The adjoint of a matrix (also called the adjugate of a matrix) is defined as the transpose of the cofactor matrix of that particular matrix. For a matrix A, the adjoint is denoted as adj (A).  On the other hand, the inverse of a matrix A is that matrix which when multiplied by the matrix A gives an identity matrix. The inverse of a Matrix A is denoted by A^-1.

Table of Contents in Matrices

Introduction to Matrices

Types of Matrices

Matrix Operations

Adjoint and Inverse of a Matrix

Rank of a Matrix and Special Matrices

Solving Linear Equations using Matrix

Adjoint of a Matrix

The determinant of a square matrix A is denoted as $|A|$.

(\begin{array}{l}If; A=\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \ \end{matrix} \right],;; then ;;\left| A \right|=\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \ \end{matrix} \right|\end{array})

The matrix formed by the cofactors of the elements is (\begin{bmatrix} A_{11} & A_{12} & A_{13} \ A_{21} & A_{22} & A_{23} \ A_{31} & A_{32} & A_{33} \end{bmatrix})

Where $$A_{11} = (-1)^{1+1} \left| \begin{matrix} a_{22} & a_{23} \ a_{32} & a_{33} \ \end{matrix} \right| = a_{22}a_{33} - a_{23}.a_{32}$$

(\begin{array}{l}{{A}_{12}}={{\left( -1 \right)}^{1+2}}\left| \begin{matrix} {{a}_{21}} & {{a}_{23}} \ {{a}_{31}} & {{a}_{33}} \ \end{matrix} \right|=-{{a}_{21}}.,{{a}_{33}}+{{a}_{23}}.,{{a}_{31}}\ {{A}_{13}}={{\left( -1 \right)}^{1+3}}\left| \begin{matrix} {{a}_{21}} & {{a}_{22}} \ {{a}_{31}} & {{a}_{32}} \ \end{matrix} \right|={{a}_{21}}{{a}_{32}}-{{a}_{22}}{{a}_{31}};\end{array} )

(\begin{array}{l}{{A}_{21}}=-{{a}_{12}}{{a}_{33}}+{{a}_{13}}.,{{a}_{32}}; {{A}_{22}}={{a}_{11}}{{a}_{33}}-{{a}_{13}}.,{{a}_{31}};\end{array} )

(\begin{array}{l}{{A}_{23}}={{\left( -1 \right)}^{2+3}}\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} \ {{a}_{31}} & {{a}_{32}} \ \end{matrix} \right|=-{{a}_{11}}{{a}_{32}}+{{a}_{12}},.,{{a}_{31}}\{{A}_{31}}={{\left( -1 \right)}^{3+1}}\left| \begin{matrix} {{a}_{12}} & {{a}_{13}} \ {{a}_{22}} & {{a}_{23}} \ \end{matrix} \right|={{a}_{12}}{{a}_{23}}-{{a}_{13}},.,{{a}_{22}};\end{array})

(\begin{array}{l}{{A}_{32}}={{\left( -1 \right)}^{3+2}}\left| \begin{matrix} {{a}_{11}} & {{a}_{13}} \ {{a}_{21}} & {{a}_{23}} \ \end{matrix} \right|=-{{a}_{11}}{{a}_{23}}+{{a}_{13}}.,{{a}_{21}}\{{A}_{33}}={{\left( -1 \right)}^{3+3}}\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} \ {{a}_{21}} & {{a}_{22}} \ \end{matrix} \right|={{a}_{11}}{{a}_{22}}-{{a}_{12}}.,{{a}_{21}};\end{array})

The adjoint of a matrix A is the transpose of the matrix of co-factors, and is written as adj A.

(\begin{array}{l}adj,A=\left[ \begin{matrix} {{A}_{33}} & {{A}_{23}} & {{A}_{13}} \ {{A}_{32}} & {{A}_{22}} & {{A}_{12}} \ {{A}_{31}} & {{A}_{21}} & {{A}_{11}} \ \end{matrix} \right]\end{array} )

The determinant of a matrix A multiplied by the unit matrix is equal to the product of A and its adjoint.

Let A be a square matrix, then (Adjoint A). A = A. (Adjoint A) = |A^T|.

