Matrices and Determinants Each Questions

Exercise 1.3

Q.1 Which of the following are comfortable for addition?

A = [\begin{matrix} 2 & 1 \\ -1 & 3 \end{matrix}], B = [\begin{matrix} 3 \\ 1 \end{matrix}]

C = [\begin{matrix} 1 & 0 \\ 2 & -1 \\ 1 & -2 \end{matrix}], D = [\begin{matrix} 2+1 \\ 3 \end{matrix}]

E = [\begin{matrix} -1 & 0 \\ 1 & 2 \end{matrix}] and , F = [\begin{matrix} 3 & 2 \\ 1+1 & -4 \\ 3+2 & 2+1 \end{matrix}]

Solution:

In the above matrices following matrices are suitable for addition.

A and E are conformable for addition because their order is same and both are square matrix.
B and D are conformable for addition because the order is same. they have two rows and 1 Columns and both are rectangular matrices.
C and F are conformable for addition because their order is same. they have three 3 rows and 2 columns and they are a rectangular matrix.

Q.2 Find the additive inverse of the following matrices.

(1)

A = [\begin{matrix} 2 & 4 \\ -2 & 1 \end{matrix}]

Solution:

Additive Inverse of a matrix is it’s negative matrix.

A = [\begin{matrix} 2 & 4 \\ -2 & 1 \end{matrix}]

Both sides multiply with (-) negative sign

-A = - [\begin{matrix} 2 & 4 \\ -2 & 1 \end{matrix}] = [\begin{matrix} -(+2) & -(+4) \\ -(-2) & -(+1) \end{matrix}]

Additive Inverse of Matrix A

-A = [\begin{matrix} -2 & -4 \\ 2 & -1 \end{matrix}]

(2)

B = [\begin{matrix} 1 & 0 & -1 \\ 2 & -1 & 3 \\ 3 & -2 & 1 \end{matrix}]

Solution:

Additive Inverse of a matrix is it’s negative matrix.

B = [\begin{matrix} 1 & 0 & -1 \\ 2 & -1 & 3 \\ 3 & -2 & 1 \end{matrix}]

Both sides multiply with (-) negative sign

-B = - [\begin{matrix} 1 & 0 & -1 \\ 2 & -1 & 3 \\ 3 & -2 & 1 \end{matrix}]

-B = - [\begin{matrix} -(+1) & -(0) & -(-1) \\ -(+2) & -(-1) & -(+3) \\ -(+3) & -(-2) & -(+1) \end{matrix}]

Additive Inverse of Matrix B

-B = [\begin{matrix} -1 & 0 & +1 \\ -2 & +1 & -3 \\ -3 & +2 & -1 \end{matrix}]

(3)

C = [\begin{matrix} 4 \\ -2 \end{matrix}]

Solution:

Additive Inverse of a matrix is it’s negative matrix.

C = [\begin{matrix} 4 \\ -2 \end{matrix}]

Both sides multiply with (-) negative sign

-C = - [\begin{matrix} 4 \\ -2 \end{matrix}] = [\begin{matrix} -(+4) \\ -(-2) \end{matrix}]

Additive Inverse of Matrix C

-C = [\begin{matrix} -4 \\ 2 \end{matrix}]

(4)

D = [\begin{matrix} 1 & 0 \\ -3 & -2 \\ 2 & 1 \end{matrix}]

Solution

Additive Inverse of a matrix is it’s negative matrix.

D = [\begin{matrix} 1 & 0 \\ -3 & -2 \\ 2 & 1 \end{matrix}]

Both sides multiply with (-) negative sign

-D = - [\begin{matrix} 1 & 0 \\ -3 & -2 \\ 2 & 1 \end{matrix}] = [\begin{matrix} -(+1) & -(0) \\ -(-3) & -(-2) \\ -(+2) & -(+1) \end{matrix}]

Additive Inverse of Matrix D

-D = [\begin{matrix} -1 & 0 \\ 3 & 2 \\ -2 & -1 \end{matrix}]

(5)

E = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]

Solution:

E = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]

Additive Inverse of a matrix is it’s negative matrix.

-E = - [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]

Both sides multiply with (-) negative sign

-E = [\begin{matrix} -1×1 & -1×0 \\ -1×0 & -1×1 \end{matrix}]

Additive Inverse of Matrix E

-E = [\begin{matrix} -1 & 0 \\ 0 & -1 \end{matrix}]

(6)

F = [\begin{matrix} √3 & 1 \\ -1 & √2 \end{matrix}]

Solution:

F = [\begin{matrix} √3 & 1 \\ -1 & √2 \end{matrix}]

Additive Inverse of a matrix is it’s negative matrix.

-F = - [\begin{matrix} √3 & 1 \\ -1 & √2 \end{matrix}]

Both sides multiply with (-) negative sign

-F = [\begin{matrix} -1×√3 & -1×1 \\ -1×-1 & -1×√2 \end{matrix}]

Additive Inverse of Matrix F

-F = [\begin{matrix} -√3 & -1 \\ 1 & -√2 \end{matrix}]

Q.3 If the given below values of A, B, C, and D, then solve.

