Determinants Test - 15

Given: A² - 2A - I = 0

⇒ A.A - 2A = I

Post multiply by A^-1, we get

AAA^-1 - 2AA^-1 = IA^-1

⇒ AI - 2I = A^-1             [∵ AA^{- 1} = A^-1 A = I]

∴ A^-1 = A - 2

The inverse of A is A - 2

Hence, the correct option is (C).

If A, B, C are matrices then A(BC) = (AB)C

A^-1 A = I

Given that, A has inverse B

AB = I = BA ....(1)

Also, A has inverse C.

AC = I = CA ....(2)

From (1) and (2),

AB = AC ....(3)

Pre multiply equation (3) by A^-1

A^-1 (AB) = A^-1 (AC)

⇒ A^-1 (AB) = A^-1 (AC)

⇒ (A^-1 A)B = (A^-1A)C

⇒ (IB) = (IC)

⇒ B = C

Thus, if a matrix A has inverses B and C then, B should be equal to C.

Hence, the correct option is (B).

The coefficient matrix for the given system of equations is:

\(\operatorname{det}(A)=1(10-4)+1(16-20)+1(2-4)=6-4-2=0\)

Therefore, either there is no solution (inconsistent) or there are infinitely many solutions (consistent and dependent).

Let, us convert the augmented matrix into the row echelon form to find its solutions. The augmented matrix is:

\(A \mid B =\left[\begin{array}{ccccc}1 & 1 & 1 & \mid & 1 \\ 2 & 1 & 4 & \mid & k \\ 4 & 1 & 10 & \mid & k ^{2}\end{array}\right]\)

\(R _{3} \rightarrow R _{3}-4 R _{1}\)

\(R _{2} \rightarrow R _{2}-2 R _{1}\)

\(A \mid B =\left[\begin{array}{ccccc}1 & 1 & 1 & \mid & 1 \\ 0 & -1 & 2 & \mid & k -2 \\ 0 & -3 & 6 & \mid & k ^{2}-4\end{array}\right]\)

\(R _{3} \rightarrow R _{3}-3 R _{2}\)

\(A \mid B =\left[\begin{array}{ccccc}1 & 1 & 1 & \mid & 1 \\ 0 & -1 & 2 & \mid & k -2 \\ 0 & 0 & 0 & \mid & k ^{2}-3 k +2\end{array}\right]\)

The rank of both the coefficient matrix and the augmented matrix must be equal for the system to be consistent.

\(\therefore k ^{2}-3 k +2=0\)

\(\Rightarrow k ^{2}-2 k - k +2=0\)

\(\Rightarrow k ( k -2)-( k -2)=0\)

\(\Rightarrow( k -2)( k -1)=0\)

\(\Rightarrow k -2=0\) OR \(k -1=0\)

\(\Rightarrow k =2\) or \(k =1\)

Hence, the correct option is (B).

Given: \(A=\left[\begin{array}{cc}2 \cos x & -2 \sin x \\ \sin x & \cos x\end{array}\right]\) and \(|\mathrm{A}|=\mathrm{k}\)

As we know that, if \(A=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right]\) is a square matrix of order 2, then determinant of \(\mathrm{A}\) is given by \(|\mathrm{A}|=\left(\mathrm{a}_{11} \times \mathrm{a}_{22}\right)-\left(\mathrm{a}_{12}-\mathrm{a}_{21}\right)\)

\(\Rightarrow|\mathrm{A}|=2 \cos ^{2} x+2 \sin ^{2} x=k\)

\(\Rightarrow k=2 \times\left(\sin ^{2} x+\cos ^{2} x\right)=2 \ldots \ldots \left(\because \sin ^{2} x+\cos ^{2} x=1\right)\)

Hence, the correct option is (A).

Given, a matrix A is such that 3A3+2A2+5A+I=0

Consider, 3A3+2A2+5A+I=0

Pre - Multiply the above polynomial by A−1.

