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Q12:
Given $3\begin{bmatrix} x & y \\ z & w \end{bmatrix} = \begin{bmatrix} x & 6 \\ -1 & 2w \end{bmatrix} + \begin{bmatrix} 4 & x+y \\ z+w & 3 \end{bmatrix}$, find the values of $x, y, z$ and $w$.

Solution :

Step 1: Scalar Multiplication of the Left-Hand Side (LHS)

According to the properties of scalar multiplication, we multiply each element of the matrix by the scalar $3$:

$3 \begin{bmatrix} x & y \\ z & w \end{bmatrix} = \begin{bmatrix} 3x & 3y \\ 3z & 3w \end{bmatrix}$ [By scalar multiplication definition: $k[A]_{ij} = [kA]_{ij}$]

Step 2: Matrix Addition on the Right-Hand Side (RHS)

Perform addition by summing corresponding elements of the two matrices:

$\begin{bmatrix} x & 6 \\ -1 & 2w \end{bmatrix} + \begin{bmatrix} 4 & x+y \\ z+w & 3 \end{bmatrix} = \begin{bmatrix} x+4 & 6+x+y \\ -1+z+w & 2w+3 \end{bmatrix}$ [By matrix addition property: $[A+B]_{ij} = A_{ij} + B_{ij}$]

Step 3: Establishing the System of Linear Equations

Equate the resulting LHS and RHS matrices. Since two matrices are equal if and only if their corresponding elements are equal [By Equality of Matrices Definition], we obtain the following system:

  • (i) $3x = x + 4$
  • (ii) $3y = 6 + x + y$
  • (iii) $3z = -1 + z + w$
  • (iv) $3w = 2w + 3$

Step 4: Solving the System of Equations

Solve (i):

$3x - x = 4 \implies 2x = 4 \implies x = 2$

Solve (ii) using $x=2$:

$3y - y = 6 + 2 \implies 2y = 8 \implies y = 4$

Solve (iv):

$3w - 2w = 3 \implies w = 3$

Solve (iii) using $w=3$:

$3z - z = -1 + 3 \implies 2z = 2 \implies z = 1$

Step 5: Verification Table

Variable Value Substitution Check
$x$ $2$ $3(2) = 2+4 \implies 6=6$ (Valid)
$y$ $4$ $3(4) = 6+2+4 \implies 12=12$ (Valid)
$z$ $1$ $3(1) = -1+1+3 \implies 3=3$ (Valid)
$w$ $3$ $3(3) = 2(3)+3 \implies 9=9$ (Valid)
x = 2, y = 4 z = 1, w = 3

Final Solution: x = 2, y = 4, z = 1, w = 3


More Questions from Class 12 Mathematics Matrices EXERCISE 3.2


CBSE Solutions for Class 12 Mathematics Matrices


Chapters in CBSE - Class 12 Mathematics


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