If , A = [\begin{matrix} -1 & 2 \\ 2 & 1 \end{matrix}] . , B = [\begin{matrix} 1 \\ -1 \end{matrix}]

C = [\begin{matrix} 1 & -1 & 2 \end{matrix}] . , D = [\begin{matrix} 1 & 2 & 3 \\ -1 & 0 & 2 \end{matrix}]

Then Find.

(1)

A + [\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}]

Solution:

Matrix A + Given Matrix

A + [\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}]

Matrix A values

A = [\begin{matrix} -1 & 2 \\ 2 & 1 \end{matrix}]

The order of matrix A and the given matrix order is same. So, they can be added easily.

= [\begin{matrix} -1 & 2 \\ 2 & 1 \end{matrix}] + [\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}]

= [\begin{matrix} -1+1 & 2+1 \\ 2+1 & 1+1 \end{matrix}]

= [\begin{matrix} 0 & 3 \\ 3 & 2 \end{matrix}]

(2)

B + [\begin{matrix} -2 \\ 3 \end{matrix}]

Solution:

Matrix B + Given Matrix

B + [\begin{matrix} -2 \\ 3 \end{matrix}]

Matrix B values

B = [\begin{matrix} 1 \\ -1 \end{matrix}]

The order of both above matrices are same. So they can be easily added.

= [\begin{matrix} 1 \\ -1 \end{matrix}] + [\begin{matrix} -2 \\ 3 \end{matrix}]

= [\begin{matrix} 1+(-2) \\ -1+3 \end{matrix}]

= [\begin{matrix} -1 \\ 2 \end{matrix}]

(3)

C + [\begin{matrix} -2 & 1 & 3 \end{matrix}]

Solution:

Matrix C + Given Matrix

C + [\begin{matrix} -2 & 1 & 3 \end{matrix}]

Matrix C values

C = [\begin{matrix} 1 & -1 & 2 \end{matrix}]

Their orders are same so they can easily added.

= [\begin{matrix} 1 & -1 & 2 \end{matrix}] + [\begin{matrix} -2 & 1 & 3 \end{matrix}]

= [\begin{matrix} 1+(-2) & -1(1) & 2+3 \end{matrix}]

= [\begin{matrix} -1 & 0 & 5 \end{matrix}]

(4)

D + [\begin{matrix} 0 & 1 & 0 \\ 2 & 0 & 1 \end{matrix}]

Solution:

Matrix D + Given Matrix

D + [\begin{matrix} 0 & 1 & 0 \\ 2 & 0 & 1 \end{matrix}]

Matrix D values

D = [\begin{matrix} 1 & 2 & 3 \\ -1 & 0 & 2 \end{matrix}]

Their orders are same. So, they can be added.

= [\begin{matrix} 1 & 2 & 3 \\ -1 & 0 & 2 \end{matrix}] + [\begin{matrix} 0 & 1 & 0 \\ 2 & 0 & 1 \end{matrix}]

= [\begin{matrix} 1+0 & 2+1 & 3+0 \\ -1+2 & 0+0 & 2+1 \end{matrix}]

= [\begin{matrix} 1 & 3 & 3 \\ 1 & 0 & 3 \end{matrix}]

(5)

Find the 2A

Solution:

Multiply Matrix A with 2

2 × Matrix A

Matrix A values

A = [\begin{matrix} -1 & 2 \\ 2 & 1 \end{matrix}]

= (2)× [\begin{matrix} -1 & 2 \\ 2 & 1 \end{matrix}]

All Matrix entries Multiply with 2

= [\begin{matrix} 2(-1) & 2×2 \\ 2×2 & 2×1 \end{matrix}]

= [\begin{matrix} -2 & 4 \\ 4 & 2 \end{matrix}]

(6)

Find (-1)B

Solution:

Multiply Matrix B with -1

(-1) × B

Matrix B values

B = [\begin{matrix} 1 \\ -1 \end{matrix}]

= -1\times [\begin{matrix} 1 \\ -1 \end{matrix}]

All Matrix Entries Multiply with -1

= [\begin{matrix} (-1)×1 \\ (-1)×-1 \end{matrix}]

= [\begin{matrix} -1 \\ 1 \end{matrix}]

(7)

Find (-2)C

Solution:

Multiply Matrix C with -2

(-2)C

Matrix C values

C = [\begin{matrix} 1 & -1 & 2 \end{matrix}]

= (-2)× [\begin{matrix} 1 & -1 & 2 \end{matrix}]

All entries multiply with -2

= [\begin{matrix} (-2)(1) & (-2)(-1) & (-2)2 \end{matrix}]

= [\begin{matrix} -2 & 2 & -4 \end{matrix}]

(8)

Find 3D

Solution:

Multiply Matrix D with 3

3 × D

Matrix D values

D = [\begin{matrix} 1 & 2 & 3 \\ -1 & 0 & 2 \end{matrix}]

= (3) [\begin{matrix} 1 & 2 & 3 \\ -1 & 0 & 2 \end{matrix}]

All entries multiply with 3

= [\begin{matrix} 3×1 & 3×2 & 3×3 \\ 3×-1 & 3×0 & 3×2 \end{matrix}]

= [\begin{matrix} 3 & 6 & 9 \\ -3 & 0 & 6 \end{matrix}]

Q.4 Performe the indicated operations and simplify the following.

