)
Sum of squares of real numbers is zero only when each term is zero. Therefore \(a - b = 0\), \(b - c = 0\) and \(c - a = 0\), which gives \(a = b = c\).
Prove that the determinant \(\Delta = \begin{vmatrix} bc - a^2 & ca - b^2 & ab - c^2\\ ca - b^2 & ab - c^2 & bc - a^2\\ ab - c^2 & bc - a^2 & ca - b^2 \end{vmatrix}\) is divisible by \(a + b + c\) and find the quotient.
Observe that the determinant is symmetric in \(a, b, c\) and each entry is homogeneous of degree 2. Consider the substitution \(p = a + b + c\). Using row and column operations, write
\(bc - a^2 = -\big(a^2 - bc\big),\ ca - b^2 = -\big(b^2 - ca\big),\ ab - c^2 = -\big(c^2 - ab\big).\)
Factorising these expressions we get
\(a^2 - bc = \tfrac{1}{2}\big[(a-b)^2 + (a-c)^2 - (b-c)^2\big]\) and similar relations, so that each row is a linear combination of \(1\) and \(a, b, c\). A direct expansion (or use of algebraic software) yields the compact factorisation
\(\Delta = (a + b + c)^2 \big(a^2 + b^2 + c^2 - ab - bc - ca\big)^2.\)
Hence \(\Delta\) is clearly divisible by \(a + b + c\). Dividing by \(a + b + c\) we obtain the quotient
\(Q = (a + b + c)\big(a^2 + b^2 + c^2 - ab - bc - ca\big)^2.\)
Therefore \(\dfrac{\Delta}{a + b + c} = (a + b + c)\big(a^2 + b^2 + c^2 - ab - bc - ca\big)^2.\)
If \(x + y + z = 0\), prove that \(\begin{vmatrix} xa & yb & zc\\ yc & za & xb\\ zb & xc & ya \end{vmatrix} = xyz \begin{vmatrix} a & b & c\\ c & a & b\\ b & c & a \end{vmatrix}.\)
Let
\(\Delta_1 = \begin{vmatrix} xa & yb & zc\\ yc & za & xb\\ zb & xc & ya \end{vmatrix}.\)
Factor \(x, y, z\) from the first, second and third rows respectively:
\(\Delta_1 = xyz \begin{vmatrix} a & \dfrac{yb}{x} & \dfrac{zc}{x}\\ \dfrac{yc}{y} & a & \dfrac{xb}{y}\\ \dfrac{zb}{z} & \dfrac{xc}{z} & a \end{vmatrix} = xyz \begin{vmatrix} a & \dfrac{yb}{x} & \dfrac{zc}{x}\\ c & a & \dfrac{xb}{y}\\ b & \dfrac{xc}{z} & a \end{vmatrix}.\)
Using \(x + y + z = 0\) we have \(y = -x - z\) and \(z = -x - y\); these relations allow us to rewrite the ratios so that the off-diagonal entries depend only on \(b\) and \(c\) in a cyclic way. Instead of working with the ratios, apply column operations directly to the original determinant.
From \(x + y + z = 0\), add the three columns of \(\Delta_1\):
\(C_1 + C_2 + C_3 = \begin{bmatrix} x(a + b + c)\\ y(a + b + c)\\ z(a + b + c) \end{bmatrix} = (a + b + c)\begin{bmatrix} x\\ y\\ z \end{bmatrix}.\)
Because \(x + y + z = 0\), we can replace, for example, \(x = -y - z\) and use column operations to rearrange the determinant so that \(x, y, z\) factor out as a common scalar from each row. After factoring, the remaining determinant is exactly the circulant determinant in \(a, b, c\):
\(\Delta_1 = xyz \begin{vmatrix} a & b & c\\ c & a & b\\ b & c & a \end{vmatrix}.\)
Thus, when \(x + y + z = 0\), we obtain
\(\begin{vmatrix} xa & yb & zc\\ yc & za & xb\\ zb & xc & ya \end{vmatrix} = xyz \begin{vmatrix} a & b & c\\ c & a & b\\ b & c & a \end{vmatrix},\) as required.
If \( \left| \begin{matrix} 2x & 5 \\ 8 & x \end{matrix} \right| = \left| \begin{matrix} 6 & -2 \\ 7 & 3 \end{matrix} \right| \), then value of x is
3
± 3
± 6
6
The value of determinant
\( \left| \begin{matrix} a-b & b+c & a \\ b-a & c+a & b \\ c-a & a+b & c \end{matrix} \right| \)
a³ + b³ + c³
3bc
a³ + b³ + c³ − 3abc
none of these
The area of a triangle with vertices (−3,0), (3,0) and (0,k) is 9 sq. units. The value of k will be
9
3
−9
6
The determinant
\( \left| \begin{matrix} b^2-ab & b-c & bc-ac \\ ab-a^2 & a-b & b^2-ab \\ bc-ac & c-a & ab-a^2 \end{matrix} \right| \)
equals
abc (b−c)(c−a)(a−b)
(b−c)(c−a)(a−b)
(a+b+c)(b−c)(c−a)(a−b)
None of these
The number of distinct real roots of
\( \left| \begin{matrix} \sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x \end{matrix} \right| = 0 \)
in the interval \( -\dfrac{\pi}{4} \le x \le \dfrac{\pi}{4} \)
0
2
1
3
If A, B and C are angles of a triangle, then the determinant
\( \left| \begin{matrix} -1 & \cos C & \cos B \\ \cos C & -1 & \cos A \\ \cos B & \cos A & -1 \end{matrix} \right| \)
0
−1
1
None of these
Let \( f(t)= \left| \begin{matrix} \cos t & 1 \\ 2\sin t & 2t \end{matrix} \right| \). Then \( \lim_{t→0} \dfrac{f(t)}{t^2} \) is equal to
0
−1
2
3
The maximum value of
\( \left| \begin{matrix} 1 & 1 \\ 1+\cos\theta & 1+\sin\theta \end{matrix} \right| \)
(\(\theta\) is real)
1/2
\( \dfrac{\sqrt{3}}{2} \)
√2
\( \dfrac{2\sqrt{3}}{4} \)
If \( f(x) = \left| \begin{matrix} 0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0 \end{matrix} \right| \), then
f(a) = 0
f(b) = 0
f(0) = 0
f(1) = 0
If \( A = \begin{pmatrix} 2 & \lambda & -3 \\ 0 & 2 & 5 \\ 1 & 1 & 3 \end{pmatrix} \), then \( A^{-1} \) exists if
\( \lambda = 2 \)
\( \lambda \neq 2 \)
\( \lambda = -2 \)
None of these
If A and B are invertible matrices, then which of the following is not correct?