Let (\begin{array}{l}A=\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \ \end{matrix} \right] ;;and ;;\mathrm{adj}; A;=;\left[ \begin{matrix} {{A}_{11}} & {{A}_{21}} & {{A}_{31}} \ {{A}_{12}} & {{A}_{22}} & {{A}_{32}} \ {{A}_{13}} & {{A}_{23}} & {{A}_{33}} \ \end{matrix} \right]\end{array} )

A. ( \begin{array}{l}=\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \ \end{matrix} \right] \cdot \left[ \begin{matrix} {{A}_{11}} & {{A}_{21}} & {{A}_{31}} \ {{A}_{12}} & {{A}_{22}} & {{A}_{32}} \ {{A}_{13}} & {{A}_{23}} & {{A}_{33}} \ \end{matrix} \right]\end{array} )

(\begin{array}{l}=\left[ \begin{matrix} a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13} & a_{11}A_{21} + a_{12}A_{22} + a_{13}A_{23} & a_{11}A_{31} + a_{12}A_{32} + a_{13}A_{33} \ a_{21}A_{11} + a_{22}A_{12} + a_{23}A_{13} & a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23} & a_{21}A_{31} + a_{22}A_{32} + a_{23}A_{33} \ a_{31}A_{11} + a_{32}A_{12} + a_{33}A_{13} & a_{31}A_{21} + a_{32}A_{22} + a_{33}A_{23} & a_{31}A_{31} + a_{32}A_{32} + a_{33}A_{33} \end{matrix} \right]\end{array})

$$\left[ \begin{matrix} \left| A \right| & 0 & 0 \ 0 & \left| A \right| & 0 \ 0 & 0 & \left| A \right| \ \end{matrix} \right] = \left| A \right| \left[ \begin{matrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \ \end{matrix} \right] = \left| A \right|I$$

Finding the Adjoint and Inverse of a Matrix

Finding the Adjoint of a Matrix: Example Problems

Example 1: If $A^T = -A$, then the elements on the diagonal of the matrix are equal to $-1$.

(a) 1

(b) -1

(c) 0

None of these

Given:

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Solution:

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AT = -A, where A is a skew-symmetric matrix and the diagonal elements of A are all zeros.

Option (c) is the answer.

Answer: If A and B are two skew-symmetric matrices of order n, then A + B is also a skew-symmetric matrix of order n.

(a) AB is a skew-symmetric matrix, meaning that its transpose is equal to its negative.

(b) AB is a symmetric matrix

(c) A and B commute if AB is a symmetric matrix

None of these

Given:

Hello

Solution:

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We are given A’ = -A and B’ = -B;

Now, $(AB) = B’A’ = (-B)(-A) = BA = AB$, if $A$ and $B$ commute.

Thus, the correct option is (c). = 0.'

If A is skew-symmetric, then AB’ + BA’ = 0 and BA’ = 0.

Symmetric

(b) Skew-Symmetric

(c) Invertible

None of these

Given:

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Solution:

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(c) we have, (A’B’) = AB = -A’B [A is skew symmetric]; = B’A’ = B(-A)

-BA =

BA^T = -BA

Example 4: Let  $\begin{array}{l}A=\left[ \begin{matrix} 1 & 2 & 3 \ 1 & 3 & 4 \ 1 & 4 & 3 \ \end{matrix} \right],\end{array}$

Then find adj A.

Given:

My Favorite Movie

Solution:

My Favorite Movie

The determinant of the remaining elements can be obtained by eliminating all the elements of the same row and column of any matrix’s co-factors.

(\left|A_{11}\right| = \begin{vmatrix} 3 & 4 \ 4 & 3 \ \end{vmatrix} = 3 \times 3 - 4 \times 4 = -7)

(\begin{array}{l}{{A}_{12}}=-\left| \begin{matrix} 1 & 4 \ 1 & 3 \ \end{matrix} \right|=1, \quad {{A}_{13}}=\left| \begin{matrix} 1 & 3 \ 1 & 4 \ \end{matrix} \right|=1; \ {{A}_{21}}=-\left| \begin{matrix} 2 & 3 \ 4 & 3 \ \end{matrix} \right|=6, \quad {{A}_{22}}=\left| \begin{matrix} 1 & 3 \ 1 & 3 \ \end{matrix} \right|=0\end{array} )

(\begin{array}{l}{{A}_{23}}=-2,,,,,{{A}_{31}}=-1;,,,,{{A}_{32}}=-1, ;;;{{A}_{33}}=1\end{array})

The transpose of the cofactor matrix is equal to adj A.