(1)

([\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] + [\begin{matrix} 0 & 2 \\ 3 & 0 \end{matrix}]) + [\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix}]

Solution:

([\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] + [\begin{matrix} 0 & 2 \\ 3 & 0 \end{matrix}]) + [\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix}]

1^st solve the inbrackets matrices

= [\begin{matrix} 1+0 & 0+2 \\ 0+3 & 1+0 \end{matrix}] + [\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix}]

Perform the Addition

= [\begin{matrix} 1 & 2 \\ 3 & 1 \end{matrix}] + [\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix}]

= [\begin{matrix} 1+1 & 2+1 \\ 3+1 & 1+0 \end{matrix}]

Answer , = [\begin{matrix} 2 & 3 \\ 4 & 1 \end{matrix}]

(2)

[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] + ([\begin{matrix} 0 & 2 \\ 3 & 0 \end{matrix}] - [\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix}])

Solution:

[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] + ([\begin{matrix} 0 & 2 \\ 3 & 0 \end{matrix}] - [\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix}])

1^st solve the inbrackets matrices

= [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] + ([\begin{matrix} 0 & 2 \\ 3 & 0 \end{matrix}] - [\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix}])

= [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] + [\begin{matrix} 0-1 & 2-1 \\ 3-1 & 0-0 \end{matrix}]

= [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] + [\begin{matrix} -1 & 1 \\ 2 & 0 \end{matrix}]

Perform the Addition

= [\begin{matrix} 1-1 & 0+1 \\ 0+2 & 1+0 \end{matrix}]

Answer , = [\begin{matrix} 0 & 1 \\ 2 & 1 \end{matrix}]

(3)

[\begin{matrix} 2 & 3 & 1 \end{matrix}] + ([\begin{matrix} 1 & 0 & 2 \end{matrix}] - [\begin{matrix} 2 & 2 & 2 \end{matrix}])

Solution:

= [\begin{matrix} 2 & 3 & 1 \end{matrix}] + ([\begin{matrix} 1 & 0 & 2 \end{matrix}] - [\begin{matrix} 2 & 2 & 2 \end{matrix}])

1^st solve the inbrackets matrices

= [\begin{matrix} 2 & 3 & 1 \end{matrix}] + [\begin{matrix} 1-2 & 0-2 & 2-2 \end{matrix}]

= [\begin{matrix} 2 & 3 & 1 \end{matrix}] + [\begin{matrix} -1 & -2 & 0 \end{matrix}]

Perform the Addition step

= [\begin{matrix} 2+(-1) & 3+(-2) & 1+(0) \end{matrix}]

= [\begin{matrix} 2-1 & 3-2 & 1+0 \end{matrix}]

Answer , = [\begin{matrix} 1 & 1 & 1 \end{matrix}]

(4)

[\begin{matrix} 1 & 2 & 3 \\ -1 & -1 & -1 \\ 0 & 1 & 2 \end{matrix}] + [\begin{matrix} 1 & 1 & 1 \\ 2 & 2 & 2 \\ 3 & 3 & 3 \end{matrix}]

Solution:

= [\begin{matrix} 1 & 2 & 3 \\ -1 & -1 & -1 \\ 0 & 1 & 2 \end{matrix}] + [\begin{matrix} 1 & 1 & 1 \\ 2 & 2 & 2 \\ 3 & 3 & 3 \end{matrix}]

Perform the Addition step

= [\begin{matrix} 1+1 & 2+1 & 3+1 \\ -1+2 & -1+2 & -1+2 \\ 0+3 & 1+3 & 2+3 \end{matrix}]

Answer , = [\begin{matrix} 2 & 3 & 4 \\ 1 & 1 & 1 \\ 3 & 4 & 5 \end{matrix}]

(5)

[\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{matrix}] + [\begin{matrix} 1 & 0 & -2 \\ -2 & -1 & 0 \\ 0 & 2 & -1 \end{matrix}]

Solution:

= [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{matrix}] + [\begin{matrix} 1 & 0 & -2 \\ -2 & -1 & 0 \\ 0 & 2 & -1 \end{matrix}]

Perform the Addition step

= [\begin{matrix} 1+1 & 2+0 & 3-2 \\ 2-2 & 3-1 & 1+0 \\ 3+0 & 1+2 & 2-1 \end{matrix}]