A−1(3A3+2A2+5A+I)=A−1(0)

⇒3A−1A3+2A−1A2+5A−1A+A−1I=0

⇒3A−1AA2+2A−1AA+5A−1A+A−1I=0

We know that A−1A=I and A−1I=A−1

⇒3∣∣A2+2∣∣A+5I+A−1=0

⇒3A2+2A+5I+A−1=0

⇒3A2+2A+5I+A−1=0

⇒A−1=−(3A2+2A+5I)

Hence, the correct option is (A).

By cramer’s rule:

If \(\Delta=0\) and \(\Delta_{1}=\Delta_{2}=\Delta_{3}=0\), then the system is consistent and has infinitely many solutions.

Given: \(x+y+z=3,2 x+2 y+2 z=6\) and \(a x+a y+a z=3 a\) As we know that,

\(\Delta=\left|\begin{array}{lll}a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3}\end{array}\right|, \Delta_{1}=\left|\begin{array}{lll}d_{1} & b_{1} & c_{1} \\ d_{2} & b_{2} & c_{2} \\ d_{3} & b_{3} & c_{3}\end{array}\right|, \Delta_{2}=\left|\begin{array}{lll}a_{1} & d_{1} & c_{1} \\ a_{2} & d_{2} & c_{2} \\ a_{3} & d_{3} & c_{3}\end{array}\right|\) and \(\Delta_{3}=\left|\begin{array}{ccc}a_{1} & b_{1} & d_{1} \\ a_{2} & b_{2} & d_{2} \\ a_{3} & b_{3} & d_{3}\end{array}\right|\)

\(\Rightarrow \Delta=\left|\begin{array}{ccc}1 & 1 & 1 \\ 2 & 4 & 1 \\ \alpha & \alpha & \alpha\end{array}\right|, \Delta_{1}=\left|\begin{array}{ccc}3 & 1 & 1 \\ 6 & 4 & 1 \\ 3 \alpha & \alpha & \alpha\end{array}\right|, \Delta_{2}=\left|\begin{array}{ccc}1 & 3 & 1 \\ 2 & 6 & 1 \\ \alpha & 3 \alpha & \alpha\end{array}\right|\) and \(\Delta_{3}=\left|\begin{array}{ccc}1 & 1 & 3 \\ 2 & 4 & 6 \\ \alpha & \alpha & 3 \alpha\end{array}\right|\)

\(\Rightarrow \Delta=0\) and \(\Delta_{1}=\Delta_{2}=\Delta_{3}=0\)

Thus, the given system has infinite solution for \(\alpha \in R\).

Hence, the correct option is (B).

Let, the system of equations be:

\(a_{1} x+b_{1} y+c_{1} z=d_{1}\)

\(a_{2} x+b_{2} y+c_{2} z=d_{2}\)

\(a_{3} x+b_{3} y+c_{3} z=d_{3}\)

\(\left[\begin{array}{lll}a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3}\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}d_{1} \\ d_{2} \\ d_{3}\end{array}\right]\) \(=\mathrm{AX}=\mathrm{B}\)

\(X=A^{-1} B=\frac{\operatorname{adj}(A)}{\operatorname{det}(A)} B\)

If \(\operatorname{det}(A) \neq 0\), system is consistent having unique solution.

If \(\operatorname{det}(A)=0\) and \((\operatorname{adj} A) . B=0\), system is consistent, with infinitely many solutions.

If \(\operatorname{det}(A)=0\) and \(( adj A) . B \neq 0\), system is inconsistent (no solution)

Given: The system of equations

\(2 x+y-3 z=5\)

\(3 x-2 y+2 z=5\) and

\(5 x-3 y-z=16\)

So, \(A=\left[\begin{array}{ccc}2 & 1 & -3 \\ 3 & -2 & 2 \\ 5 & -3 & -1\end{array}\right]\)

det \((\mathrm{A})=|\mathrm{A}|=2 \times\{(-2 \times-1)-(-3 \times 2)\}-1 \times\{(3 \times-1)-(2 \times 5)\}+(-3) \times\{(3 \times-3)-(5 \times-2)\}\)

\(\Rightarrow|\mathrm{A}|=2 \times(2+6)-1 \times(-3-10)-3 \times(-9+10)\)

\(\Rightarrow|\mathrm{A}|=16+13-3=26\)

\(\therefore|\mathrm{A}| \neq 0\)

So, system is consistent having unique solution.

