VID-M12-04-CT — Determinants — Full Chapter Test | Class 12 Mathematics

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CodeVID-M12-04-CT

Determinants — Full Chapter Test

Chapter: Determinants

Topic: Complete chapter — all topics

Maximum Marks: 35

Time: 80 minutes

Name: ____________________ Roll No.: __________ Date: ____________

General Instructions

This is a full-length, CBSE board-hard test covering the whole chapter. All questions are compulsory.
Section A carries 1 mark each (including an assertion-reason item), Section B 2 marks, Section C 3 marks and Section D 5 marks (long answer / case based).
Show all working for Sections B, C and D. Final answers are given in the Answer Key.

Section A — Multiple Choice Questions 5 × 1 = 5 marks

If $A$ is a square matrix of order $3$ with $|A|=4$, then $|\operatorname{adj}(2A)|$ equals:

A.$64$
B.$256$
C.$1024$
D.$4096$

If $A$ is a non-singular matrix of order $3$ with $|A|=-2$, then $\left|(\operatorname{adj}A)A^{-1}\right|$ equals:

A.$-2$
B.$4$
C.$-8$
D.$2$

Assertion (A): The system $x+y=2,\ 2x+2y=5$ has no solution.
Reason (R): If $|A|=0$ and $(\operatorname{adj}A)B\ne O$, then the system $AX=B$ is inconsistent.

A.Both A and R are true and R is the correct explanation of A
B.Both A and R are true but R is not the correct explanation of A
C.A is true but R is false
D.A is false but R is true

If the area of the triangle with vertices $(2,-6)$, $(5,4)$ and $(k,4)$ is $35$ square units, then $k$ is:

A.$12$ or $-2$
B.$-12$ or $2$
C.$12$ or $2$
D.$-12$ or $-2$

The value of $\begin{vmatrix}x+1&x+2&x+3\\x+2&x+3&x+4\\x+3&x+4&x+5\end{vmatrix}$ is:

A.$x$
B.$0$
C.$x^3$
D.$-2$

Section B — Short Answer I 4 × 2 = 8 marks

Without expanding, show that $\begin{vmatrix}a&b&c\\a+2x&b+2y&c+2z\\x&y&z\end{vmatrix}=0$.

If $A$ is a non-singular matrix of order $3$ and $|A|=5$, find $|\operatorname{adj}A|$ and $\left|A^{-1}\operatorname{adj}A\right|$.

Find $x$ if $\begin{vmatrix}3&x\\x&1\end{vmatrix}=\begin{vmatrix}3&2\\4&1\end{vmatrix}$.

Using determinants, find the equation of the line through $(1,2)$ and $(3,6)$.

Section C — Short Answer II 4 × 3 = 12 marks

10.

Find $A^{-1}$ for $A=\begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix}$ and hence solve $x-y+2z=1,\ 2y-3z=-1,\ 3x-2y+4z=2$.

11.

Prove that $\begin{vmatrix}1&1&1\\a&b&c\\a^3&b^3&c^3\end{vmatrix}=(a-b)(b-c)(c-a)(a+b+c)$.

12.

Show that the system $x+y+z=3,\ x+2y+3z=6,\ 2x+3y+4z=10$ is inconsistent. What value should replace $10$ on the right-hand side of the third equation to make it consistent?

13.

Show that $\begin{vmatrix}a-b-c&2a&2a\\2b&b-c-a&2b\\2c&2c&c-a-b\end{vmatrix}=(a+b+c)^3$.

Section D — Long Answer / Case Based 2 × 5 = 10 marks

14.

Case study. Three friends bought stationery from the same shop. Aman bought $2$ pens, $3$ notebooks and $1$ geometry box for ₹135. Bina bought $1$ pen, $2$ notebooks and $3$ geometry boxes for ₹180. Chetan bought $3$ pens, $1$ notebook and $2$ geometry boxes for ₹135. Let the prices of a pen, a notebook and a geometry box be ₹$x$, ₹$y$ and ₹$z$.
(a) Write the data as a matrix equation $AX=B$. [1]
(b) Find $|A|$ and $\operatorname{adj}A$. [2]
(c) Hence find the price of each item. [2]

15.

If $A=\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}$, verify that $A^3-6A^2+9A-4I=O$ and hence find $A^{-1}$.