Answer , = [\begin{matrix} 2 & 2 & 1 \\ 0 & 2 & 1 \\ 3 & 3 & 1 \end{matrix}]

(6)

([\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix}] + [\begin{matrix} 2 & 1 \\ 1 & 0 \end{matrix}]) + [\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}]

Solution:

= ([\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix}] + [\begin{matrix} 2 & 1 \\ 1 & 0 \end{matrix}]) + [\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}]

1^st solve the inbrackets matrices

= [\begin{matrix} 1+2 & 2+1 \\ 0+1 & 1+0 \end{matrix}] + [\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}]

= [\begin{matrix} 3 & 3 \\ 1 & 1 \end{matrix}] + [\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}]

Perform the Addition step

= [\begin{matrix} 3+1 & 3+1 \\ 1+1 & 1+1 \end{matrix}]

Answer , = [\begin{matrix} 4 & 4 \\ 2 & 2 \end{matrix}]

Q.5 For the following matrices A, B, and C.

A = [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}] B = [\begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 2 \\ 3 & 1 & 3 \end{matrix}]

And ,C = [\begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \end{matrix}]

Verify the following rules.

(1) Rule#1

A+C=C+A here (L.H.S is A+C) and (R.H.S is C+A)

L.H.S is A+C

Values of Matrices

A = [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}] C = [\begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \end{matrix}]

Required is A + C

= [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}] + [\begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \end{matrix}]

Perform the Addition step

= [\begin{matrix} 1-1 & 2+0 & 3+0 \\ 2+0 & 3-2 & 1+3 \\ 1+1 & -1+1 & 0+2 \end{matrix}]

Answer , = [\begin{matrix} 0 & 2 & 3 \\ 2 & 1 & 4 \\ 2 & 0 & 2 \end{matrix}]

R.H.S is C+A

Values of Matrices

C = [\begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \end{matrix}] A = [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}]

Required is C + A

= [\begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \end{matrix}] + [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}]

Perform the Addition step

= [\begin{matrix} -1+1 & 0+2 & 0+3 \\ 0+2 & -2+3 & 3+1 \\ 1+1 & 1-1 & 2+0 \end{matrix}]

Answer , = [\begin{matrix} 0 & 2 & 3 \\ 2 & 1 & 4 \\ 2 & 0 & 2 \end{matrix}]

Hence Proved

(L.H.S is A+C) = (R.H.S is C+A)

(2) Rule#2

A+B=B+A here (L.H.S is A+B) and (R.H.S is B+A)

L.H.S is A+B

Values of Matrices

A = [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}],B = [\begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 2 \\ 3 & 1 & 3 \end{matrix}]

Required is A + B

= [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}] + [\begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 2 \\ 3 & 1 & 3 \end{matrix}]

Perform the Adiition step

= [\begin{matrix} 1+1 & 2-1 & 3+1 \\ 2+2 & 3-1 & 1+2 \\ 1+3 & -1+1 & 0+3 \end{matrix}]

Answer , = [\begin{matrix} 2 & 1 & 4 \\ 4 & 2 & 3 \\ 4 & 0 & 3 \end{matrix}]

R.H.S is B+A

Values of Matrices

B = [\begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 2 \\ 3 & 1 & 3 \end{matrix}],A = [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}]

Required is B + A

= [\begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 2 \\ 3 & 1 & 3 \end{matrix}] + [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}]

Perform the Addition step

= [\begin{matrix} 1+1 & -1+2 & 1+3 \\ 2+2 & -1+3 & 2+1 \\ 3+1 & 1-1 & 3+0 \end{matrix}]

Answer ,= [\begin{matrix} 2 & 1 & 4 \\ 4 & 2 & 3 \\ 4 & 0 & 3 \end{matrix}]

Hence Proved

(L.H.S is A+B) = (R.H.S is B+A)

(3) Rule#3

B+C=C+B

(L.H.S is B+C) and (R.H.S is C+B)

L.H.S is B+C

Values of Matrices

B = [\begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 2 \\ 3 & 1 & 3 \end{matrix}],C = [\begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \end{matrix}]

Required is B + C

= [\begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 2 \\ 3 & 1 & 3 \end{matrix}] + [\begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \end{matrix}]

Perform the Addition step

= [\begin{matrix} -1+1 & -1+0 & 1+0 \\ 2+0 & -1-2 & 2+3 \\ 3+1 & 1+1 & 3+2 \end{matrix}]

Answer ,= [\begin{matrix} 0 & -1 & 1 \\ 2 & -3 & 5 \\ 4 & 2 & 5 \end{matrix}]

R.H.S is C+B

Values of the Matrices

C = [\begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \end{matrix}],B = [\begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 2 \\ 3 & 1 & 3 \end{matrix}]

Required is C + B

= [\begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \end{matrix}] + [\begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 2 \\ 3 & 1 & 3 \end{matrix}]

