krit.club logo

Matrices - Equality of matrices

Grade 12CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Two matrices A=[aij]A = [a_{ij}] and B=[bij]B = [b_{ij}] are said to be equal if they are of the same order (i.e., they have the same number of rows and columns).

•

For two matrices to be equal, each element of AA must be equal to the corresponding element of BB, which means aij=bija_{ij} = b_{ij} for all possible values of ii and jj.

•

Equality of matrices is used to solve for unknown variables by comparing corresponding entries and forming algebraic equations.

•

If two matrices have different orders, they can never be equal, regardless of their elements.

📐Formulae

A=B  ⟺  aij=bij ∀ i,jA = B \iff a_{ij} = b_{ij} \, \forall \, i, j

[a11a12a21a22]=\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} = \begin{bmatrix} b_{11} & b_{12} \ b_{21} & b_{22} \end{bmatrix}  ⟹  a11=b11,a12=b12,a21=b21,a22=b22 \implies a_{11} = b_{11}, a_{12} = b_{12}, a_{21} = b_{21}, a_{22} = b_{22}

💡Examples

Problem 1:

Find the values of x,y,zx, y, z and aa from the following equation: [x+3z+42y−7−6a−10]=[063y−2−6−30]\begin{bmatrix} x+3 & z+4 \\ 2y-7 & -6 \\ a-1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 6 \\ 3y-2 & -6 \\ -3 & 0 \end{bmatrix}

Solution:

By comparing corresponding elements:

  1. x+3=0  ⟹  x=−3x + 3 = 0 \implies x = -3
  2. z+4=6  ⟹  z=2z + 4 = 6 \implies z = 2
  3. 2y−7=3y−2  ⟹  2y−3y=−2+7  ⟹  −y=5  ⟹  y=−52y - 7 = 3y - 2 \implies 2y - 3y = -2 + 7 \implies -y = 5 \implies y = -5
  4. a−1=−3  ⟹  a=−2a - 1 = -3 \implies a = -2

Explanation:

Since the two matrices are equal and have the same order (3×23 \times 2), we equate the elements at each position (i,j)(i, j) to solve for the variables.

Problem 2:

Find xx and yy if [2x+y31x−2y]=[7311]\begin{bmatrix} 2x+y & 3 \\ 1 & x-2y \end{bmatrix} = \begin{bmatrix} 7 & 3 \\ 1 & 1 \end{bmatrix}

Solution:

Equating corresponding elements, we get a system of linear equations:

  1. 2x+y=72x + y = 7
  2. x−2y=1x - 2y = 1

Multiply equation (1) by 2: 4x+2y=144x + 2y = 14

Add this to equation (2): (4x+2y)+(x−2y)=14+1  ⟹  5x=15  ⟹  x=3(4x + 2y) + (x - 2y) = 14 + 1 \implies 5x = 15 \implies x = 3

Substitute x=3x = 3 into equation (1): 2(3)+y=7  ⟹  6+y=7  ⟹  y=12(3) + y = 7 \implies 6 + y = 7 \implies y = 1

Explanation:

Matrix equality allows us to set up simultaneous equations. Solving the equations 2x+y=72x+y=7 and x−2y=1x-2y=1 gives the values of xx and yy.

Equality of matrices Class 12 Notes & Examples | CBSE Maths