Solve the equation for x, y, z and t, if 2 x z y t + 3 1 -1 \0 2 = 3 3 5 \4 6 .

Exercise 3.2 · Q10

x, y, z

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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 treat the matrix equation as a system of four scalar equations by equating corresponding entries. After simplifying, we find $x = 3$ , $y = 6$ , $z = 9$ , and $t = 6$ .

The core idea here is that matrix equations work entry-by-entry. When two matrices are equal, every element in the same position must be equal. So instead of being intimidated by the matrix form, we can break it down into simple algebraic equations for each unknown.

Let’s go step by step.

Write the equation clearly We have:

2 \begin{bmatrix} x & z \\ y & t \end{bmatrix} + 3 \begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix} = 3 \begin{bmatrix} 3 & 5 \\ 4 & 6 \end{bmatrix}

Perform the scalar multiplications Multiply each matrix by its scalar coefficient:

2 \begin{bmatrix} x & z \\ y & t \end{bmatrix} = \begin{bmatrix} 2x & 2z \\ 2y & 2t \end{bmatrix}

3 \begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 3 & -3 \\ 0 & 6 \end{bmatrix}

3 \begin{bmatrix} 3 & 5 \\ 4 & 6 \end{bmatrix} = \begin{bmatrix} 9 & 15 \\ 12 & 18 \end{bmatrix}

Add the matrices on the left-hand side Adding entry-wise:

\begin{bmatrix} 2x + 3 & 2z - 3 \\ 2y + 0 & 2t + 6 \end{bmatrix} = \begin{bmatrix} 9 & 15 \\ 12 & 18 \end{bmatrix}

Equate corresponding entries

Since the matrices are equal, each entry gives an equation:
- Top-left: $2x + 3 = 9$
- Top-right: $2z - 3 = 15$
- Bottom-left: $2y = 12$
- Bottom-right: $2t + 6 = 18$ …

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←Previous question · 3.4.6Q9 — Find x and y, if 2 1 3 \0 x + y 0 \1 2 = 5 6 \1 8 .→Next question · 3.4.6Q11 — If x 2 \3 + y -1 \1 = 10 \5 , find the values of x and y.

Q.Solve the equation for $x, y, z$ and $t$ , if $2 \begin{bmatrix} x & z \\ y & t \end{bmatrix} + 3 \begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix} = 3 \begin{bmatrix} 3 & 5 \\ 4 & 6 \end{bmatrix}$ .

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 after simplifying both sides

Steps

Common Mistakes

Q.Solve the equation for x,y,zx, y, zx,y,z and ttt, if 2[xzyt]+3[1−102]=3[3546]2 \begin{bmatrix} x & z \\ y & t \end{bmatrix} + 3 \begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix} = 3 \begin{bmatrix} 3 & 5 \\ 4 & 6 \end{bmatrix}2[xy​zt​]+3[10​−12​]=3[34​56​].

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Q.Solve the equation for $x, y, z$ and $t$ , if $2 \begin{bmatrix} x & z \\ y & t \end{bmatrix} + 3 \begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix} = 3 \begin{bmatrix} 3 & 5 \\ 4 & 6 \end{bmatrix}$ .

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