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Calculation:

A² = A × A

⇒ A² = \(\begin{bmatrix} 2& 5 \\ 4 & 3 \end{bmatrix}\) × \(\begin{bmatrix} 2& 5 \\ 4 & 3 \end{bmatrix}\)

⇒ A2 = \(\begin{bmatrix} 24& 25 \\ 20 & 29 \end{bmatrix}\)

Given A satisfy the equation 

A2 - kA - 14I = 0

⇒ \(\begin{bmatrix} 24& 25 \\ 20 & 29 \end{bmatrix}\) - k \(\begin{bmatrix} 2& 5 \\ 4 & 3 \end{bmatrix}\) - 14 \(\begin{bmatrix} 1& 0 \\ 0 & 1 \end{bmatrix}\) = \(\begin{bmatrix} 0& 0 \\ 0 & 0 \end{bmatrix}\)

⇒ \(\begin{bmatrix} 24-2k-14&25-5k\\20-4k&29-3k-14\end{bmatrix}\) = \(\begin{bmatrix} 0& 0 \\ 0 & 0 \end{bmatrix}\)

⇒ 25 - 5k = 0

⇒ k = 5

Calculation:

Given A satisfies the equation 

A2 + 3A - 10I = 0

Multiply the equation by A^-1

⇒ A^-1A² + 3 A^-1A - 10A^-1I = 0

⇒ A + 3I - 10 A^-1 = 0

⇒ \(\begin{bmatrix} -1& 3 \\ 4 & -2 \end{bmatrix}\) + 3 \(\begin{bmatrix} 1&0 \\ 0 & 1 \end{bmatrix}\) - 10 A^-1 = \(\begin{bmatrix} 0& 0 \\ 0 & 0 \end{bmatrix}\)

⇒ \(\begin{bmatrix} -1+3&3+0\\4+0&-2+3\end{bmatrix}\) = 10 A^-1

⇒ A^-1 = \({1\over10}\begin{bmatrix} 2& 3 \\ 4 & 1 \end{bmatrix}\)

Concept:

If A is not an invertible matrix then |A| = 0

If A is a singular matrix then |A| = 0

Calculation:

Given A =  \(\begin{bmatrix} 1 & -4 & 3 \\\ 0 & 6 & -7 \\\ 2 & 4 & λ \end{bmatrix}\) is not an invertible matrix.

⇒ A is a singular matrix.

⇒ |A| = 0

⇒ \(\begin{vmatrix} 1 & -4 & 3 \\\ 0 & 6 & -7 \\\ 2 & 4 & λ \end{vmatrix}\) = 0

⇒ 1(\(\rm 6\lambda + 28 \)) + 4 (0 + 14) + 3 (0 - 12) = 0

⇒  \(\rm 6\lambda + 28 + 56 - 36 = 0\)

⇒ \(\rm 6\lambda + 48 = 0\)

⇒ \(\rm \lambda = - 8\)

Hence, if \(\begin{bmatrix} 1 & -4 & 3 \\\ 0 & 6 & -7 \\\ 2 & 4 & λ \end{bmatrix}\) is not an invertible matrix, then the value of λ = - 8

Concept:

If the matrix A is not an invertible matrix then | A | = 0

If the matrix A is the non-singular matrix then | A | ≠ 0 

Calculations:

Given, A = \(\begin{bmatrix} 1 & -3 & 2 \\\ 2 & -8 & 5 \\\ 4 & 2 & λ \end{bmatrix}\)is not an invertible matrix

As we know, If the matrix A is non invertible matrix then | A | = 0

⇒ \(\begin{vmatrix} 1 & -3 & 2 \\\ 2 & -8 & 5 \\\ 4 & 2 & λ \end{vmatrix}\) = 0

⇒ 1\(\rm (-8\lambda - 10)+3(2\lambda-20)+2(4+32)\) = 0

⇒ \(\rm -8\lambda - 10+6\lambda-60+72 = 0\)

⇒\(\rm -2\lambda +2 = 0\)

⇒\(\rm \lambda = 1\)

Hence, If \(\begin{bmatrix} 1 & -3 & 2 \\\ 2 & -8 & 5 \\\ 4 & 2 & λ \end{bmatrix}\) is not an invertible matrix, then the value of λ is 1.

