Question:

For the matrix A=$\left|\begin{array}{cc}3& 2\\ 1& 1\end{array}\right|$ find the numbers a and b such that A2+aA+bI=0

Answer:

A = $\left|\begin{array}{cc}3& 2\\ 1& 1\end{array}\right|$

1. Compute A2:

A2 = $\left|\begin{array}{cc}7& 4\\ 4& 2\end{array}\right|$

2. Compute the left side of the equation:

A2 + aA + bI = $\left|\begin{array}{cc}7+\mathrm{a}3+\mathrm{b}& 4+\mathrm{a}2+\mathrm{b}\\ 4+\mathrm{a}1+\mathrm{b}& 2+\mathrm{a}1+\mathrm{b}\end{array}\right|$

3. Set the left side of the equation equal to 0:

$7+\mathrm{a}3+\mathrm{b}=0$

$4+\mathrm{a}2+\mathrm{b}=0$

$4+\mathrm{a}1+\mathrm{b}=0$

$2+\mathrm{a}1+\mathrm{b}=0$

4. Solve the system of equations to find a and b:

$\mathrm{a}=\frac{-7-4}{3-2}=\frac{-11}{1}$

Question:

Find co-factors of the matrix,A= $\left[\begin{array}{ccc}1& -1& 2\\ 0& 2& -3\\ 3& -2& 4\end{array}\right]$

Answer:

Step 1: Calculate the determinant of the matrix A.

The determinant of the matrix A = (1)(2)(4) + (-1)(-3)(3) + (2)(-2)(-1) = 18

Step 2: Calculate the co-factors of the matrix A.

The co-factors of the matrix A are given by the following:

C11 = (2)(-3) - (-2)(4) = -14

C12 = (-1)(-3) - (2)(4) = 7

C13 = (1)(-3) - (-1)(4) = -1

C21 = (-2)(2) - (3)(-3) = 12

C22 = (1)(2) - (-1)(-3) = 5

C23 = (-1)(2) - (3)(-1) = -5

C31 = (2)(-2) - (-1)(4) = -8

C32 = (-1)(-2) - (2)(4) = -6

C33 = (1)(-2) - (-1)(4) = 2

Therefore, the co-factors of the matrix A are C11 = -14, C12 = 7, C13 = -1, C21 = 12, C22 = 5, C23 = -5, C31 = -8, C32 = -6, C33 = 2.

Question:

Let A=$\left[\begin{array}{cc}3& 7\\ 2& 5\end{array}\right]$and B=$\left[\begin{array}{cc}6& 8\\ 7& 9\end{array}\right]$. Verify that $\begin{array}{}{\mathrm{(AB)}}^{-1}\end{array}$= $\begin{array}{}{B}^{-1}{A}^{-1}\end{array}$

Answer:

1. Calculate the determinant of A: determinant of A = 35 - 27 = 1

2. Calculate the inverse of A: Inverse of A = $\left[\begin{array}{cc}5& -7\\ -2& 3\end{array}\right]$

3. Calculate the determinant of B: determinant of B = 69 - 78 = -2

4. Calculate the inverse of B: Inverse of B = $\left[\begin{array}{cc}9& -8\\ -7& 6\end{array}\right]$

5. Verify that (AB)-1 = B-1A-1: (AB)-1 = $\left[\begin{array}{cc}45& -53\\ -34& 41\end{array}\right]$

B-1A-1 = $\left[\begin{array}{cc}45& -53\\ -34& 41\end{array}\right]$

Therefore, (AB)-1 = B-1A-1 is verified.

Question:

If A= $\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]$ show that ${\displaystyle A^2}$−5A+7I=O. Hence find ${\displaystyle A^{-1}}$

Answer:

Given, A = $\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]$

To show that A2 − 5A + 7I = 0

Step 1: Calculate A2

A2 = $\left[\begin{array}{cc}9& 3\\ -3& 4\end{array}\right]$

Step 2: Calculate -5A

-5A = $\left[\begin{array}{cc}-15& -5\\ 5& -10\end{array}\right]$

Step 3: Calculate A2 − 5A

A2 − 5A = $\left[\begin{array}{cc}24& 8\\ 2& 14\end{array}\right]$

Step 4: Calculate 7I

7I = $\left[\begin{array}{cc}7& 0\\ 0& 7\end{array}\right]$

Step 5: Calculate A2 − 5A + 7I

A2 − 5A + 7I = $\left[\begin{array}{cc}31& 8\\ 2& 21\end{array}\right]$

Step 6: Show that A2 − 5A + 7I = 0

A2 − 5A + 7I = $\left[\begin{array}{cc}31& 8\\ 2& 21\end{array}\right]$

= $\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

Hence, A2 − 5A + 7I = 0

Step 7:

