Solve for x and y: x 2 \1 + y 3 \5 + -8 \-11 = O.

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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 vector equation that reduces to a system of two linear equations in $x$ and $y$ . Solving it gives $x = 1$ and $y = 2$ .

We have a linear combination of two column vectors equaling the zero vector. The key idea: a vector equation like this is really just a compact way of writing two separate equations — one for each row. Each component must independently sum to zero.

Let’s write it out. The given equation is:

x \begin{bmatrix} 2 \\ 1 \end{bmatrix} + y \begin{bmatrix} 3 \\ 5 \end{bmatrix} + \begin{bmatrix} -8 \\ -11 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}

Combine the vectors component-wise. The first component (top row) gives:

2x + 3y + (-8) = 0 \quad \Rightarrow \quad 2x + 3y = 8

The second component (bottom row) gives:

1x + 5y + (-11) = 0 \quad \Rightarrow \quad x + 5y = 11

So we have the system:

\begin{cases} 2x + 3y = 8 \\ x + 5y = 11 \end{cases}

Solve the system. From the second equation, $x = 11 - 5y$ . Substitute into the first:

2(11 - 5y) + 3y = 8

22 - 10y + 3y = 8

22 - 7y = 8

-7y = -14 \quad \Rightarrow \quad y = 2

Find $x$ . …

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Q.Solve for $x$ and $y$ : $x\begin{bmatrix} 2 \\ 1 \end{bmatrix} + y\begin{bmatrix} 3 \\ 5 \end{bmatrix} + \begin{bmatrix} -8 \\ -11 \end{bmatrix} = O$ .

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: Reducing a linear-combination vector equation to a system

Steps

Common Mistakes

Q.Solve for xxx and yyy: x[21]+y[35]+[−8−11]=Ox\begin{bmatrix} 2 \\ 1 \end{bmatrix} + y\begin{bmatrix} 3 \\ 5 \end{bmatrix} + \begin{bmatrix} -8 \\ -11 \end{bmatrix} = Ox[21​]+y[35​]+[−8−11​]=O.

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Q.Solve for $x$ and $y$ : $x\begin{bmatrix} 2 \\ 1 \end{bmatrix} + y\begin{bmatrix} 3 \\ 5 \end{bmatrix} + \begin{bmatrix} -8 \\ -11 \end{bmatrix} = O$ .

Unlock everything free for 14 days