Answer Key

Section A — Multiple Choice Questions

(C) $1024$
(A) $-2$
(A) Both A and R are true and R is the correct explanation of A
(A) $12$ or $-2$
(B) $0$

Section B — Short Answer I

Apply $R_2\to R_2-2R_3$. Row 2 becomes $(a,b,c)$, identical to Row 1. A determinant with two identical rows is $0$.
$|\operatorname{adj}A|=|A|^{2}=25$. Since $A^{-1}\operatorname{adj}A=|A|\,A^{-2}$, its determinant is $|A|^3\cdot|A|^{-2}=|A|=5$.
$3-x^2=3-8=-5\Rightarrow x^2=8\Rightarrow x=\pm2\sqrt2$.
For a point $(x,y)$ on the line, $\begin{vmatrix}x&y&1\\1&2&1\\3&6&1\end{vmatrix}=0$. Expanding along $R_1$: $x(2-6)-y(1-3)+1(6-6)=0\Rightarrow-4x+2y=0$, i.e. $y=2x$.

Section C — Short Answer II

$|A|=1(8-6)+1(0+9)+2(0-6)=-1$. $\operatorname{adj}A=\begin{bmatrix}2&0&-1\\-9&-2&3\\-6&-1&2\end{bmatrix}$, so $A^{-1}=\begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}$. $X=A^{-1}B=A^{-1}\begin{bmatrix}1\\-1\\2\end{bmatrix}=\begin{bmatrix}0\\1\\1\end{bmatrix}$, i.e. $x=0,\ y=1,\ z=1$.
Apply $C_1\to C_1-C_2$, $C_2\to C_2-C_3$: the first row becomes $(0,0,1)$ and the determinant reduces (expanding along $R_1$) to $(a-b)(b^3-c^3)-(b-c)(a^3-b^3)=(a-b)(b-c)\left[(b^2+bc+c^2)-(a^2+ab+b^2)\right]=(a-b)(b-c)\left[(c-a)(c+a)+b(c-a)\right]=(a-b)(b-c)(c-a)(a+b+c)$.
Here $|A|=1(8-9)-1(4-6)+1(3-4)=0$ and $\operatorname{adj}A=\begin{bmatrix}-1&-1&1\\2&2&-2\\-1&-1&1\end{bmatrix}$. $(\operatorname{adj}A)B$ with $B=(3,6,10)^T$ has first entry $-3-6+10=1\ne0$, so the system is inconsistent. Since $R_3=R_1+R_2$ on the left, consistency needs the right side to be $3+6=9$.
Apply $R_1\to R_1+R_2+R_3$: every entry of $R_1$ becomes $a+b+c$. Take it out: $(a+b+c)\begin{vmatrix}1&1&1\\2b&b-c-a&2b\\2c&2c&c-a-b\end{vmatrix}$. Now $C_2\to C_2-C_1$, $C_3\to C_3-C_1$: $R_1=(1,0,0)$, $R_2=(2b,-(a+b+c),0)$, $R_3=(2c,0,-(a+b+c))$. Expanding along $R_1$ gives $(a+b+c)\cdot(a+b+c)^2=(a+b+c)^3$.

Section D — Long Answer / Case Based

(a) $\begin{bmatrix}2&3&1\\1&2&3\\3&1&2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}135\\180\\135\end{bmatrix}$. (b) $|A|=2(4-3)-3(2-9)+1(1-6)=18$; $\operatorname{adj}A=\begin{bmatrix}1&-5&7\\7&1&-5\\-5&7&1\end{bmatrix}$. (c) $X=\dfrac{1}{18}\operatorname{adj}A\,B=\dfrac1{18}\begin{bmatrix}180\\450\\720\end{bmatrix}=\begin{bmatrix}10\\25\\40\end{bmatrix}$. A pen costs ₹10, a notebook ₹25 and a geometry box ₹40.
$A^2=\begin{bmatrix}6&-5&5\\-5&6&-5\\5&-5&6\end{bmatrix}$ and $A^3=\begin{bmatrix}22&-21&21\\-21&22&-21\\21&-21&22\end{bmatrix}$. Then every entry of $A^3-6A^2+9A-4I$ is $0$ (e.g. $22-36+18-4=0$ on the diagonal and $-21+30-9=0$ off it). Multiplying $A^3-6A^2+9A-4I=O$ by $A^{-1}$: $A^2-6A+9I-4A^{-1}=O$, so $A^{-1}=\dfrac14\left(A^2-6A+9I\right)=\dfrac14\begin{bmatrix}3&1&-1\\1&3&1\\-1&1&3\end{bmatrix}$.

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