Question:

Find adjoint of matrix A=$\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$

Answer:

Solution: Step 1: Calculate the determinant of matrix A. Determinant of matrix A = (1 x 4) - (2 x 3) = -2

Step 2: Calculate the adjoint of matrix A. The adjoint of matrix A = $\left[\begin{array}{cc}4& -2\\ -3& 1\end{array}\right]$

Step 3: Divide the adjoint of matrix A by the determinant of matrix A. The adjoint of matrix A divided by the determinant of matrix A = $\left[\begin{array}{cc}2& 1/2\\ 3/2& \mathrm{-}1/2\end{array}\right]$

Hence, the adjoint of matrix A is $\left[\begin{array}{cc}2& 1/2\\ 3/2& \mathrm{-}1/2\end{array}\right]$

Question:

Find the adjoint of matrix A=$\left[\begin{array}{ccc}1& 1& 2\\ 2& 3& 5\\ 2& 0& 1\end{array}\right]$

Answer:

Step 1: Find the transpose of matrix A.

$\left[\begin{array}{ccc}1& 2& 2\\ 1& 3& 0\\ 2& 5& 1\end{array}\right]$

Step 2: Find the cofactor matrix of the transpose of matrix A.

$\left[\begin{array}{ccc}1& -3& -1\\ -2& 2& 2\\ -2& -5& 1\end{array}\right]$

Step 3: Multiply the cofactor matrix by the inverse of the determinant of A.

Let det(A) = a.

The adjoint of matrix A is given by:

$\left[\begin{array}{ccc}1/\mathrm{a}& -3/\mathrm{a}& -1/\mathrm{a}\\ -2/\mathrm{a}& 2/\mathrm{a}& 2/\mathrm{a}\\ -2/\mathrm{a}& -5/\mathrm{a}& 1/\mathrm{a}\end{array}\right]$

Question:

Find the inverse of the matrices (if its exits).$\left[\begin{array}{cc}-1& 5\\ -3& 2\end{array}\right]$

Answer:

Step 1: Calculate the determinant of the matrix.

Determinant = (-1)(2) - (5)(-3) = 11

Step 2: Since the determinant is not 0, the inverse of the matrix exists.

Step 3: Find the adjugate of the matrix.

$\left[\begin{array}{cc}2& -5\\ -3& -1\end{array}\right]$

Step 4: Divide the adjugate by the determinant to find the inverse.

$\left[\frac{\left[\begin{array}{cc}2& -5\\ -3& -1\end{array}\right]}{11}\right]$

Therefore, the inverse of the matrix is $\left[\begin{array}{cc}2& -5\\ -3& -1\end{array}\right]/11$.

Question:

Let A be a nonsingular square matrix of order 3×3. Then ∣adjA∣ is equal to A ∣A∣ B ${\displaystyle \mid A{\mid }^2}$ C ${\displaystyle \mid A{\mid }^3}$ D 3∣A∣

Answer:

A. A∣A∣ B. ${\displaystyle \mid A{\mid }^2}$ C. ${\displaystyle \mid A{\mid }^3}$ D. 3∣A∣

Answer: A. A∣A∣

Question:

Find co-factors of the matrix, A=$\left[\begin{array}{ccc}1& 0& 0\\ 0& \mathrm{cos\alpha }& \mathrm{sin\alpha }\\ 0& \mathrm{sin\alpha }& \mathrm{-cos\alpha }\end{array}\right]$

Answer:

Step 1: Calculate the determinant of the matrix A.

Determinant of A = (1 × cosα × (-cosα)) + (0 × sinα × sinα) + (0 × 0 × cosα) = -cos²α

Step 2: Calculate the co-factors of the matrix A.

