2 Marks Questions

Question 1012 Marks

If A is a singular matrix, then write the value of |A|.

Answer

Since A is a singular matrix
Thus, |A| = 0

Question 1022 Marks

Evaluate the following determinant: $\begin{vmatrix}67&19&21\\39&13&14\\81&24&26 \end{vmatrix}$

Answer

Consider the determinant
$\triangle=\begin{vmatrix}67&19&21\\39&13&14\\81&24&26 \end{vmatrix}$
Applying $C_1 \rightarrow C_{1 }- 4C_3,$ We get,
$\triangle=\begin{vmatrix}4&19&21\\-3&13&14\\-3&24&26 \end{vmatrix}$
$\Rightarrow\triangle=\begin{vmatrix}4&19&21\\-3&13&14\\-3&24&26 \end{vmatrix}$
$\Rightarrow\triangle=\begin{vmatrix}1&32&35\\-3&13&14\\0&11&12\end{vmatrix} [$Applying $R_3 \rightarrow R_3 - R_2$ and $R_1 \rightarrow R_1 + R_2]$
$\Rightarrow\triangle=\begin{vmatrix}1&32&35\\0&109&119\\0&11&12\end{vmatrix} [$Applying $R_2 \rightarrow 3R_1 + R_2]$
$\Rightarrow\triangle=1(109\times12-119\times11)$
$\Rightarrow\triangle=-1$

View full question & answer→

Question 1032 Marks

Find the value of the determinant $\begin{vmatrix}243&156&300\\81&52&100\\-3&0&4\end{vmatrix}$

Answer

$[$Applying $R_1 \rightarrow R_1 - 3R_2]$
$=\begin{vmatrix}0&0&0\\81&52&100\\-3&0&4\end{vmatrix}$
$=0$

View full question & answer→

Question 1042 Marks

If $A = [A_{ij}]$ is $3 \times 3$ diaginal matrix such that $a_{11} = 1, a_{22} = 2, a_{33} = 3,$ then find $|A|.$

Answer

If $A = [A_{ij}]$ is diagonal matrix of order $n,$ then $|A| = a_{11} \times a_{22} \times a_{33} \times ...... \times a_{mn}.$
Given, $a_{11} - 1, a_{22} - 2$ and $a_{33} - 3$
$\Rightarrow |A| = 1 \times 2 \times 3 = 6 [$Applying the above property$]$

View full question & answer→

Question 1052 Marks

Find adjoint of each of the matrix.
$\begin{bmatrix}1&2\\3&4\end{bmatrix}$

Answer

Here $\text{A}=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}=\begin{bmatrix}1&2\\3&4\end{bmatrix}$
$\Rightarrow\ |\text{A}|=\begin{vmatrix}1&2\\3&4\end{vmatrix}$
$\therefore A_{11} =$ Confactor of $a_{11} = (-1)^2 (4) = 4$
$A_{12} =$ Confactor of $a_{12} = (-1)^3 (3) = -3$
$A_{21} =$ Confactor of $a_{21} = (-1)^3 (2) = -2$
$A_{22} =$ Confactor of $a_{22} = (-1)^4 (1) = 1$
$\therefore\ \text{adj.}\ \text{A}=\begin{bmatrix}\text{A}_{11}&\text{A}_{12}\\\text{A}_{21}&\text{A}_{22}\end{bmatrix}=\begin{bmatrix}4&-3\\-2&1\end{bmatrix}=\begin{bmatrix}4&-2\\-3&1\end{bmatrix}$

View full question & answer→

Question 1062 Marks

If $A$ is symmetric matrix, write whether $A^T$ is symmetric or skew$-$symmetric.

Answer

For any symmetric matrix$,A^T = A.$
Hence $,A^T$ is also symmetric.

View full question & answer→

Question 1072 Marks

Evaluate the following determinant:
$\begin{vmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta \end{vmatrix}$

Answer

$\triangle=\cos^2\theta-(-\sin^2\theta)$
$\triangle=\cos^2\theta+\sin^2\theta=1$

View full question & answer→

Question 1082 Marks

Prove that the determinant $\begin{vmatrix}x&\sin\theta&\cos\theta\\-\sin\theta&-x&1\\\cos\theta&1&x\end{vmatrix}$ is independent of θ.

Answer

$\triangle=\begin{vmatrix}x&\sin\theta&\cos\theta\\-\sin\theta&-x&1\\\cos\theta&1&x\end{vmatrix}$
$= x(-x^2-1)-\sin\theta(-x\sin\theta-\cos\theta)+\cos\theta(-\sin\theta+x\cos\theta)$
$=-x^3-x+x\sin^2\theta+\sin\theta\cos\theta-\sin\theta\cos\theta+x\cos^2\theta$
$=-x^3-x+x(\sin^2\theta+\cos^2\theta)$
$=-x^3-x+x$
$=-x^3$ (Independent of $\theta$)
Hence, $\triangle$ is independent of $\theta.$

View full question & answer→

Question 1092 Marks

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 $|\text{A}|+|\text{B}|.$

Answer

$\text{A}=\begin{bmatrix}0&\text{i}\\\text{i}&1\end{bmatrix}$$\Rightarrow|\text{A}|= 0-\text{i}^2$
$\Rightarrow|\text{A}|=-(-1)=1$
Also,
$\text{B}=\begin{bmatrix}0&1\\1&0\end{bmatrix}$
$\Rightarrow|\text{B}|=0-1=-1$
So,
$\Rightarrow|\text{A}|+|\text{B}|=1-1=0$

View full question & answer→

Question 1102 Marks

Evaluate the determinants.

$\begin{vmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{vmatrix}$
$\begin{vmatrix}x^2-x+1&x-1\\x+1&x+1\end{vmatrix}$

Answer

$\begin{vmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{vmatrix}_{\ =\ \left(\cos\theta\right)\left(\cos\theta\right)\ -\ \left(-\sin\theta\right)\left(\sin\theta\right)\ =\ \cos^2\theta\ +\ \sin^2\theta\ =\ 1}$
$\begin{vmatrix}x^2-x+1&x-1\\x+1&x+1\end{vmatrix}$

$=\left(x^2-x+1\right)\left(x+1\right)-\left(x-1\right)\left(x+1\right)$
$=x^3-x^2+x+x^2-x+1-\left(x^2-1\right)$
$=x^3+1-x^2+1$
$=x^3-x^2+2$

View full question & answer→

MATHS 2 Marks Questions