Perform the Adition step

= [\begin{matrix} -1+1 & 0-1 & 0+1 \\ 0+2 & -2-1 & 3+2 \\ 1+3 & 1+1 & 2+3 \end{matrix}]

Answer ,= [\begin{matrix} 0 & -1 & 1 \\ 2 & -3 & 5 \\ 4 & 2 & 5 \end{matrix}]

Hence Proved

(L.H.S is A+B) = (R.H.S is B+A)

(4) Rule#4

A+(B+A)=2A+B

(L.H.S is A+(B+A)) and (R.H.S is 2A+B)

L.H.S is A+(B+A)

1^st solve Inracket Matrices

= [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}]

+ [\begin{matrix} 1+1 & -1+2 & 1+3 \\ 2+2 & -1+3 & 2+1 \\ 3+1 & 1-1 & 3+0 \end{matrix}]

= [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}] + [\begin{matrix} 2 & 1 & 4 \\ 4 & 2 & 3 \\ 4 & 0 & 3 \end{matrix}]

Let’s performing the addition step

= [\begin{matrix} 1+2 & 2+1 & 3+4 \\ 2+4 & 3+2 & 1+3 \\ 1+4 & -1+0 & 0+3 \end{matrix}]

= [\begin{matrix} 3 & 3 & 7 \\ 6 & 5 & 4 \\ 5 & -1 & 3 \end{matrix}]

R.H.S is 2A+B

Values of A and B matrix

A = [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}] B = [\begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 2 \\ 3 & 1 & 3 \end{matrix}]

Required is 2A + B

= 2 [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}] + [\begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 2 \\ 3 & 1 & 3 \end{matrix}]

First A matrix multiply by 2.

= [\begin{matrix} 1×2 & 2×2 & 3×2 \\ 2×2 & 3×2 & 1×2 \\ 1×2 & -1×2 & 0×2 \end{matrix}] + [\begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 2 \\ 3 & 1 & 3 \end{matrix}]

= [\begin{matrix} 2 & 4 & 6 \\ 4 & 6 & 2 \\ 2 & -2 & 0 \end{matrix}] + [\begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 2 \\ 3 & 1 & 3 \end{matrix}]

Let’s performing the addition step

= [\begin{matrix} 2+1 & 4-1 & 6+1 \\ 4+2 & 6-1 & 2+2 \\ 2+3 & -2+1 & 0+3 \end{matrix}]

= [\begin{matrix} 3 & 3 & 7 \\ 6 & 5 & 4 \\ 5 & -1 & 3 \end{matrix}]

Hence Proved

(L.H.S is A+(B+A)) = (R.H.S is 2A+B)

(5) Rule#5

(C-B)+A=C+(A-B)

(L.H.S (C-B)+A) and (R.H.S C+(A-B))

L.H.S (C-B)+A

Values of Matrices A ,B ,and C

A = [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}] B = [\begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & 1 & 3 \end{matrix}]

and ,C = [\begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \end{matrix}]

Required is (C-B) + A

= ([\begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \end{matrix}] - [\begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & 1 & 3 \end{matrix}]) + A

Let’s performing C – B step

= [\begin{matrix} -1-1 & 0+1 & 0-1 \\ 0-2 & -2+2 & 3-2 \\ 1-3 & 1-1 & 2-3 \end{matrix}] + [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}]

= [\begin{matrix} -2 & 1 & -1 \\ -2 & 0 & 1 \\ -2 & 0 & -1 \end{matrix}] + [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}]

Let’s performing the addition step

= [\begin{matrix} -2+1 & 1+2 & -1+3 \\ -2+2 & 0+3 & 1+1 \\ -2+1 & 0-1 & -1+0 \end{matrix}]

= [\begin{matrix} -1 & 3 & 2 \\ 0 & 3 & 2 \\ -1 & -1 & -1 \end{matrix}]

R.H.S C+(A-B)

Values of Matrices A ,B ,and C

A = [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}] B = [\begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 2 \\ 3 & 1 & 3 \end{matrix}]

and ,C = [\begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \end{matrix}]

Required is C + (A – B)

= C + ([\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}] - [\begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & 1 & 3 \end{matrix}])

Let’s performing A – B step

= [\begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \end{matrix}] + [\begin{matrix} 1-1 & 2+1 & 3-1 \\ 2-2 & 3+2 & 1-2 \\ 1-3 & -1-1 & 0-3 \end{matrix}]

= [\begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \end{matrix}] + [\begin{matrix} 0 & 3 & 2 \\ 0 & 5 & -1 \\ -2 & -2 & -3 \end{matrix}]

Let’s performing the addition step

= [\begin{matrix} -1+0 & 0+3 & 0+2 \\ 0+0 & -2+5 & 3-1 \\ 1-2 & 1-2 & 2-3 \end{matrix}]

= [\begin{matrix} -1 & 3 & 2 \\ 0 & 3 & 2 \\ -1 & -1 & -1 \end{matrix}]

Hence Proved

(L.H.S is (C-B)+A) = (R.H.S is C+(A-B))

(6) Rule#6

2A+B=A+(A+B)

(L.H.S 2A+B) and (R.H.S A+(A+B))

L.H.S is 2A+B

Values of Matrices A and B

A = [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}] B = [\begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 2 \\ 3 & 1 & 3 \end{matrix}]

Required is 2A + B

= 2 [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}] + [\begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 2 \\ 3 & 1 & 3 \end{matrix}]

First A matrix multiply by 2.

