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Concept:

Let A is any square matrix, A = \(\begin{vmatrix} a_{11} &a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{vmatrix}\) 

|A| = a11(a22 a33 - a23 a32) - a12(a21 a33 - a23 a31) + a13(a21 a32 - a22 a31)

Calculation:

|A| = \(\begin{vmatrix} 2 & 5 & 8 \\ 6 & 10 & 9 \\ 4 &6& 1 \end{vmatrix}\) 

R₂ = R₂ - R₁

⇒ |A| = \(\begin{vmatrix} 2 & 5 & 8 \\ 4 & 5 & 1 \\ 4 &6& 1 \end{vmatrix}\)

R₂ = R₂ - R₃

⇒ |A| = \(\begin{vmatrix} 2 & 5 & 8 \\ 0 & -1 & 0 \\ 4 &6& 1 \end{vmatrix}\)

Expanding along  R₁

⇒ |A| = 2(-1) - 5(0) + 8(-(-4))

⇒ |A| = -2 + 32

⇒ |A| = 30

Any scalar multiplied to a row or a column can be taken out, e.g. 

\(\begin{vmatrix} ka_{11} &ka_{12} & ka_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{vmatrix}\) = k\(\begin{vmatrix} a_{11} &a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{vmatrix}\) = k[a11(a22 a33 - a23 a32) - a12(a21 a33 - a23 a31) + a13(a21 a32 - a22 a31)]

Addition or subtraction in a row or column can be written as:

\(\begin{vmatrix} a_{11}+z &a_{12} & a_{13}\\ a_{21}+y & a_{22} & a_{23}\\ a_{31}+x & a_{32} & a_{33} \end{vmatrix}\) = \(\begin{vmatrix} a_{11} &a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{vmatrix}\)+ \(\begin{vmatrix} z &a_{12} & a_{13}\\ y & a_{22} & a_{23}\\ x & a_{32} & a_{33} \end{vmatrix}\)

Area of a triangle with points (a, b), (x, y) and (p, q) = \({1\over2}\begin{vmatrix} a &b & 1\\ x & y& 1\\ p & q & 1 \end{vmatrix}\)

Value of determinant does not change with the row or column additions or subtractions within themselves.

If Two rows or columns are same or multiple of each other then the value of the determinant is zero.

The determinant can be possible only for a square matrix. 

Calculation:

A = \(\begin{bmatrix} 2 &4 \\ 1 & 5 \end{bmatrix}\)

⇒ |A| = \(\begin{vmatrix} 2 &4 \\ 1 & 5 \end{vmatrix}\)

⇒ |A| = 2 × 5 - 1 × 4

⇒ |A| = 10 - 4 

⇒ |A| = 6

Concept:

Properties of determinants:

• If each entry in any row or column of a determinant is 0, then the value of the determinant is zero.
• If we interchange any two rows (columns) of a matrix, then the determinant is multiplied by -1.
• If any two rows (columns) of a matrix are same then the value of the determinant is zero.

Calculation:

A = \(\begin{bmatrix} x^2 & y^2 & -2xy\\ x & y &0\\1&1&2\end{bmatrix}\) 

|A| = \(\begin{vmatrix} x^2 & y^2 & -2xy\\ x & y &0\\1&1&2\end{vmatrix}\)

c₃ = c₃ - c₁ - c₂

|A| = \(\begin{vmatrix} x^2 & y^2 & -(2xy+x^2+y^2)\\ x & y &0-x-y\\1&1&2-1-1\end{vmatrix}\)

|A| = \(\begin{vmatrix} x^2 & y^2 & -(x+y)^2\\ x & y &-(x+y)\\1&1&0\end{vmatrix}\)

|A| = \(\begin{vmatrix} x^2 & y^2 & 0\\ x & y &0\\1&1&0\end{vmatrix}\)

|A| = 0

Calculation: 

m sinθ + n cosθ = sin θ \(\rm \left[ \begin{array}{cc} 2 &0\\ 0&1\end{array}\right]\) + cos θ \(\rm \left[ \begin{array}{cc} 0 &1\\ -2&0\end{array}\right]\)

= \(\rm \left[ \begin{array}{cc} 2\sin θ &0\\ 0&\sin θ\end{array}\right]\)+\(\rm \left[ \begin{array}{cc} 0 &\cosθ\\ -2\cosθ&0\end{array}\right]\)

= \(\rm \left[\begin{array}{cc} 2\sin θ &\cos θ\\ -2\cos θ&\sin θ\end{array}\right]\)

Now,

|m sinθ + n cosθ| = \(\rm \left| \begin{array}{cc} 2\sin θ &\cos θ\\ -2\cos θ&\sin θ\end{array}\right|\)

= \(2\sin^2\theta \) + \(\rm 2 cos^2\theta \)

= 2(\(\sin ^2 \theta +\cos^2\theta \))

= 2

Hence, option (4) is correct. 

