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Question 512 Marks

If A is a square matrix such that |A| = 2, write the value of $\big|\text{A}\text{A}^{\text{T}}\big|.$

Answer

In a square matrix, A = A^T. Since they are of same order, AA^T = AA^T.
Given, A = 2
⇒ AA^T= 2² = 4

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Question 522 Marks

Evaluate $\begin{vmatrix}4785&4787\\4789&4791\end{vmatrix}$

Answer

Let $\triangle=\begin{vmatrix}4785&4787\\4789&4791\end{vmatrix}$
$\Rightarrow\triangle=\begin{vmatrix}4785&2\\4789&2\end{vmatrix}$ [Applying C₂ → C₂ - C₁]
$=2\times\begin{vmatrix}4785&1\\4789&1\end{vmatrix}$
$=2\times(4785-4789)$
$=2\times(-4)=-8$
$\Rightarrow\begin{vmatrix}4785&4787\\4789&4791\end{vmatrix}=-8$

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Question 532 Marks

If I₃ denotes identity matrix of order 3 × 3, write the value of its determinant.

Answer

In an identity matrix, all the diagonal elements are 1 and rest of the elements are 0.
Here,
$\text{I}_3=\begin{vmatrix}1&0&0\\0&1&0\\0&0&1\end{vmatrix}$
$\text{I}_3=1\times\begin{vmatrix}1&0\\0&1\end{vmatrix}$ [Expanding along C₁]
$\text{I}_3=1$
$\text{I}_3=1$

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Question 542 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 R₂ → R₂ - R₁]
$\Rightarrow\begin{vmatrix}\text{a}&0&1\\-\text{a}&\text{b}&0\\1-\text{a}&1&0\end{vmatrix}=0$ [Applying R₃ → R₃ - R₁]
$\Rightarrow\triangle=\begin{vmatrix}-\text{a}&\text{b}\\1-\text{a}&1\end{vmatrix}=0$
$\Rightarrow-\text{a}-\text{b}(1-\text{a})=0$
$\Rightarrow\text{a}+\text{b}=\text{ab}$

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Question 552 Marks

Evaluate the following determinant:
$\begin{vmatrix}\text{x}&-7\\\text{x}&5\text{x}+1 \end{vmatrix}$

Answer

Let $\text{A}=\begin{vmatrix}\text{x}&-7\\\text{x}&5\text{x}+1 \end{vmatrix}$
$|\text{A}|=\text{x}(5\text{x}+1)+7\times\text{x}$
$=5\text{x}^2+\text{x}+7\text{x}$
$=5\text{x}^2+8\text{x}$
Hence, $|\text{A}|=5\text{x}^2+8\text{x}$

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Question 562 Marks

If A = [A_ij] is a 3 × 3 diaginal matrix such that a₁₁ = 1, a₂₂ = 2, a₃₃ = 3, then find |A|.

Answer

If A = [A_ij] is a diagonal matrix of order n, then |A| = a₁₁ × a₂₂ × a₃₃ × ...... × a_mn.
Given, a₁₁ - 1, a₂₂ - 2 and a₃₃ - 3
⇒ |A| = 1 × 2 × 3 = 6 [Applying the above property]

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Question 572 Marks

Write the value of $\begin{vmatrix}\sin20^{\circ}&-\cos20^{\circ}\\\sin70^{\circ}&\cos70^{\circ}\end{vmatrix}$

Answer

Let $\triangle=\begin{vmatrix}\sin20^{\circ}&-\cos20^{\circ}\\\sin70^{\circ}&\cos70^{\circ}\end{vmatrix}$
$=\sin20^{\circ}\cos70^{\circ}+\cos20^{\circ}\sin70^{\circ}$
$=\sin(20^{\circ}+70^{\circ})$ [trignometric identity]
$=\sin90^{\circ}$
$=1$

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Question 582 Marks

Find the maximum value of $\begin{vmatrix}1&1&1\\1&1+\sin\theta&1\\1&1&1+\cos\theta \end{vmatrix}$

Answer

Let $\triangle=\begin{vmatrix}1&1&1\\1&1+\sin\theta&1\\1&1&1+\cos\theta \end{vmatrix}$
Applying R₂ → R₂ - R₁ and R₃ → R₃ - R₁ we get
$\triangle=\begin{vmatrix}1&1&1\\0&\sin\theta&0\\0&0&\cos\theta \end{vmatrix}$
$=\sin\theta\cos\theta$
$=\frac{\sin2\theta}{2}$
We know that $-1\leq\sin2\theta\leq1$
$\therefore$ Maximum value of $\triangle=\frac{1}{2}\times1=\frac{1}{2}$

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Question 592 Marks

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

Answer

[Applying R₁ → R₁ - 3R₂]
$=\begin{vmatrix}0&0&0\\81&52&100\\-3&0&4\end{vmatrix}$
$=0$

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Question 602 Marks

If $\text{A}=\begin{bmatrix}5&3&8\\2&0&1\\1&2&3\end{bmatrix}.$ Write the cofactor of element a₃₂.