= [\begin{matrix} 1×2 & 2×2 & 3×2 \\ 2×2 & 3×2 & 1×2 \\ 1×2 & -1×2 & 0×2 \end{matrix}] + [\begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 2 \\ 3 & 1 & 3 \end{matrix}]

= [\begin{matrix} 2 & 4 & 6 \\ 4 & 6 & 2 \\ 2 & -2 & 0 \end{matrix}] + [\begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 2 \\ 3 & 1 & 3 \end{matrix}]

Let’s performing the addition step

= [\begin{matrix} 2+1 & 4-1 & 6+1 \\ 4+2 & 6-1 & 2+2 \\ 2+3 & -2+1 & 0+3 \end{matrix}]

= [\begin{matrix} 3 & 3 & 7 \\ 6 & 5 & 4 \\ 5 & -1 & 3 \end{matrix}]

R.H.S is A+(A+B)

Values of Matrices A and B

A = [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}] B = [\begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 2 \\ 3 & 1 & 3 \end{matrix}]

Required is A+(A + B)

= A + ([\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}] + [\begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 2 \\ 3 & 1 & 3 \end{matrix}])

1^st solve the inside bracket Matrices

= [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}] + [\begin{matrix} 1+1 & 2-1 & 3+1 \\ 2+2 & 3-1 & 1+2 \\ 1+3 & -1+1 & 0+3 \end{matrix}]

= [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}] + [\begin{matrix} 2 & 1 & 4 \\ 4 & 2 & 3 \\ 4 & 0 & 3 \end{matrix}]

Let’s performing the addition step

= [\begin{matrix} 1+2 & 2+1 & 3+4 \\ 2+4 & 3+2 & 1+3 \\ 1+4 & -1+0 & 0+3 \end{matrix}]

= [\begin{matrix} 3 & 3 & 7 \\ 6 & 5 & 4 \\ 5 & -1 & 3 \end{matrix}]

Hence Proved

(L.H.S is 2A+B) = (R.H.S is A+(A+B))

(7) Rule#7

(C-B)-A=(C-A)-B

(L.H.S (C-B)-A) and (R.H.S (C-A)-B)

L.H.S is (C-B)-A

Values of Matrices A, B, and C.

A = [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}] B = [\begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 2 \\ 3 & 1 & 3 \end{matrix}]

and ,C = [\begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \end{matrix}]

Required is (C – B)-A

= ([\begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \end{matrix}] - [\begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 2 \\ 3 & 1 & 3 \end{matrix}]) - [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}]

= [\begin{matrix} -1-1 & 0+1 & 0-1 \\ 0-2 & -2+1 & 3-2 \\ 1-3 & 1-1 & 2-3 \end{matrix}] - [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}]

= [\begin{matrix} -2 & 1 & -1 \\ -2 & -1 & 1 \\ -2 & 0 & -1 \end{matrix}] - [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}]

Let’s performing the subtraction step

= [\begin{matrix} -2-1 & 1-2 & -1-3 \\ -2-2 & -1-3 & 1-1 \\ -2-1 & 0+1 & -1-0 \end{matrix}]

= [\begin{matrix} -3 & -1 & -4 \\ -4 & -4 & 0 \\ -3 & 1 & -1 \end{matrix}]

R.H.S is (C-A)-B

Values of Matrices A, B, and C.

A = [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}] B = [\begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 2 \\ 3 & 1 & 3 \end{matrix}]

and ,C = [\begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \end{matrix}]

Required is (C – A)-B

= ([\begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \end{matrix}] - [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}]) - [\begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 2 \\ 3 & 1 & 3 \end{matrix}]

First solve inbracket Matrices

= [\begin{matrix} -1-1 & 0-2 & 0-3 \\ 0-2 & -2-3 & 3-1 \\ 1-1 & 1+1 & 2-0 \end{matrix}] - [\begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 2 \\ 3 & 1 & 3 \end{matrix}]

= [\begin{matrix} -2 & -2 & -3 \\ -2 & -5 & 2 \\ 0 & 2 & 2 \end{matrix}] - [\begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 2 \\ 3 & 1 & 3 \end{matrix}]