(\begin{array}{l}\therefore Adj,,A=\left| \begin{matrix} 6 & -7 & 1 \ -1 & 1 & 2 \ -1 & 1 & -1 \ \end{matrix} \right|\end{array} )

Answer: All of the statements are true.

If $A = 0$, then $adj(A) = 0$

(b) The adjoint of a 3x3 diagonal matrix is also a diagonal matrix

(c) The product of two upper triangular matrices is an upper triangular matrix

$(d) \text{adj}(AB) = \text{adj}(A) \cdot \text{adj}(B)$

Given:

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Solution:

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(d) We have, $adj(AB) = adj(B) \cdot adj(A)$ and not $adj(AB) = adj(A) \cdot adj(B)$

Inverse of a Matrix

If $A$ and $B$ are two square matrices of the same order, such that $AB = BA = I$ ($I$ = unit matrix)

If A is a square matrix and B is its inverse, then B is called the inverse of A, i.e. B = A^-1, and A is the inverse of B. The condition for a square matrix A to possess an inverse is that the matrix A is non-singular, i.e., |A| ≠ 0. Taking the determinant of both sides, |AB| = |I| or |A| |B| = I. This relation implies that |A| ≠ 0, i.e. the matrix A is non-singular.

How to Find the Inverse of a Matrix Using the Adjoint Matrix?

We know that, $$A.(Adj,A) = |A|I \quad \text{or} \quad \frac{A.(Adj,A)}{|A|} = I \quad (\text{Provided } |A| \ne 0)$$

And $$A^{-1} = I;$$ $$A^{-1} = \frac{1}{|A|} (Adj.,A)$$

Properties of the Inverse and Adjoint of a Matrix

Property 1: For a square matrix A of order n, $$A \cdot \text{adj}(A) = \text{adj}(A) \cdot A = |A|\cdot I$$, where I is the identity matrix of order n.

Property 2: A square matrix A is invertible if and only if it is a non-singular matrix.

Problems with Finding the Inverse of a Matrix

Illustration 1: Let (\begin{array}{l}A=\left[ \begin{matrix} 1 & 0 & -1 \ 3 & 4 & 5 \ 0 & -6 & -7 \ \end{matrix} \right].\end{array} ) What is the inverse of A?

Given:

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Solution:

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From the formula, (\frac{adj,A}{\left| A \right|} = A^{-1})

We have $$\begin{array}{l}{A_{11}}=\left[ \begin{matrix} 4 & 5 \ -6 & -7 \ \end{matrix} \right]=2,,,{A_{12}}=-\left[ \begin{matrix} 3 & 5 \ 0 & -7 \ \end{matrix} \right]=21\end{array}$$

Similarly $$\begin{array}{l}{A_{13}}=-18,{A_{31}}=4,{A_{32}}=-8,{A_{33}}=4,{A_{21}}=+6,{A_{22}}=-7,{A_{23}}=6\end{array}$$

The cofactor matrix of A $$=\begin{bmatrix} 2 & 21 & -18\ 6&-7 & 6\ 4 & -8 & 4 \end{bmatrix}$$

Transpose of the cofactor matrix = adj A

(\begin{array}{l} adj A = \left[ \begin{matrix} 4 & -6 & -2 \ -8 & 7 & 21 \ 4 & -6 & -18 \ \end{matrix} \right] \end{array})

Also $$\left| A \right|=\left| \begin{matrix} 1 & 0 & -1 \ 3 & 4 & 5 \ 0 & -6 & -7 \ \end{matrix} \right|=\left{ 4\times \left( -7 \right)-\left( -6 \right)\times 5-3\times \left( -6 \right) \right}$$

-28 + 30 + 18 = 20

20 =

$\left[ \begin{matrix} \frac{1}{10} & \frac{3}{10} & \frac{2}{10} \ \frac{21}{20} & -\frac{7}{20} & -\frac{8}{20} \ -\frac{18}{20} & \frac{6}{20} & \frac{2}{10} \ \end{matrix} \right]$