Given:

The height of the triangle decreased = 30%

Formula used:

The area of the triangle = (1/2) × (b × h)     (Where b = The base of the triangle, h = The height of the triangle)

Calculation:

Let us assume the base of the triangle increased by X% and the area of the triangle is A

⇒ The area of the triangle = A = bh/2

⇒ \(b\ = {2A\over h}\)     ----(1)

⇒ When the area of the triangle increase by 5%

⇒ \({A\ \times 1.05}\ = {(h\ \times 0.7)\ \times b\over2}\)

⇒ b = \(b\ =\ {2.1A\over 0.7h}\ =\ {3A\over h}\)     ----(2)

⇒ The increment in the base of the triangle = \({({3A\over h}\ - {2A\over h})\over {2A\over h}} \times 100\) = 50%

∴ The required result will be 50%.

Concept:

Suppose two matrix A and B of order n.

Let A = kB then, |A| = kⁿ|B|.

Calculation:

Given: A = 3B and |B| = 1

To find:|A|

We know that for n order matrix, |A| = kⁿ|B|. If A = kB

A = 3B

Taking determinants both sides, we get

|A| = |3B|

So, |A| = 3²|B|

⇒ |A| = 9

Concept:

Property of determinant of a matrix:

• Let A be a matrix of order n × n then det(kA) = kn det(A)

• If A and B are two square matrices then |AB| = |A||B|

Calculation:

Given: A and B are square matrices of order 2 such that |A| = 2, |B| = 4 

Here, we have to find the value of |2 AB|

As we know that, if A and B are two determinants of order n, then |AB| = |A||B|

⇒ |2 AB| = |2A||B|

As we know that, if A is a matrix of order n, then |kA| = kn |A|, where k ∈ R.

Here n = 2 So, |2A| = 22 ⋅ |A| = 4|A|

⇒ |2 AB| = |2A||B|

= 4|A||B|

= 4 × 2 × 4

= 32

Calculation:

\(\left| {\begin{array}{*{20}{c}} {1 + x}&1&1\\ 1&{1 + y}&1\\ 1&1&{1 + z} \end{array}} \right| = 0\)

R₁ = R₁ - R₂

\(\left| {\begin{array}{*{20}{c}} {x}&-y&0\\ 1&{1 + y}&1\\ 1&1&{1 + z} \end{array}} \right| = 0\)

R₂ = R₂ - R₃

\(\left| {\begin{array}{*{20}{c}} {x}&-y&0\\ 0&{y}&-z\\ 1&1&{1 + z} \end{array}} \right| = 0\)

Now, Expanding along R₁, we get

⇒ x [y(1 + z) - (-z)] - (-y) [0 - (-z)] + 0 = 0

⇒ x [y + yz + z] + y [z] = 0

⇒ xy + xyz + xz + yz = 0

Dividing by xyz

⇒ \(\rm {1\over z}+ 1 +{1\over y}+ {1\over x} = 0\)

⇒ \(\boldsymbol{\rm x^{-1}+ y^{-1}+ z^{-1} = -1}\)

Concept:

If \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} {{{\rm{a}}_{11}}}&{{{\rm{a}}_{12}}}\\ {{{\rm{a}}_{21}}}&{{{\rm{a}}_{22}}} \end{array}} \right]\) then determinant of A is given by:

|A| = a11 × a22 – a21 × a12

Calculation:

Let A = \(\begin{vmatrix} 5 & \rm k\\ 3 & 3 \end{vmatrix}\)

Given that, the value of the determinant is 24.

|A| = 24

⇒ 15 - 3k = 24

⇒15 - 24 = 3k

⇒ 3k = -9

∴ k = -3

Concept:

Properties of Determinant of a Matrix:

• If each entry in any row or column of a determinant is 0, then the value of the determinant is zero.
• For any square matrix say A, |A| = |AT|.
• If we interchange any two rows (columns) of a matrix, the determinant is multiplied by -1.
• If any two rows (columns) of a matrix are same then the value of the determinant is zero.