Hence, the correct option is (B).

From the definition of the inverse of a matrix, \(A^{-1}=\frac{\operatorname{adj}(A)}{|A|}\)

Therefore, in the given question, \(k=|A|\).

Using the properties of the determinant of inverse of a matrix, we have:

\(k =| A |=\frac{1}{\left| A ^{-1}\right|}\)

Now, \(\left|A^{-1}\right|=1(24-3)+2(9-12)+3(2-12)=21-6-30=-15\)

Therefore, \(k =-15\)

Hence, the correct option is (B).

Given determinant is \(\left|\begin{array}{lll}0 & c & b \\ c & 0 & a \\ b & a & 0\end{array}\right|^{2}\), simplifying the determinant based on first row we get,

\(=\left\{0 \times\left(0-a^{2}\right)-c \times(c \times 0-b \times a)+b(c \times a-b \times 0)\right\}^{2}\)

\(=(a b c+a b c)^{2}\)

\(=4 a^{2} b^{2} c^{2}\)

So, the right answer is \(4 \mathrm{a}^{2} \mathrm{~b}^{2} \mathrm{c}^{2}\)

Hence, the correct option is (B).

We have \(\left[\begin{array}{ccc}1 & 2 & 3 \\ 4 & 5 & 6 \\ 5 & 8 & 9\end{array}\right]\)

Eliminate row and column containing 4 we get:

\(\left[\begin{array}{ll}2 & 3 \\ 8 & 9\end{array}\right]\)

Now, determinant \(=2 \times 9-(8 \times 3)\)

\(=-6\)

Now from sign convention of \(3 \times 3\), matrix

For element 4, sign is -ve:   \(\quad \quad\quad\left(\because\left[\begin{array}{lll}+ & - & + \\ - & + & - \\ + & - & +\end{array}\right]\right)\)

\(\therefore\) Cofactor of element 4 is 6.

Hence, the correct option is (C).

\(\mathrm{A}=\left[\begin{array}{ll}1 & 3 \\ 4 & 2\end{array}\right]\)

\(\operatorname{det} \mathrm{A}=2-12\)

\(=-10\)

\(\mathrm{~B}=\left[\begin{array}{cc}2 & -1 \\ 1 & 2\end{array}\right]\)

\(\operatorname{det} \mathrm{B}=4-(-1)\)

\(=5\)

\(\mathrm{~B}^{\prime}=\left[\begin{array}{cc}2 & 1 \\ -1 & 2\end{array}\right]\)

\(\operatorname{det} \mathrm{B}^{\prime}=4-(-1)\)

\(=5\)

\(\left|\mathrm{ABB}^{\prime}\right|=|\mathrm{A}||\mathrm{B} \|| \mathrm{B}^{\prime} \mid\)

\(=(-10)(5)(5)\)

\(=-250\)

Hence, the correct option is (B).

Given: \(A=\left[\begin{array}{ccc}-1 & 2 & 3 \\ -2 & 3 & 5 \\ 4 & -2 & 1\end{array}\right]\)

Here, we have to find the cofactor matrix for the given matrix A As we know that, cofactor of an element \(\mathrm{a}_{\mathrm{ij}}\) is given by:

\(C_{i j}=\left\{\begin{array}{c}M_{i j}, \text { when } i+j \text { is even } \\ -M_{i j}, \text { when } i+j \text { is odd }\end{array}\right.\)

\(C_{11}=(-1)^{2} \times\left|\begin{array}{cc}3 & 5 \\ -2 & 1\end{array}\right|=13\)

\(C_{12}=(-1)^{3} \times\left|\begin{array}{cc}-2 & 5 \\ 4 & 1\end{array}\right|=22\)

\(C_{13}=(-1)^{4} \times\left|\begin{array}{cc}-2 & 3 \\ 4 & -2\end{array}\right|=-8\)

Similarly, we can say that \(\mathrm{C}_{21}=-8, \mathrm{C}_{22}=-13\) and \(\mathrm{C}_{23}=6\)

Similarly, we can also say that, \(\mathrm{C}_{31}=1, \mathrm{C}_{32}=-1\) and \(\mathrm{C}_{33}=1\)

So, the required cofactor matrix is \(\left[\begin{array}{ccc}13 & 22 & -8 \\ -8 & -13 & 6 \\ 1 & -1 & 1\end{array}\right]\)

Hence, the correct option is (B).