Illustration 2: If the product of a matrix A and $\left[ \begin{matrix} 1 & 1 \ 2 & 0 \ \end{matrix} \right]$ is the matrix $\left[ \begin{matrix} 3 & 2 \ 1 & 1 \ \end{matrix} \right]$,

Then A-1 is given by:

(\begin{array}{l}(a) \left[ \begin{matrix} 0 & -1 \ -2 & 4 \ \end{matrix} \right];;;; \ (b) \left[ \begin{matrix} 0 & 1 \ -2 & 4 \ \end{matrix} \right] ;;;; \ (c)\left[ \begin{matrix} 0 & -1 \ 2 & 4 \ \end{matrix} \right]\end{array} )

None of these

Given:

Hello World

Solution:

Hello World

If (AB = C), then by using the formula (\Rightarrow {{A}^{-1}}=B{{C}^{-1}}), we can get the value of (A^{-1}).

Here, $$A\left[ \begin{matrix} 1 & 1 \ 2 & 0 \ \end{matrix} \right]=\left[ \begin{matrix} 3 & 2 \ 1 & 1 \ \end{matrix} \right]\Rightarrow {{A}^{-1}}=\left[ \begin{matrix} 1 & 1 \ 2 & 0 \ \end{matrix} \right]{{\left[ \begin{matrix} 3 & 2 \ 1 & 1 \ \end{matrix} \right]}^{-1}}=\left[ \begin{matrix} 1 & 1 \ 2 & 0 \ \end{matrix} \right]\left[ \begin{matrix} 1 & -2 \ -1 & 3 \ \end{matrix} \right]=\left[ \begin{matrix} 0 & 1 \ 2 & -4 \ \end{matrix} \right]$$

Illustration 3:

Let $$A =\left[ \begin{matrix} 2 & 1 & -1 \ 0 & 1 & 0 \ 1 & 3 & -1 \ \end{matrix} \right];and; B =\left[ \begin{matrix} 1 & 2 & 5 \ 2 & 3 & 1 \ -1 & 1 & 1 \ \end{matrix} \right].$$ Prove that $${{\left( AB \right)}^{-1}}={{B}^{-1}}{{A}^{-1}}.$$

Given:

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Solution:

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By using the formula (\begin{array}{l}{{\left( AB \right)}^{-1}}=\frac{adj,AB}{\left| AB \right|}.\end{array} ), we can obtain (\left( AB \right)^{-1}) by obtaining (\left| AB \right|) and (adj,AB). Similarly, we can also obtain the values of (\left( B \right)^{-1}) and (\left( A \right)^{-1}). Then, by multiplying (\left( B \right)^{-1}) and (\left( A \right)^{-1}), we can prove the given problem.

Here, (\begin{array}{l}AB=\left[ \begin{matrix} 2 & 1 & -1 \ 0 & 1 & 0 \ 1 & 3 & -1 \ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 & 5 \ 2 & 3 & 1 \ -1 & 1 & 1 \ \end{matrix} \right]=\left[ \begin{matrix} 2+2+1 & 4+3-1 & 10+1-1 \ 0+2+0 & 0+3+0 & 0+1+0 \ 1+6+1 & 2+9-1 & 5+3-1 \ \end{matrix} \right]=\left[ \begin{matrix} 5 & 6 & 10 \ 2 & 3 & 1 \ 8 & 10 & 7 \ \end{matrix} \right]\end{array} )

Now, $$\left| AB \right|=\left| \begin{matrix} 5 & 6 & 10 \ 2 & 3 & 1 \ 8 & 10 & 7 \ \end{matrix} \right|=5(21-10)-6(14-8)+10(20-24)=55-36-40=-21.$$

The matrix of cofactors of $\left|\ AB \right|$ is equal to

(\begin{array}{l}\left[ \begin{matrix} 3\left( 7 \right)-1\left( 10 \right) & -2\left( 7 \right)+8\left( 1 \right) & 2\left( 10 \right)-3\left( 8 \right) \ -6\left( 7 \right)+10\left( 10 \right) & 5\left( 7 \right)-8\left( 10 \right) & 5\left( 10 \right)-6\left( 8 \right) \ 6\left( 1 \right)-10\left( 3 \right) & 5\left( 1 \right)-2\left( 10 \right) & 5\left( 3 \right)-6\left( 2 \right) \ \end{matrix} \right]=\left[ \begin{matrix} 11 & 6 & -4 \ -58 & -45 & 2 \ -24 & -15 & 3 \ \end{matrix} \right]\end{array} )

$$\left(AB\right)^{-1}=\frac{adj,AB}{\left| AB \right|}=\frac{-1}{21}\left[ \begin{matrix} 11 & 58 & -24 \ -6 & -45 & 15 \ -4 & -2 & 3 \ \end{matrix} \right]$$