Calculation:

\(\begin{vmatrix} \rm x-y & \rm y-z & \rm z-x\\ \rm y-z & \rm z-x & \rm x-y \\ \rm z-x & \rm x-y & \rm y-z \end{vmatrix}\)

Apply R1 → R1 + R₂ + R₃, We get

= \(\begin{vmatrix} \rm 0 & \rm 0 & \rm 0\\ \rm y-z & \rm z-x & \rm x-y \\ \rm z-x & \rm x-y & \rm y-z \end{vmatrix}\)

As we can see that the entry of the first row is zero. 

We know that,

If each entry in any row or column of a determinant is 0, then the value of the determinant is zero.

∴ \(\begin{vmatrix} \rm x-y & \rm y-z & \rm z-x\\ \rm y-z & \rm z-x & \rm x-y \\ \rm z-x & \rm x-y & \rm y-z \end{vmatrix}\) = 0

Concept:

Property of determinant of a matrix:

Let A be a matrix of order n × n then det(kA) = kn det(A)

Calculation:

Given: \({\rm{A}} = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} 2&3&1\\ { - 1}&0&2\\ { - 3}&1&2 \end{array}} \right]\)

Here the order of the matrix is 3.

Now,

det(A) = 2(0 - 2) - 3(-2 + 6) + 1(-1 + 0)

= 2(-2) - 3(4) + 1(-1)

= - 4 - 12 - 1

= -17

Now using the property the value of det(2²A) is:

det(2²A) = det(4A) = 43 det(A)

= 64 × -17

= -1088

Concept:

For a 2 x 2 matrix there is a short-cut formula for inverse as given 

\(\rm {\begin{bmatrix} a & & b\\ c& & d \end{bmatrix}}^{-1} = \frac{1}{(ad-bc)}\begin{bmatrix} d & & -b\\ -c& & a \end{bmatrix}\)

Calculation:

As we know that inverse of matrix \(\rm {\begin{bmatrix} a & & b\\ c& & d \end{bmatrix}}^{-1} = \frac{1}{(ad-bc)}\begin{bmatrix} d & & -b\\ -c& & a \end{bmatrix}\)

⇒\({\begin{bmatrix} 2 & 3\\ 4 & 1 \end{bmatrix}}^{-1}= \frac{1}{(2.1-4.3)}\begin{bmatrix} 1 & -3\\ -4 & 2 \end{bmatrix}\)

⇒ \({\begin{bmatrix} 2 & 3\\ 4 & 1 \end{bmatrix}}^{-1}= \frac{1}{-10}\begin{bmatrix} 1 & -3\\ -4 & 2 \end{bmatrix}\)

⇒ \({\begin{bmatrix} 2 & 3\\ 4 & 1 \end{bmatrix}}^{-1}= \frac{1}{10}\begin{bmatrix} -1 & 3\\ 4 & -2 \end{bmatrix}\)

Hence option 4 is the correct answer.

Concept:

Calculation:

Here, det A = 1/9

det |(3A)^-1| = (3^-1)⁵ det (A^-1) .....(∵ Order = 5)

= \(\frac {1}{3^5}\)× 1/detA

= \(\frac {1}{3^5}\)× 9

= 1/3³

= 1/27

Hence, option (2) is correct.

Calculation:

Let Δ = \(\left |\begin{array}{ccc} \rm x+2 & \rm x+3 & \rm x+5 \\ \rm x+4 &\rm x+6 &\rm x+9 \\ \rm x+8 &\rm x+11 &\rm x+15 \end{array}\right|\)

\(\begin{aligned} \\ &\text { By applying } \rm C_{2} \rightarrow C_{2}-C_{1}, C_{3} \rightarrow C_{3}-C_{1}\\ &=\left|\begin{array}{ccc} \rm x+1 & 1 & 3 \\ \rm x+3 & 2 & 5 \\ \rm x+7 & 3 & 7 \end{array}\right|\\ & \end{aligned}\)

\(\text { By applying } \rm R_{2} \rightarrow R_{2}-R_{1}, \text { and } R_{3} \rightarrow R_{3}-R_{2}\\ \)

\(= \left|\begin{array}{ccc} \rm x+1 & 1 & 3 \\ 2 & 1 & 2 \\ 4 & 1 & 2 \end{array}\right|\)

= (x + 1) (0) - 1 (4 - 8) + 3(2 - 4)

= 4 - 6

= -2

Hence, option (2) is correct.