Given $A = \begin{pmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{pmatrix}$, $B = \begin{pmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{pmatrix}$, find $BA$ and use this to solve the system of equations $y + 2z = 7$, $x - y = 3$, $2x + 3y + 4z = 17$.

Given A = 2 2 -4 \-4 2 -4 \2 -1 5 , B = 1 -1 0 \2 3 4 \0 1 2 , find BA and use this to solve the system of equ

NCERT Exemplar · Q20

A = \begin{pmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{pmatrix}

Andaman Nicobar CbseLong· 5mImportance★★★★★

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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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$BA=6I$ , so $B^{-1}=\tfrac16A$ ; the system is $BX=C$ with $C=(3,17,7)^T$ , giving $X=\tfrac16AC=(2,-1,4)^T$ , i.e. $x=2,\,y=-1,\,z=4$ .

The idea

If the product of two matrices is a scalar times the identity, each is (a scalar multiple of) the other's inverse. Computing $BA$ first hands us $B^{-1}$ for free — and it is $B$ , not $A$ , that is the coefficient matrix of the given system.

Step 1 — Compute $BA$

BA=\begin{pmatrix}1&-1&0\\2&3&4\\0&1&2\end{pmatrix}\begin{pmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{pmatrix}.

Entry by entry (row of $B$ times column of $A$ ):

Row 1: $2+4+0=6$ , $\;2-2+0=0$ , $\;-4+4+0=0$ .
Row 2: $4-12+8=0$ , $\;4+6-4=6$ , $\;-8-12+20=0$ .
Row 3: $0-4+4=0$ , $\;0+2-2=0$ , $\;0-4+10=6$ .

BA=\begin{pmatrix}6&0&0\\0&6&0\\0&0&6\end{pmatrix}=6I\quad\Longrightarrow\quad B^{-1}=\frac16A.

Step 2 — Match the system to $B$

The equations $y+2z=7,\;x-y=3,\;2x+3y+4z=17$ , reordered as

x-y=3,\qquad 2x+3y+4z=17,\qquad y+2z=7,

have coefficient rows equal to the rows of $B$ :

$B\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}3\\17\\7\end{pmatrix}=C.$ …

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Q.Given $A = \begin{pmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{pmatrix}$ , $B = \begin{pmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{pmatrix}$ , find $BA$ and use this to solve the system of equations $y + 2z = 7$ , $x - y = 3$ , $2x + 3y + 4z = 17$ .

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 $\det(A)=0$

The idea

Step 1 — Compute $BA$

Step 2 — Match the system to $B$

Method: Using a Given Matrix Product Equal to a Scalar Multiple of $I$ to Find an Inverse

Steps

Common Mistakes

The idea

Step 1 — Compute BABABA

Step 2 — Match the system to BBB

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Step 1 — Compute $BA$

Step 2 — Match the system to $B$

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