Next, $$\left| B \right|=\left| \begin{matrix} 1 & 2 & 5 \ 2 & 3 & 1 \ -1 & 1 & 1 \ \end{matrix} \right|=1\left( 3-1 \right)-2\left( 2+1 \right)+5\left( 2+3 \right)=21$$

Cofactor matrix of B  (\begin{array}{l}=\begin{bmatrix} 2 & -3 & 5 \ 3 & 6 & -3 \ -13 & 9 & -1 \end{bmatrix}\end{array})

(\begin{array}{l}adj B = \begin{bmatrix} 2 & 3 & -13 \ -3 & 6 & 9 \ 5 & -3 & -1 \end{bmatrix}\end{array})

∴(\begin{array}{l}{B^{-1}}=\frac{adj,B}{\left| B \right|}=\frac{1}{21}\left[ \begin{matrix} 2 & 3 & -13 \ -3 & 6 & 9 \ 5 & -3 & -1 \ \end{matrix} \right];;;\left| A \right|=\left[ \begin{matrix} 2 & 1 & -1 \ 0 & 1 & 0 \ 1 & 3 & -1 \ \end{matrix} \right]=1\left(-2+1\right)=-1\end{array})

The cofactor matrix of A (\begin{bmatrix} -1 & 0 & -1\ -2&-1 & -5\ 1 & 0 & 2 \end{bmatrix})

$$\begin{array}{l},,A^{-1}=\frac{adj,A}{\left| A \right|}=\frac{1}{-1}\left[ \begin{matrix} 1 & 2 & -1 \ 0 & 1 & 0 \ 1 & 5 & -2 \ \end{matrix} \right]\end{array} $$

$$\therefore \left[ \begin{matrix} -1 & -2 & 1 \ 0 & -1 & 0 \ -1 & -5 & 2 \ \end{matrix} \right]^{-1}{\left[ \begin{matrix} 2 & 3 & -13 \ -3 & 6 & 9 \ 5 & -3 & -1 \ \end{matrix} \right]^{-1}}=-\frac{1}{21}\left[ \begin{matrix} -1 & -2 & 1 \ 0 & -1 & 0 \ -1 & -5 & 2 \ \end{matrix} \right]$$

Thus, (\left( AB \right)^{-1}=\left[ \begin{matrix} -\frac{3}{2} & \frac{5}{14} & \frac{1}{14} \ \frac{1}{2} & \frac{1}{7} & \frac{1}{21} \ \frac{1}{7} & -\frac{1}{14} & -\frac{1}{21} \ \end{matrix} \right]), where (A = \left[ \begin{matrix} 11 & 58 & -24 \ -6 & -45 & 15 \ -4 & -2 & 3 \ \end{matrix} \right]) and (B = \left[ \begin{matrix} -\frac{1}{21} & 0 & 0 \ 0 & \frac{1}{7} & 0 \ 0 & 0 & \frac{1}{3} \ \end{matrix} \right]).

Illustration 4: If (\begin{array}{l}A=\left[ \begin{matrix} 0 & 2y & z \ x & y & -z \ x & -y & z \ \end{matrix} \right]\end{array}) satisfies (A’ = A^{-1}),

After that,

Then

(\begin{array}{l}(a)\ x=\pm \frac{1}{\sqrt{6}},y=\pm \frac{1}{\sqrt{6}},z=\pm \frac{1}{\sqrt{3}};; ;;;;;;; (b)\ x=\pm \frac{1}{\sqrt{2}},y=\pm \frac{1}{\sqrt{6}},z=\pm \frac{1}{\sqrt{3}}\end{array} )

(c) x = ±1/√6, y = ±1/√2, z = ±1/√3; (d) x = ±1/√2, y = ±1/3, z = ±1/√2

Given:

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Solution:

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We can use the multiplication method to obtain values of x, y, and z by noting that (AA’ = I), where (A’ = A^{-1}) and (A A^{-1} = I).

