2 Marks Questions

Question 1012 Marks

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

Answer

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

Question 1022 Marks

If $\text{A}=\begin{bmatrix} 1 & -3 \\ 2 & 0 \end{bmatrix},$ write adj A.

Answer

$|\text{A}|=\begin{bmatrix} 1 & -3 \\ 2 & 0 \end{bmatrix}=6\neq0$
A is a non-singular matrix. Therefore, it is invertible.
Let $C_{ij}$ be a cofactor of $a_{ij}$ in A.
The cofactors of element A are given by
$C_{11} = 0$
$C_{12} = -2$
$C_{21} = 3$
$C_{22} = 1$
$\therefore\ \text{adj A}=\text{A}=\begin{bmatrix} 0 & -2 \\ 3 & 1 \end{bmatrix}^\text{T}=\begin{bmatrix} 0 & 3 \\ -2 & 1 \end{bmatrix}$

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Question 1032 Marks

Find the adjoint of the following matrices: $\begin{bmatrix}1 & \frac{\tan\alpha}{2} \\ -\frac{\tan\alpha}{2} & 1 \end{bmatrix}$Verify that (adjoint A) A = |A|I = A (adjoint A) for the above matrices.

Answer

$\text{D}=\begin{bmatrix}1 & \frac{\tan\alpha}{2} \\ -\frac{\tan\alpha}{2} & 1 \end{bmatrix}$
$\text{adjoint D}=\begin{bmatrix}1 & -\frac{\tan\alpha}{2} \\ \frac{\tan\alpha}{2} & 1 \end{bmatrix}$
$(\text{adjoint D)D}=\begin{bmatrix} 1+\tan^2\frac{\alpha}{2} & 0 \\ 0 &1+\tan^2\frac{\alpha}{2} \end{bmatrix}$
$|\text{D}|=1+\tan^2\frac{\alpha}{2}$
$|\text{D}|\text{I}=\begin{bmatrix} 1+\tan^2\frac{\alpha}{2} & 0 \\ 0 &1+\tan^2\frac{\alpha}{2} \end{bmatrix}$

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Question 1042 Marks

If $\text{A}=\begin{bmatrix}2&4\\4&3\end{bmatrix},\text{X}=\begin{bmatrix}\text{n}\\1\end{bmatrix},\text{B}=\begin{bmatrix}8\\11\end{bmatrix}$and AX = B, then find n.

Answer

Here,
$\begin{bmatrix}2&4\\4&3\end{bmatrix}\begin{bmatrix}\text{n}\\1\end{bmatrix}=\begin{bmatrix}8\\11\end{bmatrix}$
$\Rightarrow\begin{bmatrix}2\text{n}+4\\4\text{n}+3\end{bmatrix}=\begin{bmatrix}8\\11\end{bmatrix}$
$\Rightarrow2\text{n}+4=8$
$\Rightarrow2\text{n}=4$
$\Rightarrow\text{n}=2$

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Question 1052 Marks

If $C_{ij}$ is the cofactor of the element $a_{ij}$ of the matrix $\text{A}=\begin{bmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{bmatrix},$ then write the value of $a_{32}C_{32}$.

Answer

In the given matrix $\text{A}=\begin{bmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{bmatrix}$
$C_{32} = (-1)^{3+2} (8 - 30) = 22$
Therefore, $a_{32}C_{32} = 5 \times 22 = 110$.
Hence, the value of $a_{32}C_{32}$ is $110$.

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Question 1062 Marks

$A, B, C$ are three non-null square matrices of the same order, write the condition on A such that $AB = AC ⇒ B = C$.

Answer

Consider AB = AC. On multiplying both sides by $A^{-1}$, we get $AA^{-1}B = AA^{-1}$
$\Rightarrow IB = IC$ [Because $AA^{-1} = I$ where I is the identity matrix] ⇒ B = C Therefore, the required condition is A must be invertible or $|\text{A}|\neq0$.

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Question 1072 Marks

Let $A$ be a square matrix such that $A^2 - A + I = 0,$ then write $A^{-1}$ interms of $A.$

Answer

$A^2 - A + I = 0$
$Px-$ multiplying with $A^{-1},$
$(A^{-1} A) - (A^{-1} A) + A^{-1}I = 0$
$IA - I + A^{-1} = 0$
$A^{-1} = I - A$
Hence, $A^{-1} = I - A$

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Question 1082 Marks

Show that the following systems of linear equations is inconsistent:

3x + y = 5,

-6x - 2y = 9

Answer

$\text{D}=\begin{vmatrix}3&1\\-6&-2\end{vmatrix}=-6+6=0$
$\text{D}_1=\begin{vmatrix}5&1\\9&-2\end{vmatrix}=-10-9=-19\neq0$
Since D = 0 but $\text{D}_1\neq0$
Hence the given system of equations is inconsistent.

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Question 1092 Marks

If $\text{A}=\begin{bmatrix}5&3&8\\2&0&1\\1&2&3\end{bmatrix}.$ Write the cofactor of element $a_{32}$.

Answer

Minor of $\text{a}_{32}=\text{M}_{32}=\begin{vmatrix}5&8\\2&1\end{vmatrix}=5-16=-11$
Cofactor of $\text{a}_{\text{n}}=\text{A}_{32}=(-1)^{3+2}\text{M}_{32}=11$
Hence, the cofactor of the elements $a_n$ is $11$.

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Question 1102 Marks

If the points $(a, 0), (0, b)$ and $(1, 1)$ are collinear, prove that $a + b = ab$.

Answer

If the points $(a, 0), (0, b)$ and $(1, 1)$ are collinear, then
$\begin{vmatrix}\text{a}&0&1\\0&\text{b}&1\\1&1&1\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{a}&0&1\\-\text{a}&\text{b}&0\\1&1&1\end{vmatrix}=0$ [Applying $R_2 → R_2 - R_1$]
$\Rightarrow\begin{vmatrix}\text{a}&0&1\\-\text{a}&\text{b}&0\\1-\text{a}&1&0\end{vmatrix}=0$ [Applying $R_3 → R_3 - R_1$]
$\Rightarrow\triangle=\begin{vmatrix}-\text{a}&\text{b}\\1-\text{a}&1\end{vmatrix}=0$
$\Rightarrow-\text{a}-\text{b}(1-\text{a})=0$
$\Rightarrow\text{a}+\text{b}=\text{ab}$

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MATHS 2 Marks Questions