Given determinant is \(\left|\begin{array}{ccc}\mathrm{i} & \mathrm{i}^{2} & \mathrm{i}^{3} \\ \mathrm{i}^{4} & \mathrm{i}^{6} & \mathrm{i}^{8} \\ \mathrm{i}^{9} & \mathrm{i}^{12} & \mathrm{i}^{15}\end{array}\right|\)

Since, we have,

\(\mathrm{i}=\sqrt{-1}\)

\(\mathrm{i}^{2}=-1, \mathrm{i}^{3}=-\mathrm{i}, \mathrm{i}^{4}=1,\mathrm{i}^{6}=-1, \mathrm{i}^{8}=1, \mathrm{i}^{9}=\mathrm{i}, \mathrm{i}^{12}=1 \text {, and } \mathrm{i}^{15}=-\mathrm{i}\)

\(=\left|\begin{array}{ccc}\mathbf{i} & -1 & -\mathrm{i} \\ 1 & -1 & 1 \\ \mathrm{i} & 1 & -\mathrm{i}\end{array}\right|\)

\(=\mathrm{i}(\mathrm{i}-1)+1(-\mathrm{i}-\mathrm{i})-\mathrm{i}(1+\mathrm{i})\)

\(=\mathrm{i}^{2}-\mathrm{i}-2\mathrm{i}-\mathrm{i}-\mathrm{i}^{2}\)

\(=-4 \mathrm{i}\)

Hence, the correct option is (D).

The system of linear non-homogeneous equations in matrix form is written as:

\(A X=B\)

Let, A be a \(3 \times 3\) matrix:

\(A =\left[\begin{array}{lll} a _{11} & a _{12} & a _{13} \\ a _{21} & a _{22} & a _{23} \\ a _{31} & a _{32} & a _{33}\end{array}\right], X =\left[\begin{array}{l} x _{1} \\ x _{2} \\ x _{3}\end{array}\right], B =\left[\begin{array}{l} b _{1} \\ b _{2} \\ b _{3}\end{array}\right]\)

\(a_{11} x_{1}+a_{12} x_{2}+a_{13} x_{3}=b_{1}\)

\(a_{21} x_{1}+a_{22} x_{2}+a_{23} x_{3}=b_{2}\)

\(a_{31} x_{1}+a_{32} x_{2}+a_{33} x_{3}=b_{3}\)

The set of values \(x_{1}, x_{2}, x_{3}\) which satisfies the above equations are called the solutions of the system.

\(\left[\begin{array}{ccc}2 & -1 & 3 \\ 1 & 1 & 1 \\ 1 & -1 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}9 \\ 6 \\ 2\end{array}\right]\)

\(2 x-y+3 z=9 \quad \quad \ldots\)(1)

\(x+y+z=6 \quad \quad \ldots\)(2)

\(x-y+z=2 \quad \quad \ldots\)(3)

On solving equation (1), (2), and (3) using the elimination method we will get the values 

\(x = 1, y = 2, z = 3\)

Hence, the correct option is (A).