Let’s performing the subtraction step

= [\begin{matrix} -2-1 & -2+1 & -3-1 \\ -2-2 & -5+1 & 2-2 \\ 0-3 & 2-1 & 2-3 \end{matrix}]

= [\begin{matrix} -3 & -1 & -4 \\ -4 & -4 & 0 \\ -3 & 1 & -1 \end{matrix}]

Hence Proved (L.H.S is (C-B)-A) = (R.H.S is (C-A)-B)

(8) Rule#8

(A+B)+C=A+(B+C)

(L.H.S is (A+B)+C) and (R.H.S is A+(B+C))

L.H.S is (A+B)+C

Values of Matrices A ,B and ,C

A = [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}] B = [\begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 2 \\ 3 & 1 & 3 \end{matrix}]

and ,C = [\begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \end{matrix}]

Required is (A + B)+C

= ([\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}] + [\begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 2 \\ 3 & 1 & 3 \end{matrix}]) + [\begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \end{matrix}]

First solve inbracket Matrices.

= [\begin{matrix} 1+1 & 2-1 & 3+1 \\ 2+2 & 3-1 & 1+2 \\ 1+3 & -1+1 & 0+3 \end{matrix}] + [\begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \end{matrix}]

= [\begin{matrix} 2 & 1 & 4 \\ 4 & 2 & 3 \\ 4 & 0 & 3 \end{matrix}] + [\begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \end{matrix}]

Let’s performing the addition step

= [\begin{matrix} 2-1 & 1+0 & 4+0 \\ 4+0 & 2-2 & 3+3 \\ 4+1 & 0+1 & 3+2 \end{matrix}]

= [\begin{matrix} 1 & 1 & 4 \\ 4 & 0 & 6 \\ 5 & 1 & 5 \end{matrix}]

R.H.S is A+(B+C)

Values of Matrices A ,B and ,C

A = [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}] B = [\begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 2 \\ 3 & 1 & 3 \end{matrix}]

and ,C = [\begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \end{matrix}]

Required is A+ (B + C)

= [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}] + ([\begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 2 \\ 3 & 1 & 3 \end{matrix}] + [\begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \end{matrix}])

First solve inbracket Matrices.

= [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}] + [\begin{matrix} 1-1 & -1+0 & 1+0 \\ 2+0 & -1-2 & 2+3 \\ 3+1 & 1+1 & 3+2 \end{matrix}]

= [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}] + [\begin{matrix} 0 & -1 & 1 \\ 2 & -3 & 5 \\ 4 & 2 & 5 \end{matrix}]

Let’s performing the addition step

= [\begin{matrix} 1+0 & 2-1 & 3+1 \\ 2+2 & 3-3 & 1+5 \\ 1+4 & -1+2 & 0+5 \end{matrix}]

= [\begin{matrix} 1 & 1 & 4 \\ 4 & 0 & 6 \\ 5 & 1 & 5 \end{matrix}]

Hence Proved (L.H.S is (A+B)+C) = (R.H.S is A+(B+C))

(9) Rule#9

A+(B-C)=(A-C)+B

(L.H.S is A+(B-C)) and (R.H.S is (A-C)+B)

L.H.S is A+(B-C)

Values of Matrices A, B, and C.

Required is A +(B – C)

= [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}] + ([\begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 2 \\ 3 & 1 & 3 \end{matrix}] - [\begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \end{matrix}])

1^st solve bracket Matrices.

= [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}] + [\begin{matrix} 1+1 & -1-0 & 1-0 \\ 2-0 & -1+2 & 2-3 \\ 3-1 & 1-1 & 3_2 \end{matrix}]

= [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}] + [\begin{matrix} 2 & -1 & 1 \\ 2 & 1 & -1 \\ 2 & 0 & 1 \end{matrix}]

Let’s performing the addition step

= [\begin{matrix} 1+2 & 2-1 & 3+1 \\ 2+2 & 3+1 & 1-1 \\ 1+2 & -1+0 & 0+1 \end{matrix}]

= [\begin{matrix} 3 & 1 & 4 \\ 4 & 4 & 0 \\ 3 & -1 & 1 \end{matrix}]

R.H.S is (A-C)+B

Values of Matrices A, B, and C.

Required is (A – C)+ B

= ([\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}] - [\begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \end{matrix}]) + [\begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 2 \\ 3 & 1 & 3 \end{matrix}]

1^st solve bracket Matrices.