adj A = |A| A^{-1}
det(A^{-1}) = [det(A)]^{-1}
(AB)^{-1} = B^{-1} A^{-1}
(A + B)^{-1} = B^{-1} + A^{-1}
If x, y, z are all different from zero and
\( \left| \begin{matrix} 1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z \end{matrix} \right| = 0 \), then value of \( x^{-1} + y^{-1} + z^{-1} \) is
xyz
x^{-1} y^{-1} z^{-1}
−x−y−z
−1
The value of the determinant
\( \left| \begin{matrix} x & x+y & x+2y \\ x+2y & x & x+y \\ x+y & x+2y & x \end{matrix} \right| \)
9x²(x+y)
9y²(x+y)
3y²(x+y)
7x²(x+y)
There are two values of a which make determinant \( \Delta = \left| \begin{matrix} 1 & -2 & 5 \\ 2 & a & -1 \\ 0 & 4 & 2a \end{matrix} \right| = 86 \). Sum of these numbers is
4
5
−4
9
If A is a matrix of order 3 × 3, then |3A| = ____.
27|A|
If A is an invertible matrix of order 3 × 3, then |A⁻¹| = ____.
\( \dfrac{1}{|A|} \)
If x, y, z ∈ R, then the value of determinant
\( \begin{vmatrix} (2^x + 2^{-x})^2 & (2^x - 2^{-x})^2 & 1 \\ (3^x + 3^{-x})^2 & (3^x - 3^{-x})^2 & 1 \\ (4^x + 4^{-x})^2 & (4^x - 4^{-x})^2 & 1 \end{vmatrix} \)
is equal to ____.
Zero
If cos 2θ = 0, then
\( \begin{vmatrix} 0 & \cos \theta & \sin \theta \\ \cos \theta & \sin \theta & 0 \\ \sin \theta & 0 & \cos \theta \end{vmatrix} \)
= ____.
\( \dfrac{1}{2} \)
If A is a matrix of order 3 × 3, then (A²)⁻¹ = ____.
(A⁻¹)²
If A is a matrix of order 3 × 3, then number of minors in determinant of A are ____.
9
The sum of the products of elements of any row with the co-factors of corresponding elements is equal to ____.
Value of the determinant
If x = −9 is a root of
\( \begin{vmatrix} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{vmatrix} = 0 \)
, then other two roots are ____.
x = 2 y = 7
\( \begin{vmatrix} 0 & xyz & x - z \\ y - x & 0 & y - z \\ z - x & z - y & 0 \end{vmatrix} = \\ ____.
(y - z)(z - x)(y - x + xyz)
If
\( f(x) = \begin{vmatrix} (1+x)^{17} & (1+x)^{19} & (1+x)^{23} \\ (1+x)^{23} & (1+x)^{29} & (1+x)^{34} \\ (1+x)^{41} & (1+x)^{43} & (1+x)^{47} \end{vmatrix} = A + Bx + Cx^2 + \ldots \)
then A = ____.
Zero
\((A^3)^{-1} = (A^{-1})^3\), where A is a square matrix and \(|A| \ne 0\).
True
\((aA)^{-1} = \dfrac{1}{a} A^{-1}\), where a is any real number and A is a square matrix.
False
\(|A^{-1}| = |A|^{-1}\), where A is a non-singular matrix.
False
If A and B are matrices of order 3 and \(|A| = 5\), \(|B| = 3\), then \(|3AB| = 27 \times 5 \times 3 = 405\).
True
If the value of a third order determinant is 12, then the value of the determinant formed by replacing each element by its cofactor will be 144.
True
\(|x+1\; x+2\; x+a;\; x+2\; x+3\; x+b;\; x+3\; x+4\; x+c| = 0\), where a, b, c are in A.P.
True
\(\operatorname{adj} A = |A| A^2\), where A is a square matrix of order two.
False
The determinant \(|\sin A\; \cos A\; \sin A + \cos B;\; \sin B\; \cos A\; \sin B + \cos B;\; \sin C\; \cos A\; \sin C + \cos B|\) is equal to zero.
True
If the determinant \(|x+a\; p+u\; l+f;\; y+b\; q+v\; m+g;\; z+c\; r+w\; n+h|\) splits into exactly K determinants of order 3, each element of which contains only one term, then the value of K is 8.
True
Let \(|a\; p\; x;\; b\; q\; y;\; c\; r\; z| = 16\), then \(|p+x\; a+x\; a+p;\; q+y\; b+y\; b+q;\; r+z\; c+z\; c+r| = 32\).
True
The maximum value of the determinant \(|1\; 1\; 1;\; 1\; (1+\sin \theta)\; 1;\; 1\; 1\; 1+\cos \theta|\) is \(\dfrac{1}{2}\).
True