Let \({r}\) be the common ratio of the given GP

Now,

\(\left|\begin{array}{lll}\ln a_{1} & \ln a_{2} & \ln a_{3} \\ \ln a_{4} & \ln a_{5} & \ln a_{6} \\ \ln a_{7} & \ln a_{8} & \ln a_{9}\end{array}\right|\)

Apply, \(\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1}\) and \(\mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{1}\)

\(\begin{aligned}&=\left|\begin{array}{lll}\ln a_{1} & \ln a_{2}-\ln a_{1} & \ln a_{3}-\ln a_{2} \\ \ln a_{4} & \ln a_{5}-\ln a_{4} & \ln a_{6}-\ln a_{5} \\ \ln a_{7} & \ln a_{8}-\ln a_{7} & \ln a_{9}-\ln a_{8}\end{array}\right| \\&=\left|\begin{array}{lll}\ln a_{1} & \ln r & \ln r \\ \ln a_{4} & \ln r & \ln r \\ \ln a_{7} & \ln r & \ln r\end{array}\right| \quad\left[\text { Since, } \frac{a_{p+1}}{a_{p}}=r \text { and } \log a_{p+1}-\log a_{p}=\log \frac{a_{p+1}}{a_{p}}=\log \right]\end{aligned}\)

\(=0\qquad  \qquad{\left[\text { Since, } \text{C}_{3}=\text{C}_{2}\right]}\)

Hence, the correct option is (A).

Given that,

\(A=\left[\begin{array}{ll}x & 2 \\ 4 & x\end{array}\right]\) and \(\operatorname{det}\left\{A^{2}\right\}=64\)

\(\Rightarrow|\mathrm{A}|={x}^{2}-8 \quad \quad \ldots(1)\)

Given \(\left|\mathrm{A}^{2}\right|=64\)

\(\Rightarrow|\mathrm{A}|^{2}=64 \quad \quad\) \(\left[\because\left|\mathrm{A}^{\mathrm{n}}\right|=|\mathrm{A}|^{\mathrm{n}}\right]\)

\(\Rightarrow|\mathrm{A}|=(64)^{\frac{1}{2}}=8\quad \quad \ldots(2)\)

From equation (1) and (2),

\(x^{2}-8=8\)

\(\Rightarrow x^{2}=16\)

\(\Rightarrow x=\pm 4\)

Hence, the correct option is (C).

Let, A is the second-order matrix,

As we know, AA^-1 = I

Taking determinants both sides, we get:

det (AA^-1) = det I

det(A^-1) × det (A) = 1              [∵ det (AB) = det A × det B]

∴ det(A^-1) = 1det(A)=1|A|" role="presentation">\(\frac{1}{\operatorname{det}(\mathrm{A})}\)=\(\frac{1}{|\mathrm{~A}|}\)

Hence, the correct option is (B).

Given vertices are A (1, 1) ,B ( 6, 0) and C ( 3, 2).

We know that area of triangle ABC is given by,

\(\Delta=\frac{1}{2}\left|\begin{array}{lll}{x}_{1} & {y}_{1} & 1 \\ {x}_{2} & {y}_{2} & 1 \\ {x}_{3} & {y}_{3} & 1\end{array}\right|\)

\(\Rightarrow \Delta=\frac{1}{2}\left|\begin{array}{ccc}1 & 1 & 1 \\ 6 & 0 & 1 \\ 3 & 2 & 1\end{array}\right|\)

\(\Rightarrow \Delta=\frac{1}{2}[1(0-2)-1(6-3)+1(12-0)]\)

\(\Rightarrow \Delta=\frac{7}{2}\)

Hence, the correct option is (A).

Given: \(A=\left[\begin{array}{ccc}2 & -3 & 4 \\ 1 & 0 & -1 \\ 3 & 1 & -2\end{array}\right]\)

Here, we have to find the value of \(M_{13} \times C_{21}+C_{33} \times C_{31}\) for the given matrix

\(M_{13}=\left|\begin{array}{cc}1 & 0 \\ 3 & 1\end{array}\right|=1-0=1\)

\(C_{21}=(-1)^{3} \times\left|\begin{array}{cc}-3 & 4 \\ 1 & -2\end{array}\right|=-1 \times(6-4)=-2\)

\(C_{31}=(-1)^{4} \times\left|\begin{array}{cc}-3 & 4 \\ 0 & -1\end{array}\right|=1 \times(3-0)=3\)

\(C_{33}=(-1)^{6} \times\left|\begin{array}{cc}2 & -3 \\ 1 & 0\end{array}\right|=1 \times(0+3)=3\)

\(M_{13} \times C_{21}+C_{33} \times C_{31}=1 \times(-2)+3 \times 3=7\)

Hence, the correct option is (A).