Explanation:

\(Δ = \begin{vmatrix} Ax & x^2 & 1 \\\ By & y^2 & 1 \\\ Cz & z^2 & 1 \end{vmatrix}\)  and  \(Δ_1 = \begin{vmatrix} A & B & C \\\ x & y & z \\\ zy & zx & xy \end{vmatrix}\)

Calculating the determinant of Δ₁ :

\(Δ_1 = \begin{vmatrix} A & B & C \\\ x & y & z \\\ zy & zx & xy \end{vmatrix}\)

Δ₁ = A × (x × y² - x × z²) - B × (x² × y - z² × y) + C × (x² × z - y² × z)

     = A × x (y² - z²) - B × y (x² - z²) + C × z (x² - y²).............(1)

Calculating the determinant of Δ about first column:

\(Δ = \begin{vmatrix} Ax & x^2 & 1 \\\ By & y^2 & 1 \\\ Cz & z^2 & 1 \end{vmatrix}\)

Δ = A × x (y2 - z2) - B × y (x2 - z2) + C × z (x2 - y2)............(2)

If we subtract (1) and (2) 

Δ₁ - Δ = 0

Hence option 3 is the correct answer.

The value of n will be 5

Explanation:

We know that I x A = A

A² = I - A
A³ = A(I - A) = IA - A² = A - (I - A) = 2A - I 
A⁴ = A x A³ = A(2A - I) = 2A² -A = 2(I - A) - A = 2I - 3A
A⁵ = A(2I - 3A) = 2A - 3A² = 2A - 3(I-A) = 5A - 3I

Thus, A⁵ = 5A - 3I

Concept:

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

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

Calculation:

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

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

⇒ \({\begin{bmatrix} 2+3i & -i\\ i & 2-3i \end{bmatrix}}^{-1}= \frac{1}{4+9-1}\begin{bmatrix} 2-3i & i\\ -i & 2+3i \end{bmatrix}\)

⇒ \({\begin{bmatrix} 2+3i & -i\\ i & 2-3i \end{bmatrix}}^{-1}= \frac{1}{12}\begin{bmatrix} 2-3i & i\\ -i & 2+3i \end{bmatrix}\)

Hence, option 2 is correct.

Concept:

A matrix is singular if and only if its determinant is zero.

Calculation:

If A′ = A; where A′ is the transpose of A then |A| = |A'|

But it is not necessary that |A| = 0, so A is not a  singular matrix.

Hence, Statement 1 is wrong.

Given, A3 = I

Taking determinants both sides, we get

⇒|A3 | = |I| = 1

⇒ |A| = 1

Here, |A|≠ 0 so, A is a non-singular matrix

Hence, option (2) is correct.  

Concept:

Symmetric matrix is a square matrix that is equal to its transpose.

i.e. A = AT.

A skew-symmetric (or antisymmetric or antimetric) matrix is a square matrix whose transpose equals its negative.

i.e.  AT = - A

(A + B)T = (A)T  + (B)T 

(AT)T  = A

Calculations:

Given, A is any square matrix and AT is transpose matrix A

Consider the statement "A + AT is always symmetric."

We know that "a symmetric matrix is a square matrix that is equal to its transpose."

i.e. A = AT.

Consider, the matrix A + AT ....(1)

Taking transpose of the matrix A + AT 

(A + AT)^T = (A)^T  + (AT)T 

⇒(A + AT)^T = (A)T + A

⇒(A + AT)T = A + (A)T 

⇒ A + AT is symmetric.

Hence, the statement "A + AT is always symmetric is true.

Consider the statement "A - AT is always anti-symmetric" 

A skew-symmetric (or antisymmetric or antimetric) matrix is a square matrix whose transpose equals its negative.

i.e.  AT = - A

Consider, the matrix A - AT ....(1)

Taking transpose of the matrix A - AT 

(A - AT)T = (A)T  - (AT)T 

⇒(A - AT)T = (A)T - A

⇒(A - AT)T = - (A - AT)

⇒ A - AT is symmetric.

Hence, the statement "A - AT is always symmetric is true.

Hence, the following statements in respect of a square matrix A and its transpose AT :

1. A + AT is always symmetric.

2. A - AT is always anti-symmetric.

both are correct

Concept:

We have various matrix identities as,

• For symmetric matrices, A = A' and B = B'
• For skew-symmetric matrices, A = - A'
• (A ± B)' = A' ± B'
• (AB)' = B'A'

Calculation:

A and B are symmetric matrices

As we know that for symmetric matrices, we have A = A' and B = B'

(AB' - BA') = AB - BA

∵ (A ± B)' = A' ± B'

(AB - BA)' = (AB)' - (BA)'

∵ (AB)' = B'A'

⇒ (AB - BA)' = B'A' - A'B'

∵ A = A' and B = B

⇒ (AB - BA)' = BA - AB

⇒ (AB - BA)' = - (AB - BA)

Since for skew-symmetric matrices, A = - A'

Hence (AB' - BA') is Skew symmetric matrix.

