If $\begin{bmatrix}xy & 4\\ z+6 & x+y\end{bmatrix} = \begin{bmatrix}8 & w\\ 0 & 6\end{bmatrix}$, then find values of $x$, $y$, $z$ and $w$.

> [!TLDR] > We equate corresponding entries of two equal matrices to get a system of equations. Solving gives $x=2$, $y=4$, $z=-6$, $w=4$ (or the swapped pair $x=4$, $y=2$). Two matrices are equal **if and only if** every entry in the same position is equal. That’s the core idea here — no shortcuts, no tricks. Once you write down the equalities, you’re just solving a few simple equations. 1. **Equate the (1,1) entry:** The top-left entry of the left matrix is $xy$, and of the right matrix is $8$. So $$xy = 8.$$ 2. **Equate the (1,2) entry:** Top-right: left has $4$, right has $w$. So $$4 = w.$$ 3. **Equate the (2,1) entry:** Bottom-left: left has $z+6$, right has $0$. So $$z + 6 = 0 \quad\Rightarrow\quad z = -6.$$ 4. **Equate the (2,2) entry:** Bottom-right: left has $x+y$, right has $6$. So $$x + y = 6.$$ Now we have $xy = 8$ and $x + y = 6$. These are the classic “sum and product” equations. The two numbers whose sum is $6$ and product is $8$ are $2$ and $4$. So either $x=2$, $y=4$ or $x=4$, $y=2$. …

If xy 4z+6 x+y = 8 w\0 6 , then find values of x, y, z and w.

NCERT Exemplar · Q37

\begin{bmatrix}xy & 4\\ z+6 & x+y\end{bmatrix} = \begin{bmatrix}8 & w\\ 0 & 6\end{bmatrix}

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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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We equate corresponding entries of two equal matrices to get a system of equations. Solving gives $x=2$ , $y=4$ , $z=-6$ , $w=4$ (or the swapped pair $x=4$ , $y=2$ ).

Two matrices are equal if and only if every entry in the same position is equal. That’s the core idea here — no shortcuts, no tricks. Once you write down the equalities, you’re just solving a few simple equations.

Equate the (1,1) entry: The top-left entry of the left matrix is $xy$ , and of the right matrix is $8$ . So

xy = 8.

Equate the (1,2) entry: Top-right: left has $4$ , right has $w$ . So

4 = w.

Equate the (2,1) entry: Bottom-left: left has $z+6$ , right has $0$ . So

z + 6 = 0 \quad\Rightarrow\quad z = -6.

Equate the (2,2) entry: Bottom-right: left has $x+y$ , right has $6$ . So

x + y = 6.

Now we have $xy = 8$ and $x + y = 6$ . These are the classic “sum and product” equations. The two numbers whose sum is $6$ and product is $8$ are $2$ and $4$ . So either $x=2$ , $y=4$ or $x=4$ , $y=2$ . …

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Q.If $\begin{bmatrix}xy & 4\\ z+6 & x+y\end{bmatrix} = \begin{bmatrix}8 & w\\ 0 & 6\end{bmatrix}$ , then find 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: Solving for unknowns using matrix equality

Steps

Common Mistakes

Q.If [xy4z+6x+y]=[8w06]\begin{bmatrix}xy & 4\\ z+6 & x+y\end{bmatrix} = \begin{bmatrix}8 & w\\ 0 & 6\end{bmatrix}[xyz+6​4x+y​]=[80​w6​], then find values of xxx, yyy, zzz and www.

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Q.If $\begin{bmatrix}xy & 4\\ z+6 & x+y\end{bmatrix} = \begin{bmatrix}8 & w\\ 0 & 6\end{bmatrix}$ , then find values of $x$ , $y$ , $z$ and $w$ .

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