If the system of equation \(a_{1} x+b_{1} y+c_{1}=0, a_{2} x+b_{2} y+c_{2}=0, a_{3} x+b_{3} y+c_{3}=\) 0 be consistent, then:

\(\left|\begin{array}{lll}a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3}\end{array}\right|=0\)

Given, the system of equation \(2 x+3 y+5=0, x+k y+5=0, k x-12 y-14=0\) be consistent.

\(\left|\begin{array}{ccc}2 & 3 & 5 \\ 1 & k & 5 \\ k & -12 & -14\end{array}\right|=0\)

\(2(-14 k+60)-3(-14-5 k)+5\left(-12-k^{2}\right)=0\)

\(\Rightarrow-28 k+120+42+15 k-60-5 k^{2}=0\)

\(\Rightarrow-5 k^{2}-13 k+102=0\)

\(\Rightarrow 5 k^{2}+13 k-102=0\)

\(\Rightarrow 5 k^{2}-30 k+17 k-102=0\)

\(\Rightarrow 5 k(k-6)+17(k-6)=0\)

\(\Rightarrow(k-6)(5 k+17)=0\)

\(\Rightarrow k=6, \frac{-17}{5}\)

Thus, if the system of equation \(2 x+3 y+5=0, x+k y+5=0, k x-12 y-14=0\) be consistent, then the values of \(k\) are \(6, \frac{-17}{5}\)

Hence, the correct option is (A).

Let, A be any square matrix.

(adj A^T) = |A^T| (A^T) ^-1

(adj A)^T = (|A| A^-1)^T

(adj A^T) = (adj A)^T

Given, A is a square matrix.

Consider, (adj A^T) - (adj A)^T

= |A^T| (A^T)^-1 - (|A| A^-1)^T

= |A|^T (A^T)^-1 - (|A|^T(A^-1)^T

= |A|^T (A^-1)^T - (|A| ^T(A^-1)^T

= 0

= Null Matrix

Thus, If A is a square matrix, then what is adj A^T - (adj A)^T equal to null matrix.

Hence, the correct option is (C).

The system of equations \(A X=0\) is said to be homogenous system of equations, then If \(|A| \neq 0\), then its solution \(X=0\), is called trivial solution.

If \(|A|=0\). Then \(A X=0\) has a non-trivial solution which means the system will have infinitely many solutions.

Given: \(2 x-3 y=0\) and \(2 x+a y=0\)

These equations can be written as: \(A X=B\) where,

\(A=\left[\begin{array}{cc}2 & -3 \\ 2 & \alpha\end{array}\right], X=\left[\begin{array}{l}x \\ y\end{array}\right]\) and \(B=\left[\begin{array}{l}0 \\ 0\end{array}\right]\)

As we know that, the given system is a homogenous system of equation. So, in order to say that the system has infinitely many solutions: \(|A|=0\).

\(| A |=2 \alpha+6=0\)

\(\Rightarrow \alpha=-3\)

Hence, the correct option is (B).

If the \(1, \omega\) and \(\omega^{2}\) are the cube roots of unity, then \(1+\omega+\omega^{2}=0\)

If any row and column have all the elements as zero in the square matrix, then the determinant is zero.

\(\mathrm{D}=\left|\begin{array}{ccc}1 & \omega & \omega^{2} \\ \omega & \omega^{2} & 1 \\ \omega^{2} & 1 & \omega\end{array}\right|\)

\(\mathrm{R}_{3}=\mathrm{R}_{3}+\mathrm{R}_{1}+\mathrm{R}_{2}\)

\(\mathrm{D}=\left|\begin{array}{ccc}1 & \omega & 1+\omega+\omega^{2} \\ \omega & \omega^{2} & 1+\omega+\omega^{2} \\ \omega^{2} & 1 & 1+\omega+\omega^{2}\end{array}\right|\)

\(\mathrm{D}=\left|\begin{array}{ccc}1 & \omega & 0 \\ \omega & \omega^{2} & 0 \\ \omega^{2} & 1 & 0\end{array}\right|=0\)

Hence, the correct option is (C).