(\begin{array}{l}A’A = 1 \Leftrightarrow A’ = {{A}^{-1}}\end{array} ) he said

He said, “Now.”

(\begin{array}{l}AA’ = \left[ \begin{matrix} 0 & 2y & z \ x & y & -z \ x & -y & z \end{matrix} \right] \left[ \begin{matrix} 0 & x & x \ 2y & y & -y \ z & -z & z \end{matrix} \right] = \left[ \begin{matrix} 4y^2 + z^2 & 2y^2 - z^2 & -2y^2 + z^2 \ 2y^2 - z^2 & x^2 + y^2 + z^2 & x^2 - y^2 - z^2 \ -2y^2 + z^2 & x^2 - y^2 - z^2 & x^2 + y^2 + z^2 \end{matrix} \right] \end{array})

Thus, $$\begin{array}{l} AA’=I \Rightarrow 4y^2+z^2=1, ;; 2y^2-z^2=0, ;; x^2+y^2+z^2=1, ;; x^2-y^2-z^2=0 \end{array}$$

(\begin{array}{l}x=\pm \frac{1}{\sqrt{2}},y=\pm \frac{1}{\sqrt{6}},z=\pm \frac{1}{\sqrt{3}}\end{array})

Illustration 5: If $$A=\left[ \begin{matrix} 0 & 1 & 2 \ 1 & 2 & 3 \ 3 & x & 1 \ \end{matrix} \right] ;;and ;;{{A}^{-1}}=\left[ \begin{matrix} 1/2 & -1/2 & 1/2 \ -4 & 3 & y \ 5/2 & -3/2 & 1/2 \ \end{matrix} \right],$$

After that,

Then

x = 1, y = -1

(b) x = -1, y = 1

(c) \ \ x = 2, \ \ y = -\frac{1}{2}

x = 1/2, y = 1/2

Given:

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Solution:

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We know that $$A{{A}^{-1}}=I$$, so by solving it, we can obtain the values of $x$ and $y$.

We have

(\begin{array}{l}A{{A}^{-1}}=\left[ \begin{matrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \ \end{matrix} \right]\left[ \begin{matrix} 0 & 1 & 2 \ 1 & 2 & 3 \ 3 & x & 1 \ \end{matrix} \right]\left[ \begin{matrix} 1/2 & -1/2 & 1/2 \ -4 & 3 & y \ 5/2 & -3/2 & 1/2 \ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & y+1 \ 0 & 1 & 2\left( y+1 \right) \ 4\left( 1-x \right) & 3\left( x-1 \right) & 2+xy \ \end{matrix} \right]\end{array} )

(\begin{array}{l}1-x=0, \ x+1=0 \ y-1=0, \ y+1=0 \ 2+xy=1\end{array})

x = 1, y = -1

Frequently Asked Questions

The formula to find the Inverse of a matrix A is A^-1 = (1/det(A)) * adj(A)

The Inverse of a matrix A is given by $$A^{-1} = \frac{\text{adj }A}{\text{det }A}.$$

What is the Adjoint of a matrix?

The transpose of the cofactor matrix of a matrix is its Adjoint.

The cofactor formula is: C_{ij} = (-1)^{i+j} \det A_{ij}

The cofactor formula is given by A_{ij} = (-1)^{i+j} \det M_{ij}, where \det M_{ij} is the minor of a_{ij}.

What is an (adj A) n x n Matrix?

If $A$ is a square matrix of order $n$, then $A\cdot adj(A) = adj(A)\cdot A = |A|I$, where $I$ is the identity matrix of order $n$.

A minor of a determinant is the determinant of the matrix formed by deleting one row and one column of the original matrix.

The minor of an element aij of a determinant is denoted by Mij and is calculated by deleting its ith row and jth column in which the element aij lies.

A non-singular matrix is a square matrix that has an inverse.

A square matrix X is said to be non-singular if its determinant, |X|, is a non-zero value. i.e. |X| ≠ 0.

A singular matrix is a square matrix in which the determinant is equal to 0.

A square matrix B is said to be singular if |B| = 0.

What is the Adjoint of a 2x2 Matrix?

Swap the elements on the main diagonal (a11 and a22). Then give negative sign for the elements at a12 and a21 position. The resulting matrix is the adjoint of the given 2x2 matrix.