Determinants — Class 12 Mathematics PYQs

If $A$ is a square matrix of order 2 such that $\det (A)=4$, then $\det (4\cdot \text{adj }A)$ is equal to :

(A) $16$
(B) $64$
(C) $256$
(D) $512$

If $A$ is a square matrix of order 3 such that $\det (A)=9$, then $\det (9A^{-1})$ is equal to

(A) $9$
(B) $9^2$
(C) $9^3$
(D) $9^4$

Let $P$ be a skew-symmetric matrix of order 3. If $\det (P)=\alpha$, then $(2025{)}^{\alpha }$ is

(A) $0$
(B) $1$
(C) $2025$
(D) $(2025{)}^3$

If $\left|\begin{array}{ccc}-a& b& c\\ a& -b& c\\ a& b& -c\end{array}\right|=kabc$, then the value of $k$ is :

If $a_{ij}$ and $A_{ij}$ represent the $(ij{)}^{\text{th}}$ element and its cofactor of $\left[\begin{array}{ccc}2& -3& 5\\ 6& 0& 4\\ 1& 5& -7\end{array}\right]$ respectively, then the value of $a_{11}{A}_{21}+a_{12}{A}_{22}+a_{13}{A}_{23}$ is :

$\left|\begin{array}{cc}x+1& x-1\\ x^2+x+1& x^2-x+1\end{array}\right|$ is equal to :

Assertion (A): For matrix $A=\left[\begin{array}{ccc}1& \cos \theta & 1\\ -\cos \theta & 1& \cos \theta \\ -1& -\cos \theta & 1\end{array}\right]$, where $\theta \in [0,2\pi ]$, $|A|\in [2,4]$.

Reason (R): $\cos \theta \in [-1,1]$, for all $\theta \in [0,2\pi ]$.

(a) If $A=\left[\begin{array}{ccc}1& -2& 0\\ 2& -1& -1\\ 0& -2& 1\end{array}\right]$, find $A^{-1}$ and use it to solve the system of equations :

$$x-2y=10$$
$$2x-y-z=8$$
$$-2y+z=7$$

OR

(b) If $A=\left[\begin{array}{ccc}3& 1& 1\\ 1& 2& x\\ -1& a& 2\end{array}\right]$ and $A^{-1}=\left[\begin{array}{ccc}1& -1& 1\\ -8& 7& -5\\ b& y& 3\end{array}\right]$, find $(a+x)-(b+y)$.

(a) If $A=\left[\begin{array}{ccc}1& 2& 1\\ 2& 3& 2\\ 5& 0& 4\end{array}\right]$ and $B^{-1}=\left[\begin{array}{ccc}1& 3& 3\\ 1& 4& 3\\ 1& 3& 4\end{array}\right]$, find $(AB{)}^{-1}$. Also, find $|(AB{)}^{-1}|$.

OR

(b) Given $A=\left[\begin{array}{ccc}1& 1& 1\\ 2& 3& 2\\ 1& 1& 2\end{array}\right]$, find $A^{-1}$. Use it to solve :

$$x+y+z=1$$
$$2x+3y+2z=2$$
$$x+y+2z=4$$

(a) Solve the following system of equations using matrices :

$$\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4$$
$$\frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1$$
$$\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2$$

where $x,y,z\ne 0$.

OR

(b) If $A=\left[\begin{array}{cc}1& \cot x\\ -\cot x& 1\end{array}\right]$, show that $A^{\prime }{A}^{-1}=\left[\begin{array}{cc}-\cos 2x& -\sin 2x\\ \sin 2x& -\cos 2x\end{array}\right]$.