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Find the (adj A), if
A = \(\rm \begin{bmatrix} 3& 7& 1 \\ 2& 1& 8\\ 4& 5& 0\end{bmatrix}\)
1. \(\rm \begin{bmatrix} -40& 32& 55 \\ 5& -4& 13\\ 55& -22& -11\end{bmatrix}\)
2. \(\rm \begin{bmatrix} 3& 7& 1 \\ 2& 1& 8\\ 4& 5& 0\end{bmatrix}\)
3. \(\rm \begin{bmatrix} -40& 32& 6 \\ 5& -4& 13\\ 55& -22& -11\end{bmatrix}\)
4. \(\rm \begin{bmatrix} -40& 5& 55\\ 32& -4& -22\\ 6& 13& -11\end{bmatrix}\)
Correct Answer - Option 4 : \(\rm \begin{bmatrix} -40& 5& 55\\ 32& -4& -22\\ 6& 13& -11\end{bmatrix}\)Concept:If matrix A = [aij]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 \\ 2Read more
Correct Answer – Option 4 : \(\rm \begin{bmatrix} -40& 5& 55\\ 32& -4& -22\\ 6& 13& -11\end{bmatrix}\)
Concept:
If matrix A = [aij]
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 = \(\rm \begin{bmatrix} -40& 5& 55\\ 32& -4& -22\\ 6& 13& -11\end{bmatrix}\)
See lessIf the matrix \(\rm A = \left [ \begin{array}{cc} \alpha & \beta \\ \beta & \alpha \end{array}\right ]\) is such that A2 = I, then which one of the following is correct?
1. α = 0, β = 1 OR α = 1, β = 0
2. α = 1, β = 1 OR α = 1, β = 0
3. α ≠ 1, β = 0
4. α = β = 0
Correct Answer - Option 1 : α = 0, β = 1 OR α = 1, β = 0Calculation: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 \\Read more
Correct Answer – Option 1 : α = 0, β = 1 OR α = 1, β = 0
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.
See lessThe inverse of the matrix \(\rm \begin{bmatrix} 2+3i & -i\\ i & 2-3i \end{bmatrix}\)is
1. \( \frac{1}{12}\begin{bmatrix} 2-3i & i\\ -i & -2+3i \end{bmatrix}\)
2. \( \frac{1}{12}\begin{bmatrix} 2-3i & i\\ -i & 2+3i \end{bmatrix}\)
3. \( \frac{1}{12}\begin{bmatrix} 2+3i & i\\ -i & 2-3i \end{bmatrix}\)
4. \( \frac{1}{12}\begin{bmatrix} 2-3i & -i\\ i & 2+3i \end{bmatrix}\)
Correct Answer - Option 2 : \( \frac{1}{12}\begin{bmatrix} 2-3i & i\\ -i & 2+3i \end{bmatrix}\)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} dRead more
Correct Answer – Option 2 : \( \frac{1}{12}\begin{bmatrix} 2-3i & i\\ -i & 2+3i \end{bmatrix}\)
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.
See lessIf A and B are symmetric matrices of the same order, then (AB’ – BA’) is:
1. Skew symmetric matrix
2. Symmetric matrix
3. Null matrix
4. Identity matrix
Correct Answer - Option 1 : Skew symmetric matrixConcept: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 matricesAs we know that for symmetric matrices, we have A = A'Read more
Correct Answer – Option 1 : Skew symmetric matrix
Concept:
We have various matrix identities as,
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.
See lessLet \(\Delta = \begin{vmatrix} Ax & x^2 & 1 \\\ By & y^2 & 1 \\\ Cz & z^2 & 1 \end{vmatrix}\) and \(\Delta_1 = \begin{vmatrix} A & B & C \\\ x & y & z \\\ zy & zx & xy \end{vmatrix}\), then
1. Δ1 = -Δ
2. Δ1 ≠ Δ
3. Δ1 – Δ = 0
4. Δ1 = Δ = 0
Correct Answer - Option 3 : Δ1 - Δ = 0Explanation:\(Δ = \begin{vmatrix} Ax & x^2 & 1 \\\ By & y^2 & 1 \\\ Cz & z^2 & 1 \end{vmatrix}\) and \(Δ_1 = \begin{vmatrix} A & B & C \\\ x & y & z \\\ zy & zx & xy \end{vmatrix}\)Calculating the determinant of Read more
Correct Answer – Option 3 : Δ1 – Δ = 0
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.
See lessA square matrix A satisfies A2=I-A , where I is the identity matrix. If An =5A – 3I, find the value of n.
The value of n will be 5Explanation:We know that I x A = AA2 = I - AA3 = A(I - A) = IA - A2 = A - (I - A) = 2A - I A4 = A x A3 = A(2A - I) = 2A2 -A = 2(I - A) - A = 2I - 3AA5 = A(2I - 3A) = 2A - 3A2 = 2A - 3(I-A) = 5A - 3IThus, A5 = 5A - 3I
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
See lessConsider 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.
Which of the statements given above is / are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Correct Answer - Option 3 : Both 1 and 2Concept: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 = ACalculations:Read more
Correct Answer – Option 3 : Both 1 and 2
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
See lessIf \(\begin{bmatrix} 1 & -3 & 2 \\\ 2 & -8 & 5 \\\ 4 & 2 & λ \end{bmatrix}\) is not an invertible matrix, then what is the value of λ ?
1. -1
2. 0
3. 1
4. 2
Correct Answer - Option 3 : 1Concept:If the matrix A is not an invertible matrix then | A | = 0If 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 matrRead more
Correct Answer – Option 3 : 1
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.
See lessConsider the following statements:
1. If A′ = A; then A is a singular matrix, where A′ is the transpose of A.
2. If A is a square matrix such that A3 = I, then A is non-singular.
Which of the statements given above is/are correct ?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Correct Answer - Option 2 : 2 only 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 = ITaking determinaRead more
Correct Answer – Option 2 : 2 only
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.
See lessFind the inverse of the matrix \(\begin{bmatrix} 1& -3 \\ 4& 2 \end{bmatrix}\)
1. \({1\over14}\begin{bmatrix} 2& -3 \\ 4& -1 \end{bmatrix}\)
2. \({1\over14}\begin{bmatrix} 2& 3 \\ -4& 1 \end{bmatrix}\)
3. \({1\over14}\begin{bmatrix} 2& -3 \\ -4& 1 \end{bmatrix}\)
4. \({1\over14}\begin{bmatrix} 2& 3 \\ 4& 1 \end{bmatrix}\)
Correct Answer - Option 2 : \({1\over14}\begin{bmatrix} 2& 3 \\ -4& 1 \end{bmatrix}\)Concept: For any matrix A, If matrix A = [aij]A-1 = \(\rm \text{(adj A)}\over|A|\)The adj A matrix is the transpose of the matrix [Aij] is the cofactor of the element aij.Calculation:A = \(\begin{bmatrix} 1Read more
Correct Answer – Option 2 : \({1\over14}\begin{bmatrix} 2& 3 \\ -4& 1 \end{bmatrix}\)
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 = \(\boldsymbol{{1\over14}\begin{bmatrix} 2& 3 \\ -4& 1 \end{bmatrix}}\)
See less