\(A=\left[\begin{array}{lll}3 & 7 & 1 \\ 2 & 1 & 8 \\ 4 & 5 & 0\end{array}\right]\)

\(\operatorname{Cof}\left(a_{11}\right)=0-40=-40\)

\(\operatorname{Cof}\left(a_{12}\right)=-(0-32)=32\)

\(\operatorname{Cof}\left(a_{13}\right)=10-4=6\)

\(\operatorname{Cof}\left(a_{21}\right)=-(0-5)=5\)

\(\operatorname{Cof}\left(a_{22}\right)=0-4=-4\)

\(\operatorname{Cof}\left(a_{23}\right)=-(15-28)=13\)

\(\operatorname{Cof}\left(a_{31}\right)=56-1=55\)

\(\operatorname{Cof}\left(a_{32}\right)=-(24-2)=-22\)

\(\operatorname{Cof}\left(a_{33}\right)=3-14=-11\)

\(\operatorname{cof} A=\left[\begin{array}{ccc}-40 & 32 & 6 \\ 5 & -4 & 13 \\ 55 & -22 & -11\end{array}\right]\)

\(\operatorname{Adj} A=(\operatorname{cof} A)^{\top}=\left[\begin{array}{ccc}-40 & 32 & 6 \\ 5 & -4 & 13 \\ 55 & -22 & -11\end{array}\right]^{T}\)

\(\Rightarrow \operatorname{Adj} A=\left[\begin{array}{ccc}-40 & 5 & 55 \\ 32 & -4 & -22 \\ 6 & 13 & -11\end{array}\right]\)

Hence, the correct option is (D).

\(A=\left[\begin{array}{ccc}8 & 7 & 0 \\ 6 & 5 & 4 \\ 3 & 2 & 1\end{array}\right]\)

\(\Rightarrow|A|=\left|\begin{array}{lll}8 & 7 & 0 \\ 6 & 5 & 4 \\ 3 & 2 & 1\end{array}\right|\)

\(\Rightarrow|\mathrm{A}|=8(5-8)-7(6-12)+0(12-15)\)

\(\Rightarrow|\mathrm{A}|=-24+42+0=18\)

Therefore, from the properties of determinants \(\left|A^{-1}\right|=\frac{1}{|A|}=\frac{1}{18}\)

Hence, the correct option is (B).

Given: A = 3B and |B| = 1

To find: |A|

We know that for n order matrix, |A| = kⁿ|B|. If A = kB

A = 3B

Taking determinants both sides, we get:

|A| = |3B|

So, |A| = 3²|B|

⇒ |A| = 9

Hence, the correct option is (B).

Given: the matrix \(A\) is a non singular matrix of order 3.

Since, \(A\) is non-singular \(A^{-1}\) exist.

\(A ^{-1}=\frac{1}{| A |} \cdot \operatorname{adj} \cdot A \quad \quad \ldots \)(1)

Pre-multiply equation (1) by A,

\(AA ^{-1}=\frac{ A }{| A |} \cdot adj \cdot A\)

\(\Rightarrow| A \mid= A \cdot( adj \cdot A)\)

\(\Rightarrow A \cdot( adj \cdot A )=| A | I\quad \quad \ldots \)(2)

Now, post multiply equation (1) by A,

\(A^{-1} A=\frac{1}{|A|}(a d j \cdot A) A\)

\(\Rightarrow \| A \mid=(\operatorname{adj} \cdot A) A\)

\(\Rightarrow(\operatorname{adj} \cdot A) A=|A| I\quad \quad \ldots \)(3)

From (2) and (3), we have:

The statement \(A (\operatorname{adj} A )=(\operatorname{adj} A)A\) is correct.

Also, A is a non-singular matrix.

Thus, \(A ^{-1}\) exist.