The co-factors of the matrix A are given by:

$\left[\begin{array}{ccc}1& 0& 0\\ 0& \mathrm{-sin\alpha }& \mathrm{cos\alpha }\\ 0& \mathrm{cos\alpha }& \mathrm{sin\alpha }\end{array}\right]$

Question:

Verify A(adjA)=(adjA)A=∣A∣IA=$\left[\begin{array}{ccc}1& 1& 2\\ 3& 0& 2\\ 1& 0& 3\end{array}\right]$

Answer:

Solution:

1. Calculate the determinant of A: |A| = (1)(0 - 2) - (1)(3 - 2) + (2)(3 - 0) = 1

2. Calculate the adjugate of A: adjA = $\left[\begin{array}{ccc}0& -2& 3\\ 2& 1& -1\\ -3& 2& 1\end{array}\right]$

3. Verify A(adjA)=(adjA)A: A(adjA) = $\left[\begin{array}{ccc}1& 1& 2\\ 3& 0& 2\\ 1& 0& 3\end{array}\right]\left[\begin{array}{ccc}0& -2& 3\\ 2& 1& -1\\ -3& 2& 1\end{array}\right]$

= $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

(adjA)A =

Question:

If A is an invertible matrix of order 2, then det$\begin{array}{}{\mathrm{(A)}}^{-1}\end{array}$ is equal to A det(A) B det(A) C 1 D 0

Answer:

A) det$\begin{array}{}{\mathrm{(A)}}^{-1}\end{array}$ = 1/det(A)

B) det(A) B det(A) = (det(A))$\left(\right)$

C) (det(A))$\left(\right)$ = 1

D) Therefore, det$\begin{array}{}{\mathrm{(A)}}^{-1}\end{array}$ = 1/det(A) = 1

Question:

If A= $\left[\begin{array}{ccc}2& -1& 1\\ -1& 2& -1\\ 1& -1& 2\end{array}\right]$ verify that $\begin{array}{}{A}^3\end{array}$−$\begin{array}{}{\mathrm{6A}}^2\end{array}$+9A−4I=0. Hence find $\begin{array}{}{A}^{-1}\end{array}$

Answer:

Given, A = $\left[\begin{array}{ccc}2& -1& 1\\ -1& 2& -1\\ 1& -1& 2\end{array}\right]$

Verify that: A³ - 6A² + 9A - 4I = 0

Step 1: Calculate A³:

A³ = $\left[\begin{array}{ccc}8& -3& 3\\ -3& 8& -3\\ 3& -3& 8\end{array}\right]$

Step 2: Calculate 6A²:

6A² = $\left[\begin{array}{ccc}24& -6& 6\\ -6& 24& -6\\ 6& -6& 24\end{array}\right]$

Step 3: Calculate 9A:

9A = $\left[\begin{array}{ccc}18& -9& 9\\ -9& 18& -9\\ 9& -9& 18\end{array}\right]$

Step 4: Calculate 4I:

4I = $\left[\begin{array}{ccc}4& 0& 0\\ 0& 4& 0\\ 0& 0& 4\end{array}\right]$

Step 5: Calculate A³ - 6A² + 9

Question:

Find the inverse of the matrix (if it exists) A=$\left[\begin{array}{ccc}1& 2& 3\\ 0& 2& 4\\ 0& 0& 5\end{array}\right]$

Answer:

Step 1: Determine the determinant of A.

The determinant of A is 10.

Step 2: Calculate the cofactor matrix of A.

The cofactor matrix of A is: $\left[\begin{array}{ccc}5& -4& 3\\ -2& 2& -1\\ 0& 0& 1\end{array}\right]$

Step 3: Calculate the transpose of the cofactor matrix.

The transpose of the cofactor matrix is: $\left[\begin{array}{ccc}5& -2& 0\\ -4& 2& 0\\ 3& -1& 1\end{array}\right]$

Step 4: Divide the transpose of the cofactor matrix by the determinant of A.

The inverse of A is: $\left[\begin{array}{ccc}0.5& -0.2& 0\\ -0.4& 0.2& 0\\ 0.3& -0.1& 0.1\end{array}\right]$

Question:

For the matrix A= $\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& -3\\ 2& -1& 3\end{array}\right]$ show that $\begin{array}{}{A}^3\end{array}$−6$\begin{array}{}{A}^2\end{array}$+5A+11I=0. Hence find $\begin{array}{}{A}^{-1}\end{array}$

Answer:

Step 1: Calculate A^3 A^3 = $\left[\begin{array}{ccc}6& 6& -9\\ 6& -9& -21\\ -9& -21& 30\end{array}\right]$

