Find the values of a, b, c and d, if 3 a bc d = a 6\-1 2d + 4 a+bc+d 3 .

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Puducherry CbseShort· 3mImportance★★★★★

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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Equating corresponding entries of the two equal matrices gives $a=2,\ b=4,\ c=1,\ d=3$ .

The governing rule is: two matrices are equal only when their corresponding entries are equal. So this single matrix equation splits into four ordinary equations, one per position.

Set up both sides

Given

3\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a & 6 \\ -1 & 2d \end{bmatrix} + \begin{bmatrix} 4 & a+b \\ c+d & 3 \end{bmatrix}.

Add the two matrices on the right entry by entry, and multiply the left matrix by the scalar $3$ :

\begin{bmatrix} 3a & 3b \\ 3c & 3d \end{bmatrix} = \begin{bmatrix} a+4 & a+b+6 \\ c+d-1 & 2d+3 \end{bmatrix}.

Match entries and solve

Top-left: $3a = a+4 \Rightarrow 2a = 4 \Rightarrow a = 2.$
Bottom-right: $3d = 2d+3 \Rightarrow d = 3.$
Top-right: $3b = a+b+6$ . With $a=2$ : $3b = b+8 \Rightarrow 2b = 8 \Rightarrow b = 4.$ …

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Q.Find the values of $a$ , $b$ , $c$ and $d$ , if $3\begin{bmatrix}a & b\\ c & d\end{bmatrix} = \begin{bmatrix}a & 6\\ -1 & 2d\end{bmatrix} + \begin{bmatrix}4 & a+b\\ c+d & 3\end{bmatrix}$ .

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 $\det(A)=0$

Set up both sides

Match entries and solve

Method: Solving a matrix equation by simplifying both sides then equating entries

Steps

Common Mistakes

Q.Find the values of aaa, bbb, ccc and ddd, if 3[abcd]=[a6−12d]+[4a+bc+d3]3\begin{bmatrix}a & b\\ c & d\end{bmatrix} = \begin{bmatrix}a & 6\\ -1 & 2d\end{bmatrix} + \begin{bmatrix}4 & a+b\\ c+d & 3\end{bmatrix}3[ac​bd​]=[a−1​62d​]+[4c+d​a+b3​].

Set up both sides

Match entries and solve

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Q.Find the values of $a$ , $b$ , $c$ and $d$ , if $3\begin{bmatrix}a & b\\ c & d\end{bmatrix} = \begin{bmatrix}a & 6\\ -1 & 2d\end{bmatrix} + \begin{bmatrix}4 & a+b\\ c+d & 3\end{bmatrix}$ .

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