\(AA ^{-1}= I\)

\(A \left(\frac{1}{| A |} \cdot adj \cdot A \right)= I\)

\(A (\operatorname{adj} A )=| A | I\)

Take determinants Both side:

\(| A (\operatorname{adj} A )|=| A\mid I\)

We know that if \(A\) be an \(n\)-rowed square matrix and '\(k\)' be any scalar then| \(KA \mid=\) \(K ^{n}| A |\)

Now,

\(|A||\operatorname{adj} A|=|A|^{n}\left|I_{n}\right|\)

\(|A||\operatorname{adj} A|=|A|^{n}\)

\(|\operatorname{adj} A|=|A|^{n-1}\)

The statement \(|\operatorname{adj} A |=| A |\) is not correct.

Hence, the correct option is (A).

\(\left|\begin{array}{ccc}1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z\end{array}\right|=0\)

\(R_{1}=R_{1}-R_{2}\)

\(\left|\begin{array}{ccc}x & -y & 0 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z\end{array}\right|=0\)

\(R_{2}=R_{2}-R_{3}\)

\(\left|\begin{array}{ccc}x & -y & 0 \\ 0 & y & -z \\ 1 & 1 & 1+z\end{array}\right|=0\)

Now, Expanding along \(R_{1}\), we get:

\(x[y(1+z)-(-z)]-(-y)[0-(-z)]+0=0\)

\(\Rightarrow x[y+y z+z]+y[z]=0\)

\(\Rightarrow x y+x y z+x z+y z=0\)

Dividing by \({xyz}\),

\(\frac{1}{\mathrm{z}}+1+\frac{1}{\mathrm{y}}+\frac{1}{\mathrm{x}}=0\)

\(\Rightarrow {x}^{-1}+{y}^{-1}+{z}^{-1}=-{1}\)

Hence, the correct option is (A).

Area of the triangle \(A=\frac{1}{2}\left|\begin{array}{lll} x _{1} & y _{1} & 1 \\ x _{2} & y _{2} & 1 \\ x _{3} & y _{3} & 1\end{array}\right|\)

\(\Rightarrow A =\frac{1}{2}\left|\begin{array}{ccc}1 & -2 & 1 \\ 3 & 1 & 1 \\ 2 & 4 & 1\end{array}\right|\)

\(\Rightarrow 2 A =1(1-4)-(-2)(3-2)+1(12-2)\)

\(\Rightarrow 2 A=-3+2+10\)

\(\Rightarrow A =\frac{9}{2}=4.5\) sq. unit

Hence, the correct option is (B).

Since, the linear equations are inconsistent:

\(D =0\)

\(\left|\begin{array}{ccc}3 & 3 & 3 \\ 3 & 6 & 12 \\ 3 & 12 & 30\end{array}\right|\)

Applying \(C 1= C 1- C 2\)

\(\left|\begin{array}{ccc}0 & 3 & 3 \\ -3 & 6 & 12 \\ -9 & 12 & 30\end{array}\right|\)

Applying C2 \(=C 2-C 3\)

\(\left|\begin{array}{ccc} 0 & 0 & 3 \\ -3 & -6 & 12 \\ -9 & -18 & 30\end{array}\right|\)

Expanding along \(a13\)

\(3\left|\begin{array}{cc}-3 & -6\\-9 & -18\end{array}\right|=54-54=0 \)

\(D_{1}=2 p_{2}-6 p+4=0 \)

\(p=1,2 \)

\(D_{1}=\left|\begin{array}{ccc}3 & 3 & 3 \\3 p & 6 & 12 \\3 p^{2} & 12 & 30\end{array}\right| \)

\(p=1,2 \)

\(D_{2}=\left|\begin{array}{ccc}3 & 3 & 3 \\3 & 3 p & 12 \\3 & 3 p^{2} & 30\end{array}\right|\)

\(p=1,2\)

\(D_{3}=\left|\begin{array}{ccc}3 & 3 & 3 \\3 & 6 & 3 p \\3 & 12 & 3 p^{2} \end{array}\right|\)

\(p=1,2\)

So,

The solution set of \(p\) is:

\(R -\{1,2\}\)

Hence, the correct option is (B).