Q.Given $A=\begin{bmatrix}1 & 3\\2 & 2\end{bmatrix}$, $I=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$ if $A-\lambda I$ is a singular matrix then ......... (A) $\lambda=0$ (B) $\lambda^2-3\lambda-4=0$ (C) $\lambda^2+3\lambda-4=0$ (D) $\lambda^2-3\lambda-6=0$

A standard property of determinants: $\det A=\det A'$ for any square matrix $A$.

One of the fundamental properties of determinants is that the determinant of a matrix and its transpose are always equal, i.e.

$\det A=\det A'$

This is because expanding $\det A$ along its rows is identical in value to expanding $\det A'$ along its columns (which were the rows of $A$) — cofactor expansion gives the same sum of terms either way. This holds for every square matrix $A$, regardless of order.

Use the identity $\det(A^{-1}) = 1/\det(A)$ and check it against the claim $\det(A^{-1})=\det(A)$.

For an invertible matrix $A$, $A A^{-1} = I$, so taking determinants:

$\det(A)\det(A^{-1}) = \det(I) = 1 \implies \det(A^{-1}) = \dfrac{1}{\det A}$

For this to equal $\det A$, we'd need $\det A = \dfrac{1}{\det A}$, i.e. $(\det A)^2 = 1$, i.e. $\det A = \pm1$. This is not true for every invertible $2\times2$ matrix (e.g. $A = \begin{bmatrix}2&0\\0&1\end{bmatrix}$ has $\det A = 2$ but $\det(A^{-1}) = 1/2\ne 2$). So the statement is false in general.

"Singular" is defined as $|A|=0$.

A square matrix $A$ is called singular if and only if $|A| = 0$ (equivalently, it has no inverse). This is the standard definition, so the statement is true.

Scaling every entry of a $2\times2$ matrix by 5 scales its determinant by $5^2=25$.

For a square matrix $A$ of order $n$ and scalar $k$, the rule is $|kA|=k^n|A|$ (each of the $n$ rows contributes one factor of $k$). Here $A$ is $2\times2$ so $n=2$ and $k=5$:

$|5A|=5^2|A|=25|A|.$

For an $n\times n$ matrix, $|kA| = k^n|A|$.

Here $n=3$, so $|kA| = k^3|A|$.

Multiplying every entry of an $n\times n$ matrix by $K$ scales the determinant by $K^n$.

If $A$ is a square matrix of order $n$, then $|KA| = K^n |A|$, because $K$ can be pulled out of each of the $n$ rows, contributing a factor $K$ each time. For $A$ of order $3\times 3$, $n=3$, so

$|KA| = K^3|A|$

$\det(A^{-1})=\dfrac{1}{\det(A)}$.

For any invertible matrix $A$, $AA^{-1}=I$. Taking determinants of both sides and using the property $\det(PQ)=\det(P)\det(Q)$:

$\det(A)\det(A^{-1})=\det(I)=1.$

Since $A$ is invertible, $\det(A)\ne0$, so we can divide:

$\det(A^{-1})=\dfrac{1}{\det(A)}.$

This holds for a square invertible matrix of any order, including order $2$.