= [\begin{matrix} 1+1 & 2-0 & 3-0 \\ 2-0 & 3+2 & 1-3 \\ 1-1 & -1-1 & 0-2 \end{matrix}] + [\begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 2 \\ 3 & 1 & 3 \end{matrix}]

= [\begin{matrix} 2 & 2 & 3 \\ 2 & 5 & -2 \\ 0 & -2 & -2 \end{matrix}] + [\begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 2 \\ 3 & 1 & 3 \end{matrix}]

Let’s performing the addition step

= [\begin{matrix} 2+1 & 2-1 & 3+1 \\ 2+2 & 5-1 & -2+2 \\ 0+3 & -2+1 & -2+3 \end{matrix}]

= [\begin{matrix} 3 & 1 & 4 \\ 4 & 4 & 0 \\ 3 & -1 & 1 \end{matrix}]

Hence Proved (L.H.S is A+(B-C)) = (R.H.S is (A-C)+B)

(10) Rule#10

2A+2B=2(A+B)

(L.H.S is 2A+2B) and (R.H.S is 2(A+B))

L.H.S is 2A+2B

Values of Matrices A and B

Required is 2A + 2B

= 2 [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}] + 2 [\begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 2 \\ 3 & 1 & 3 \end{matrix}]

1^st both Matrices Multiply with 2.

= [\begin{matrix} 2×1 & 2×2 & 2×3 \\ 2×2 & 2×3 & 2×1 \\ 2×1 & 2×-1 & 2×0 \end{matrix}] + [\begin{matrix} 2×1 & 2×-1 & 2×1 \\ 2×2 & 2×-1 & 2×2 \\ 2×3 & 2×1 & 2×3 \end{matrix}]

= [\begin{matrix} 2 & 4 & 6 \\ 4 & 6 & 2 \\ 2 & -2 & 0 \end{matrix}] + [\begin{matrix} 2 & -2 & 2 \\ 4 & -2 & 4 \\ 6 & 2 & 6 \end{matrix}]

Let’s performing the addition step

= [\begin{matrix} 2+2 & 4-2 & 6+2 \\ 4+4 & 6-2 & 2+4 \\ 2+6 & -2+2 & 0+6 \end{matrix}]

= [\begin{matrix} 4 & 2 & 8 \\ 8 & 4 & 6 \\ 8 & 0 & 6 \end{matrix}]

R.H.S is 2(A+B)

Values of Matrices A and B

Required is 2(A + B)

= 2 ([\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \end{matrix}] + [\begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 2 \\ 3 & 1 & 3 \end{matrix}])

First solve bracket Matrices.

= 2 [\begin{matrix} 1+1 & 2-1 & 3+1 \\ 2+2 & 3-1 & 1+2 \\ 1+3 & -1+1 & 0+3 \end{matrix}]

= 2 [\begin{matrix} 2 & 1 & 4 \\ 4 & 2 & 3 \\ 4 & 0 & 3 \end{matrix}]

Matrix elements multiply with 2.

= [\begin{matrix} 2×2 & 2×1 & 2×4 \\ 2×4 & 2×2 & 2×3 \\ 2×4 & 2×0 & 2×3 \end{matrix}]

= [\begin{matrix} 4 & 2 & 8 \\ 8 & 4 & 6 \\ 8 & 0 & 6 \end{matrix}]

Hence Proved (L.H.S is 2A+2B = (R.H.S is 2(A+B))

Q.7

If , 2 [\begin{matrix} 2 & 4 \\ -3 & a \end{matrix}] +3 [\begin{matrix} 1 & b \\ 8 & -4 \end{matrix}] = [\begin{matrix} 7 & 10 \\ 18 & 1 \end{matrix}]

Find the values of (a) and (b).

Solution:

Given equation

= 2 [\begin{matrix} 2 & 4 \\ -3 & a \end{matrix}] +3 [\begin{matrix} 1 & b \\ 8 & -4 \end{matrix}] = [\begin{matrix} 7 & 10 \\ 18 & 1 \end{matrix}]

Required is find values of (a) and (b)

= 2 [\begin{matrix} 2 & 4 \\ -3 & a \end{matrix}] +3 [\begin{matrix} 1 & b \\ 8 & -4 \end{matrix}] = [\begin{matrix} 7 & 10 \\ 18 & 1 \end{matrix}]

Perform the multiplication step

[\begin{matrix} 2×2 & 2×4 \\ 2×-3 & 2×a \end{matrix}] + [\begin{matrix} 3×1 & 3×b \\ 3×8 & 3×-4 \end{matrix}] = [\begin{matrix} 7 & 10 \\ 18 & 1 \end{matrix}]

[\begin{matrix} 4 & 8 \\ -6 & 2a \end{matrix}] + [\begin{matrix} 3 & 3b \\ 24 & -12 \end{matrix}] = [\begin{matrix} 7 & 10 \\ 18 & 1 \end{matrix}]

Perform the addition step

[\begin{matrix} 4+3 & 8+3b \\ -6+24 & 2a-12 \end{matrix}] = [\begin{matrix} 7 & 10 \\ 18 & 1 \end{matrix}]

8+3b is equal to 10 (8+3b=10)

2a-12 is equal to 1 (2a-12=1)

1^st find the value of (a).

2a-12=1

2a=1+12=13

So the value of (a) is , a= \frac{13}{2}

2^nd find the value of (b).

8+3b=10

3b=10-8

3b=2

So the value of (b) is , b= \frac{2}{3}