Given 3 x y z w = x 6 \-1 2w + 4 x+y z+w 3 , find the values of x, y, z and w.

Exercise 3.2 · Q12

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}

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Appeared in past exams:AP EAPCET 2021· Set eng-2021-08-23-AN· 1mreworded

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Concept understanding — Matrix Equation Solving

Solving a System of Equations by the Matrix Method

A system of linear equations can be written as a single matrix equation and solved in one clean step using the inverse of a matrix. This is the Class-12 "matrix method" for simultaneous equations.

Writing the system as $AX=B$

Take the system

a_1x+b_1y+c_1z=d_1,\quad a_2x+b_2y+c_2z=d_2,\quad a_3x+b_3y+c_3z=d_3.

Collect the coefficients, the unknowns, and the constants into matrices:

A=\begin{pmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{pmatrix},\quad X=\begin{pmatrix}x\\y\\z\end{pmatrix},\quad B=\begin{pmatrix}d_1\\d_2\\d_3\end{pmatrix}.

Then the whole system is just

AX=B.

Solving when $A$ is invertible

If $\det(A)\neq0$ , then $A^{-1}$ exists, and multiplying both sides on the left by $A^{-1}$ gives

X=A^{-1}B,\qquad\text{where }\;A^{-1}=\frac{1}{\det(A)}\operatorname{adj}(A).

So you compute $\det(A)$ , then $\operatorname{adj}(A)$ , form $A^{-1}$ , and multiply by $B$ . The single column $X=A^{-1}B$ hands you $x$ , $y$ , $z$ at once, and because $A^{-1}$ is unique, the solution is unique.

Watch out

Multiply in the correct order: $X=A^{-1}B$ , not $BA^{-1}$ . Matrix multiplication is not commutative, and $BA^{-1}$ is not even defined here.

When $\det(A)=0$

If $\det(A)=0$ , $A^{-1}$ does not exist and the inverse method fails. The system is then either inconsistent (no solution) or has infinitely many solutions. Decide which by computing $(\operatorname{adj}A)\,B$ :

$(\operatorname{adj}A)\,B\neq O$ → no solution (inconsistent).
$(\operatorname{adj}A)\,B=O$ → infinitely many solutions (consistent, dependent). …

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This is a matrix equation where corresponding entries must be equal. By equating entries after simplifying, we get four simple equations that yield $x = 2$ , $y = 4$ , $z = 1$ , and $w = 3$ .

The key idea: a matrix equation like this is really just a compact way of writing several ordinary equations at once. Two matrices are equal exactly when every entry in the same position is equal. So we can "unpack" the matrix equation into four separate scalar equations — one for each position — and solve them.

Let’s work through it.

Write the equation clearly. We have

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}.

Simplify the right-hand side. Adding two matrices means adding corresponding entries:

\begin{bmatrix} x+4 & 6+(x+y) \\ -1+(z+w) & 2w+3 \end{bmatrix} = \begin{bmatrix} x+4 & x+y+6 \\ z+w-1 & 2w+3 \end{bmatrix}.

Simplify the left-hand side. Multiplying a matrix by a scalar multiplies every entry:

3 \begin{bmatrix} x & y \\ z & w \end{bmatrix} = \begin{bmatrix} 3x & 3y \\ 3z & 3w \end{bmatrix}.

Equate corresponding entries.

Since the two matrices are equal, we get four equations:
- Top-left: $3x = x + 4$
- Top-right: $3y = x + y + 6$
- Bottom-left: $3z = z + w - 1$
- Bottom-right: $3w = 2w + 3$
Solve each equation.
- From $3x = x + 4$ , subtract $x$ : $2x = 4$ , so $x = 2$ .
- From $3w = 2w + 3$ , subtract $2w$ : $w = 3$ .
- Now plug $x = 2$ into $3y = x + y + 6$ : …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

GUJCET 2023Set 091 markMCQ
$\begin{bmatrix} x+y & -2 \\ 7+z & x-y \end{bmatrix} = \begin{bmatrix} -7 & -2 \\ 5 & 0 \end{bmatrix}$
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Equate corresponding entries of the two matrices and solve.

Concept: Matching entries:
$x+y=-7,\quad x-y=0,\quad 7+z=5.$
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GUJCET 2021Set 151 markMCQ
$AB = \begin{bmatrix} -6 & 26 \\ -1 & 19 \end{bmatrix}$
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Concept:

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$A = \begin{bmatrix} \sin\alpha & -\cos\alpha \\ \cos\alpha & \sin\alpha \end{bmatrix}$
›Reveal solutionSolution

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Concept. $A'$ is the transpose; add it to $A$ and match with the identity.

Steps. With $A=\begin{bmatrix}\sin\alpha&-\cos\alpha\\\cos\alpha&\sin\alpha\end{bmatrix}$ , …
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GUJCET 2026Set x1 markMCQ
$A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}$
›Reveal solutionSolution

Compute $A^2$ , add $I$ , and compare with each option.

$A^2=\begin{bmatrix}3&-2\\4&-2\end{bmatrix}\begin{bmatrix}3&-2\\4&-2\end{bmatrix}=\begin{bmatrix}9-8&-6+4\\12-8&-8+4\end{bmatrix}=\begin{bmatrix}1&-2\\4&-4\end{bmatrix}.$
$A^2+I=\begin{bmatrix}2&-2\\4&-3\end{bmatrix}.$ …
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GUJCET 2026Set x1 markMCQ
$A = \begin{bmatrix} a & b \\ c & -a \end{bmatrix}$
›Reveal solutionSolution

Compute $A^2$ for the given matrix and set it equal to $I$ . …

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GUJCET 2025Set 031 markMCQ
$A = \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}$
›Reveal solutionSolution

Form $A^2-2I$ and compare each entry with $KA$ .

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$A = \begin{bmatrix} a & b \\ c & -a \end{bmatrix}$
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Squaring $A$ gives a scalar matrix $(a^2+bc)I$ ; setting it to $I$ yields $a^2+bc=1$ , i.e. $1-a^2-bc=0$ .

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Given $4x+3y=6xy$ ... (1) and $8x+6y=9xy$ ... (2).

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…

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Q.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 $w$ .

Concept understanding — Matrix Equation Solving

Solving a System of Equations by the Matrix Method

Writing the system as $AX=B$

Solving when $A$ is invertible

When $\det(A)=0$

Method: Equating entries when the unknown also appears on the right

Steps

Common Mistakes

Q.Given 3[xyzw]=[x6−12w]+[4x+yz+w3]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}3[xz​yw​]=[x−1​62w​]+[4z+w​x+y3​], find the values of x,y,zx, y, zx,y,z and www.

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Q.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 $w$ .

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