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 612 Marks

Evaluate the following determinant:
$\begin{vmatrix}1&4&9\\4&9&16\\9&16&25 \end{vmatrix}$

Answer

Let $\triangle$ be the determinant.
$\triangle=\begin{vmatrix}1&4&9\\4&9&16\\9&16&25 \end{vmatrix}$
Applying R₃ → R₃ - R₂, we get
$\Rightarrow\triangle=\begin{vmatrix}1&4&9-4\\4&9&16-9\\9&16&25-16 \end{vmatrix}$
$\Rightarrow\triangle=\begin{vmatrix}1&4&5\\4&9&7\\9&16&9\end{vmatrix}$
$\Rightarrow\triangle=\begin{vmatrix}1&5&5\\4&13&7\\9&25&9 \end{vmatrix}$ [Applying C₂ → C₁ + C₂]
$\Rightarrow\triangle=\begin{vmatrix}1&0&0\\4&-7&-13\\9&-20&-36 \end{vmatrix}$ [Applying C₂ → 5C₁ - C₂ and C₃ → 5C₁ - C₃]
$\Rightarrow\triangle=1(7\times36-13\times20)$
$\Rightarrow\triangle=252-260=-8$

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Question 622 Marks

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

Answer

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

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Question 632 Marks

Write rthe value of the determinant $\begin{vmatrix}\text{p}&\text{p}+1\\\text{p}-1&\text{p}\ \end{vmatrix}$

Answer

$\begin{vmatrix}\text{p}&\text{p}+1\\\text{p}-1&\text{p}\ \end{vmatrix}=\text{p}^2-(\text{p}+1)(\text{p}-1)$
$=\text{p}^2-(\text{p}^2-1)$
$=\text{p}^2-\text{p}^2+1$
$=1$

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Question 642 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 652 Marks

Write the value of the determinant $\begin{vmatrix}2&3&4\\5&6&8\\6\text{x}&9\text{x}&12\text{x}\end{vmatrix}$

Answer

$\begin{vmatrix}2&3&4\\5&6&8\\6\text{x}&9\text{x}&12\text{x}\end{vmatrix}$
$=\begin{vmatrix}2&3&4\\5&6&8\\2&3&4\end{vmatrix}$ [Taking 2x common from R₃]
$=0$

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Question 662 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 C₁ → C₁- 4C₃, 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 R₃ → R₃ - R₂ and R₁ → R₁ + R₂]
$\Rightarrow\triangle=\begin{vmatrix}1&32&35\\0&109&119\\0&11&12\end{vmatrix}$ [Applying R₂ → 3R₁ + R₂]
$\Rightarrow\triangle=1(109\times12-119\times11)$
$\Rightarrow\triangle=-1$

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Question 672 Marks

If |A| = 2, where A is 2 × 2 matrix, find |adj A|.

Answer

For any square matrix A of order n, |adj A| = |A|^n-1
Given, |A| = 2
Here, order is 2
⇒ |adj A| = |2|²^-1 = 2

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Question 682 Marks

Find the value of x, if:
$\begin{vmatrix}3&\text{x}\\\text{x}&1\end{vmatrix}=\begin{vmatrix}3&2\\4&1\end{vmatrix}$

Answer

Given, $\begin{vmatrix}3&\text{x}\\\text{x}&1\end{vmatrix}=\begin{vmatrix}3&2\\4&1\end{vmatrix}$

$\Rightarrow3-\text{x}^2=3-8$

$\Rightarrow-\text{x}^2=-8$

$\Rightarrow\text{x}^2=8$

$\Rightarrow\text{x}=\pm2\sqrt{2}$

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Question 692 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,

$\Rightarrow|\text{A}|+|\text{B}|=1-1=0$

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Question 702 Marks

Find the value of x, if:
$\begin{vmatrix}2&3\\4&5\end{vmatrix}=\begin{vmatrix}\text{x}&3\\2\text{x}&5\end{vmatrix}$

Answer

Given, $\begin{vmatrix}2&3\\4&5\end{vmatrix}=\begin{vmatrix}\text{x}&3\\2\text{x}&5\end{vmatrix}$

$\Rightarrow2\times5-3\times4=\text{x}\times5-3\times2\text{x}$

$\Rightarrow10-12=5\text{x}-6\text{x}$

$\Rightarrow-2=-\text{x}$

$\Rightarrow\text{x}=2$

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