Suppose the vectors x1, x2 and x3 are the solutions of the system of linear equations, Ax = b when the vector b on the right side is equal to b1, b2 and b3 respectively. if x_1 = [ \beginmatrix 1 1 1 \endmatrix ] , x_2 = [ \beginmatrix 0 2 1 \endmatrix ] , x_3 = [ \beginmatrix 0 0 1 \endmatrix ]b_1 = [ \beginmatrix 1 0 0 \endmatrix ] , b_2 = [ \beginmatrix 0 2 0 \endmatrix ] and b_3 = [ \beginmatrix 0 0 2 \endmatrix ] , then the determinant of A is equal to :

Matrices and Determinants PYQ — JEE Main 2020 · 4 Sep · Shift 2 · Q32 with Solution

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Correct answer: C

We are given that for the system $A x=b,$ when the right-hand side is $b_1,b_2,b_3$ , the corresponding solutions are $x_1,x_2,x_3$ respectively.

So, $A x_1=b_1,\quad A x_2=b_2,\quad A x_3=b_3.$

Form matrices using these column vectors: $X=[x_1\ x_2\ x_3], \qquad B=[b_1\ b_2\ b_3].$ Then the above relations combine into $AX=B.$ Hence, $A=BX^{-1}$ and therefore $\det(A)=\frac{\det(B)}{\det(X)}.$
Now write the matrices explicitly.

Given

\quad x_2=\begin{bmatrix}0\\2\\1\end{bmatrix}, \quad x_3=\begin{bmatrix}0\\0\\1\end{bmatrix}.$$ So $$X=\begin{bmatrix} 1&0&0\\ 1&2&0\\ 1&1&1 \end{bmatrix}.$$ Also, $$b_1=\begin{bmatrix}1\\0\\0\end{bmatrix}, \quad b_2=\begin{bmatrix}0\\2\\0\end{bmatrix}, \quad b_3=\begin{bmatrix}0\\0\\2\end{bmatrix}.$$ Thus $$B=\begin{bmatrix} 1&0&0\\ 0&2&0\\ 0&0&2 \end{bmatrix}.$$ 4. Compute the determinants. Since $B$ is diagonal, $$\det(B)=1\cdot 2\cdot 2=4.$$ Now, $$X=\begin{bmatrix} 1&0&0\\ 1&2&0\\ 1&1&1 \end{bmatrix}.$$ This is lower triangular, so its determinant is the product of diagonal entries: $$\det(X)=1\cdot 2\cdot 1=2.$$ 5. Therefore, $$\det(A)=\frac{\det(B)}{\det(X)}=\frac{4}{2}=2.$$ 6. Check options: - A: $\frac{3}{2}$ ❌ - B: $4$ ❌ - C: $2$ ✅ - D: $\frac{1}{2}$ ❌ Hence the correct answer is **Option C**.

View written solutionFree

Correct answer: C

We are given that for the system $A x=b,$ when the right-hand side is $b_1,b_2,b_3$ , the corresponding solutions are $x_1,x_2,x_3$ respectively.

So, $A x_1=b_1,\quad A x_2=b_2,\quad A x_3=b_3.$

Form matrices using these column vectors: $X=[x_1\ x_2\ x_3], \qquad B=[b_1\ b_2\ b_3].$ Then the above relations combine into $AX=B.$ Hence, $A=BX^{-1}$ and therefore $\det(A)=\frac{\det(B)}{\det(X)}.$
Now write the matrices explicitly.

Given

\quad x_2=\begin{bmatrix}0\\2\\1\end{bmatrix}, \quad x_3=\begin{bmatrix}0\\0\\1\end{bmatrix}.$$ So $$X=\begin{bmatrix} 1&0&0\\ 1&2&0\\ 1&1&1 \end{bmatrix}.$$ Also, $$b_1=\begin{bmatrix}1\\0\\0\end{bmatrix}, \quad b_2=\begin{bmatrix}0\\2\\0\end{bmatrix}, \quad b_3=\begin{bmatrix}0\\0\\2\end{bmatrix}.$$ Thus $$B=\begin{bmatrix} 1&0&0\\ 0&2&0\\ 0&0&2 \end{bmatrix}.$$ 4. Compute the determinants. Since $B$ is diagonal, $$\det(B)=1\cdot 2\cdot 2=4.$$ Now, $$X=\begin{bmatrix} 1&0&0\\ 1&2&0\\ 1&1&1 \end{bmatrix}.$$ This is lower triangular, so its determinant is the product of diagonal entries: $$\det(X)=1\cdot 2\cdot 1=2.$$ 5. Therefore, $$\det(A)=\frac{\det(B)}{\det(X)}=\frac{4}{2}=2.$$ 6. Check options: - A: $\frac{3}{2}$ ❌ - B: $4$ ❌ - C: $2$ ✅ - D: $\frac{1}{2}$ ❌ Hence the correct answer is **Option C**.