Concept:

If matrix A = [a_ij]

The adj A matrix is the transpose of the matrix [Aij] is the cofactor of the element aij.

Calculation:

A = \(\rm \begin{bmatrix} 3& 7& 1 \\ 2& 1& 8\\ 4& 5& 0\end{bmatrix}\) 

Cof(a₁₁) = 0 - 40 = -40

Cof(a12) = -(0 - 32) = 32

Cof(a13) = 10 - 4 = 6

Cof(a₂₁) = -(0 - 5) = 5

Cof(a22) = 0 - 4 = -4

Cof(a23) = -(15 - 28) = 13

Cof(a31) = 56 - 1 = 55

Cof(a32) = -(24 - 2) = -22

Cof(a33) = 3 - 14 = -11

cof A = \(\rm \begin{bmatrix} -40& 32& 6 \\ 5& -4& 13\\ 55& -22& -11\end{bmatrix}\)

Adj A = (cof A)^T = \(\rm \begin{bmatrix} -40& 32& 6 \\ 5& -4& 13\\ 55& -22& -11\end{bmatrix}^T\)

⇒ Adj A = \(\rm \begin{bmatrix} -40& 5& 55\\ 32& -4& -22\\ 6& 13& -11\end{bmatrix}\)

 \(A = \;\left[ {\begin{array}{*{20}{c}} 1&0&0&0&1\\ 0&1&1&1&0\\ 0&1&1&1&0\\ 0&1&1&1&0\\ 1&0&0&0&1 \end{array}} \right]\)  

\(\left| {\rm{A}} \right| = {\rm{\;}}\left| {\begin{array}{*{20}{c}} 1&0&0&0&1\\ 0&1&1&1&0\\ 0&1&1&1&0\\ 0&1&1&1&0\\ 1&0&0&0&1 \end{array}} \right|\)

R5 → R5 – R1

R3 → R3 – R2

R4 → R4 – R2

\(\left| {\rm{A}} \right| = {\rm{\;}}\left| {\begin{array}{*{20}{c}} 1&0&0&0&1\\ 0&1&1&1&0\\ 0&0&0&0&0\\ 0&0&0&0&0\\ 0&0&0&0&0 \end{array}} \right|\)

 Rank of the matrix is 2

Concept: 

For any matrix A, If matrix A = [aij]

A-1 = \(\rm \text{(adj A)}\over|A|\)

Calculation:

A = \(\begin{bmatrix} 1& -3 \\ 4& 2 \end{bmatrix}\) = [aij]

|A| = \(\begin{vmatrix} 1& -3 \\ 4& 2 \end{vmatrix}\) 

⇒ |A| = 2 × 1 - (-3) × 4

⇒ |A| = 14

Now, finding cofactor of matrix

a₁₁ = 1, cof(a11) = 2

a₁₂ = -3, cof(a12) = -4

a₂₁ = 4, cof(a21) = 3

a₂₂ = 2, cof(a22) = 1

Cof(A) = \(\begin{bmatrix} 2& -4 \\ 3& 1 \end{bmatrix}\)

adj(A) = [cof(A)]^T 

adj(A) = \(\begin{bmatrix} 2& 3 \\ -4& 1 \end{bmatrix}\)

A^-1 = \(\rm \text{(adj A)}\over|A|\) 

⇒ A^-1 = \(\boldsymbol{{1\over14}\begin{bmatrix} 2& 3 \\ -4& 1 \end{bmatrix}}\)

Calculation:

Here, \(\rm A = \left [ \begin{array}{cc} α & β \\ β & α \end{array}\right ]\)

\(\begin{array}{l} \therefore \rm A^{2}=A=\left[\begin{array}{ll} α & β \\ β & α \end{array}\right]\left[\begin{array}{ll} α & β \\ β & α \end{array}\right] \\ \end{array}\)

\(=\left[\begin{array}{l} α^{2}+β^{2} & 2 α β\\ 2 α β &α^{2}+β^{2} \end{array}\right] \\ \text { Now } \rm A^{2}=I \\ \)

\(\Rightarrow \left[\begin{array}{l} α^{2}+β^{2} & 2 α β \\ 2 α β& α^{2}+β^{2} \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ \\ \Rightarrow α^{2}+β^{2}=1, \quad α β=0\)
So, α = 0, β = 1 OR  α = 1, β = 0 

Hence, option (1) is correct.