Step 2: Calculate A^2 A^2 = $\left[\begin{array}{ccc}3& 3& -5\\ 3& 0& -12\\ -5& -12& 15\end{array}\right]$

Step 3: Substitute A^3 and A^2 in the given equation A^3 − 6A^2 + 5A + 11I = 0 $\left[\begin{array}{ccc}6& 6& -9\\ 6& -9& -21\\ -9& -21& 30\end{array}\right]$ − 6$\left[\begin{array}{ccc}3& 3& -5\\ 3& 0& -12\\ -5& -12& 15\end{array}\right]$ + 5$\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& -3\\ 2& -1& 3\end{array}\right]$ + 11

Question:

Find cofactors of A= $\left[\begin{array}{ccc}2& 1& 3\\ 4& -1& 0\\ -7& 2& 1\end{array}\right]$

Answer:

Step 1: Calculate the determinant of the matrix A.

Det A = 2×(-1)×1 + 1×0×(-7) + 3×2×4 = -18

Step 2: Calculate the cofactors of each element in the matrix.

Cofactor of 2 = (-1)×1 = -1 Cofactor of 1 = 0×(-7) = 0 Cofactor of 3 = 2×4 = 8 Cofactor of 4 = (-1)×(-7) = 7 Cofactor of -1 = 1×3 = 3 Cofactor of 0 = 4×(-7) = -28 Cofactor of -7 = 2×1 = 2 Cofactor of 2 = (-1)×3 = -3 Cofactor of 1 = 0×4 = 0

Step 3: Construct the matrix of cofactors of A.

$\left[\begin{array}{ccc}-1& 0& 8\\ 7& 3& -28\\ 2& -3& 0\end{array}\right]$

Question:

Find the inverse of the matrix (if it exists) $\left|\begin{array}{cc}2& -2\\ 4& 3\end{array}\right|$

Answer:

Step 1: Calculate the determinant of the matrix.

The determinant of the matrix is 14.

Step 2: Calculate the adjoint of the matrix.

The adjoint of the matrix is $\left|\begin{array}{cc}3& 2\\ -4& 2\end{array}\right|$.

Step 3: Divide the adjoint of the matrix by the determinant.

The inverse of the matrix is $\left|\begin{array}{cc}3& 2\\ -4& 2\end{array}\right|/14$.

Question:

Find the inverse of the matrix (if it exists) A= $\left[\begin{array}{ccc}1& 0& 0\\ 3& 3& 0\\ 5& 2& -1\end{array}\right]$

Answer:

Step 1: Calculate the determinant of the matrix A.

The determinant of A is -15.

Step 2: Calculate the adjoint of matrix A.

The adjoint of A is $\left[\begin{array}{ccc}3& -3& 0\\ 0& 1& 0\\ -5& 2& 1\end{array}\right]$

Step 3: Divide the adjoint of A by the determinant of A.

The inverse of A is $\left[\begin{array}{ccc}\mathrm{1/15}& \mathrm{-1/5}& 0\\ 0& \mathrm{1/15}& 0\\ \mathrm{1/3}& \mathrm{2/15}& \mathrm{-1/15}\end{array}\right]$

Question:

Verify A(adjA)=(adjA)A=∣A∣I A=$\left[\begin{array}{cc}2& 3\\ 4& 6\end{array}\right]$

Answer:

Step 1: Calculate the determinant of matrix A.

∣A∣ = (2 * 6) - (3 * 4) = 6 - 12 = -6

Step 2: Calculate the adjugate of matrix A.

adjA = $\left[\begin{array}{cc}6& -3\\ -4& 2\end{array}\right]$

Step 3: Calculate (adjA)A.

(adjA)A = $\left[\begin{array}{cc}6& -3\\ -4& 2\end{array}\right]\left[\begin{array}{cc}2& 3\\ 4& 6\end{array}\right]$

= $\left[\begin{array}{cc}12& -6\\ -8& 4\end{array}\right]$

Step 4: Calculate A(adjA).

A(adjA) = $\left[\begin{array}{cc}2& 3\\ 4& 6\end{array}\right]\left[\begin{array}{cc}6& -3\\ -4& 2\end{array}\right]$

= $\left[\begin{array}{cc}12& -6\\ -8& 4\end{array}\right]$

Step 5: Calculate ∣A∣I.

∣A∣I =

Step 6: Verify A(adjA)=(adjA)A=∣A∣I.